More about Monty Hall Problem

The previous post is on the Monty Hall Problem. This post adds to the discussion by looking at three pieces from New York Times. Anyone who is not familiar with the problem should read the previous post or other online resources of the problem.

The first piece describes a visit by John Tierney at the home of Monty Hall, who was the host of the game show Let’s Make a Deal, the show on which the Monty Hall Problem was loosely based. The visit took place in Beverly Hills back in 1991, one year after the firestorm caused by the publication of Monty Hall Problem by Marilyn vos Savant in Parade Magazine. The purpose of the visit is to obtain more confirmation that the counter intuitive answer (switching door) is correct and to get more insight about the game from Monty Hall himself.

Before discussing the visit, here’s the statement of Monty Hall Problem. The previous post actually uses five pictures to describe the problem.

    “Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the other doors, opens another door, say No. 3, which has a goat. He then says to you, ‘Do you want to pick door No. 2?’ Is it to your advantage to take the switch?”

In front of Tierney, Monty Hall performed a simulation in his dining room. Monty Hall put three miniature cardboard doors on a table and represented the car with an ignition key. The goats were played by a package of raisins and a roll of Life Savers. Monty Hall conducted 10 rounds of the game as the contestant (John Tierney) tried the non-switching strategy. The result was that the contestant won four cars and six goats.

Then in the next 10 rounds, the contestant tried switching doors. The result was markedly different – the contestant won eight cars and two goats. The simulation gives strong indication that it is more advantageous to switch door.

The standard solution: with the switching door strategy, the contestant wins the car 2/3 of the time. On the other hand, the contestant only wins the car 1/3 of the time if he/she stick with the original pick.

It turns out that the optimal solution of switching door should come with a caveat, which is given in the previous post. The caveat is that the strategy of switching door works only if the moves of the game are determined by random chance – the contestant picks a door at random, then the host picks a door at random (if the contestant’s pick happens to be the winning door). After the choices are made at random, the host will be required to offer the chance to switch. Once all these conditions are met, the switching door strategy will be optimal.

The description in the preceding paragraph works like a computer program, in which all the rules are clear cut. There is no psychological factor involved. But Monty Hall did not run his show like a computer program. He sometimes did things that are meant to trick the contestants.

Month Hall did not always followed the standard assumptions. According to the NY Times piece, sometimes Monty Hall simply opened the contestant’s original pick (if it is a goat). So the picking a door showing a goat routine did not have to be done. Sometimes, he conned the contestant into switching door (if the contestant’s original door is a car), in which case, the switching strategy would have a zero chance of winning.

So there is psychological factor to consider in the actual Monty Hall game. The host may try to trick you into making a wrong move.

In fact, the statement of the problem given above (the same wording as in Marilyn vos Savant’s column) has a loop hole of sort. It did not explicitly make clear that the host will always open a door with a goat and then offer a switch. Monty Hall said, “if the host is required to open a door all the time and offer you a switch, then you should take the switch.” Furthermore, he said “but if he has the choice whether to allow a switch or not, beware. Caveat emptor. It all depends on his mood.” Here’s his advice, “if you can get me to offer you $5,000 not to open the door, take the money and go home.”

So the actual way in which the game is played may render the “academic” solution of the Monty Hall problem useless or meaningless. In any case, whenever the standard solution of Monty Hall Problem is given, the caveat should also be given. Indeed, if the game is played according to a random chances and if the contestant always switches door, then the company that hosts the game would lose too many cars! For financial reasons, they cannot run the show like clock work.

The second piece from NY Times is on the cognitive dissonance that some people experienced with the Monty Hall problem.

The third piece is a more recent discussion and is concerned with variations of the standard Monty Hall Problem. This piece consider tweaks to the basic game in two ways. The question: does each way of tweaking reduce the probability of the host losing a car? Or what is the probability of the host losing a car? Here’s the two tweaks.

First challenge: Suppose Monty Hall’s game show is losing money, and they want to tweak the game so they don’t lose the car as frequently. They announce to the audience that once you choose the door, the host will flip a fair coin to decide which of the other doors will be opened. If the car is behind the door that they open, you lose. If it is not behind the door that they open, you get to choose whether to stick with your door or to change to the other closed door. What do you do if that happens? And now, going into the game, assuming that the contestant behaves optimally, what is the probability that the game show will lose the car?

Second challenge: Now suppose that the game show gets even greedier, and secretly tweaks the game even more. But news of their tweaking is leaked, and the contestants find out that the host is actually using a weighted coin when they are determining which other door will be opened, and the coin is weighted such that the door with the car behind it will be opened by the host with ¾ probability. (When the car is behind the door that you choose initially, the host uses a fair coin to determine which of the other doors to open.) Now, if the door that they open doesn’t have the car behind it – should you stick with your original choice or should you switch? And, assuming that the contestant behaves optimally, did the nefarious game show succeed in reducing the probability that they lose the car?

In each tweak, the problem assumes that the contestant behaves optimally in the game. I suppose that means the contestant always switches door if the switching is possible. An even more interesting exercise would be to calculate the probability under the switching scenario and under the sticking scenario.

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\copyright 2017 – Dan Ma

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The gambler’s ruin problem

This post discusses the problem of the gambler’s ruin. We start with a simple illustration.

Two gamblers, A and B, are betting on the tosses of a fair coin. At the beginning of the game, player A has 1 coin and player B has 3 coins. So there are 4 coins between them. In each play of the game, a fair coin is tossed. If the result of the coin toss is head, player A collects 1 coin from player B. If the result of the coin toss is tail, player A pays player B 1 coin. The game continues until one of the players has all the coins (or one of the players loses all his/her coins). What is the probability that player A ends up with all the coins?

Intuitively, one might guess that player B is more likely to win all the coins since B begins the game with more coin. For example, player A in the example has only 1 coin to start. Thus there is a 50% chance that he/she will lose everything on the first toss. On the other hand, player B has a cushion since he/she can afford to lose some tosses at the beginning.

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Simulated Plays of the Game

One way to get a sense of the winning probability of a player is to play the game repeatedly. Rather than playing with real money, why not simulate a series of coin tosses using random numbers generated in a computer? For example, the function =RAND() in Excel generates random numbers between 0 and 1. If the number is less than 0.5, we take it as a tail (T). Otherwise it is a head (H). As we simulate the coin tosses, we add or subtract to each player’s account accordingly. If the coin toss is a T, we subtract 1 from A and add 1 to B. If the coin toss is an H, we add 1 to A and subtract 1 from B. Here’s a simulation of the game.

    \begin{array}{ccccccc}    \text{Game} &  \text{ } & \text{Coin Tosses} &  \text{ } & \text{Coins of Player A}  & \text{ } & \text{Coins of Player B}  \\    \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   \text{ } &   \text{ }& \text{ } &   \text{ } & 1 &  & 3  \\   1 &   \text{ }& H &   \text{ } & 2 &  & 2  \\   \text{ } &   \text{ }& T &   \text{ } & 1 &  & 3  \\  \text{ } &   \text{ }& H &   \text{ } & 2 &  & 2  \\   \text{ } &   \text{ }& H &   \text{ } & 3 &  & 1  \\   \text{ } &   \text{ }& H &   \text{ } & 4 &  & 0  \\       \end{array}

In the first simulation, player A got lucky with 4 heads in 5 tosses. The following shows the next simulation.

    \begin{array}{ccccccc}    \text{Game} &  \text{ } & \text{Coin Tosses} &  \text{ } & \text{Coins of Player A}  & \text{ } & \text{Coins of Player B}  \\    \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   \text{ } &   \text{ }& \text{ } &   \text{ } & 1 &  & 3  \\   2 &   \text{ }& H &   \text{ } & 2 &  & 2  \\   \text{ } &   \text{ }& H &   \text{ } & 3 &  & 1  \\  \text{ } &   \text{ }& T &   \text{ } & 2 &  & 2  \\   \text{ } &   \text{ }& H &   \text{ } & 3 &  & 1  \\   \text{ } &   \text{ }& H &   \text{ } & 4 &  & 0  \\        \end{array}

Player A seems to be on a roll. Here’s the results of the next two simulated games.

    \begin{array}{ccccccc}    \text{Game} &  \text{ } & \text{Coin Tosses} &  \text{ } & \text{Coins of Player A}  & \text{ } & \text{Coins of Player B}  \\    \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   \text{ } &   \text{ }& \text{ } &   \text{ } & 1 &  & 3  \\   3 &   \text{ }& T &   \text{ } & 0 &  & 4  \\    \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   \text{ } &   \text{ }& \text{ } &   \text{ } & 1 &  & 3  \\   4 &   \text{ }& H &   \text{ } & 2 &  & 2  \\  \text{ } &   \text{ }& T &   \text{ } & 1 &  & 3  \\   \text{ } &   \text{ }& T &   \text{ } & 0 &  & 4  \\           \end{array}

Now player A loses both of these games. For good measure, we show one more simulation.

    \begin{array}{ccccccc}    \text{Game} &  \text{ } & \text{Coin Tosses} &  \text{ } & \text{Coins of Player A}  & \text{ } & \text{Coins of Player B}  \\    \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   \text{ } &   \text{ }& \text{ } &   \text{ } & 1 &  & 3  \\   5 &   \text{ }& H &   \text{ } & 2 &  & 2  \\   \text{ } &   \text{ }& H &   \text{ } & 3 &  & 1  \\  \text{ } &   \text{ }& T &   \text{ } & 2 &  & 2  \\   \text{ } &   \text{ }& H &   \text{ } & 3 &  & 1  \\   \text{ } &   \text{ }& T &   \text{ } & 2 &  & 2  \\    \text{ } &   \text{ }& H &   \text{ } & 3 &  & 1  \\  \text{ } &   \text{ }& T &   \text{ } & 2 &  & 2  \\   \text{ } &   \text{ }& T &   \text{ } & 1 &  & 3  \\   \text{ } &   \text{ }& T &   \text{ } & 0 &  & 4  \\      \end{array}

This one is a little more drawn out. Several tosses of tail in a row spells bad news for player A. When a coin is tossed long enough, there bounds to be some runs of tails. Since it is a fair coin, there bounds to be runs of head too, which benefit player A. Wouldn’t the runs of H cancel out the runs of T so that each player wins about half the time? To get an idea, we continue with 95 more simulations (to get a total of 100). Player A wins 23 of the simulated games and player B wins 77 of the games. So player A wins about 25% of the times. Note that A owns about 25% of the coins at the start of the game.

The 23 versus 77 in the 100 simulated games may be just due to luck for player B. We simulate 9 more runs of 100 games each (for a total of 10 runs with 100 games each). The following shows the results.

    Ten Simulated Runs of 100 Games Each
    \begin{array}{ccccccc}    \text{Simulated} &  \text{ } & \text{Player A} &  \text{ } & \text{Player B}  & \text{ } & \text{Total}  \\  \text{Runs} &  \text{ } & \text{Winning} &  \text{ } & \text{Winning}  & \text{ } & \text{ }  \\  \text{ } &  \text{ } & \text{Frequency} &  \text{ } & \text{Frequency}  & \text{ } & \text{ }  \\    \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   1 &   \text{ }& 23 &   \text{ } & 77 &  & 100  \\   2 &   \text{ }&34 &   \text{ } & 66 &  & 100  \\   3 &   \text{ }& 19 &   \text{ } & 81 &  & 100  \\  4 &   \text{ }& 24 &   \text{ } & 76 &  & 100  \\   5 &   \text{ }& 27 &   \text{ } & 73 &  & 100  \\   6 &   \text{ }& 22 &   \text{ } & 78 &  & 100  \\   7 &   \text{ }& 21 &   \text{ } & 79 &  & 100  \\  8 &   \text{ }& 24 &   \text{ } & 76 &  & 100  \\  9 &   \text{ }& 28 &   \text{ } & 72 &  & 100  \\  10 &   \text{ }& 24 &   \text{ } & 76 &  & 100  \\  \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\  \text{Total} &   \text{ }& 246 &   \text{ } & 754 &  & 1000  \\    \end{array}

The simulated runs fluctuate quite a bit. Player A wins any where from 19% to 34% of the times. The results are no where near even odds of winning for player A. In the 1,000 simulated games, player A wins 24.6% of the times, roughly identical to the player A’s share of the coins at the beginning. If the simulation is carried many more times (say 10,000 times or 100,000 times), the overall winning probability for player A would be very close to 25%.

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Gambler’s Ruin Problem

Here’s a slightly more general description of the problem of gambler’s ruin.

Gambler’s Ruin Problem

Two gamblers, A and B, are betting on the tosses of a fair coin. At the beginning of the game, player A has n_1 coins and player B has n_2 coins. So there are n_1+n_2 coins between them. In each play of the game, a fair coin is tossed. If the result of the coin toss is head, player A collects 1 coin from B. If the result of the coin toss is tail, player A pays B 1 coin. The game continues until one of the players has all the coins. What is the probability that player A ends up with all the coins? What is the probability that player B ends up with all the coins?

Solution: Gambler’s Ruin Problem

The long run probability of player A winning all the coins is \displaystyle P_A=\frac{n_1}{n_1+n_2}.
The long run probability of player B winning all the coins is \displaystyle P_B=\frac{n_2}{n_1+n_2}.

In other words, the probability of winning for a player is the ratio of the number of coins the player starts with to the total numbers of coins. Turning it around, the probability of a player losing all his/her coins is the ratio of the number of coins of the other player to the total number of coins. Thus the player with the fewer number of coins at the beginning has a greater chance of losing everything. Think a casino. Say, it has 10,000,000 coins. You, as a gambler, have 100 coins. Then you would have a 99.999% chance of losing all your coins.

The 99.999% chance of losing everything is based on the assumption that all bets are at even odds. This is certainly not the case in a casino. The casino enjoys a house edge. So the losing probabilities would be worse for a gambler playing in a casino.

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Deriving the Probabilities

We derive the winning probability for a more general case. The coin that is tossed does not have to be a fair coin. Let p be the probability that the coin toss results in a head. Let q=1-p. Suppose that there are n coins initially between player A and player B.

Let A_i be the probability that player A will own all the n coins when player A starts the game with i coins and player B starts with n-i coins. On the other hand, let B_i be the probability that player B will own all the n coins when player A starts the game with i coins and player B starts with n-i coins.

The goal here is to derive expressions for both A_i and B_i for i=1,2,\cdots,n-1 and to show that A_i+B_i=1. The last point, though seems like common sense, is not entirely trivial. It basically says that the game will end in finite number of moves (it cannot go on indefinitely).

Another point to keep in mind is that we can assume A_0=0 and A_n=1 as well as B_0=0 and B_n=1. In the following derivation, H is the event that the first toss is a head and T is the event that the first toss is a tail.

Let E be the event that player A owning all the coins when player A starts the game with i coins and player B starts with n-i coins, i.e. A_i=P(E). First condition on the outcome of the first toss of the coin.

    \displaystyle \begin{aligned} A_i&=P(E | H) \ P(H)+P(E | T) \ P(T) \\&=P(E | H) \ p+P(E | T) \ q  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \end{aligned}

Consider the conditional probabilities P(E | H) and P(E | T). Note that P(E | H)=A_{i+1}. If the first toss is a head, then player A has i+1 coins and player B would have n-(i+1) coins. At that point, the probability of player A owning all the coins would be the same as the probability of winning as if the game begins with player A having i+1 coins. Similarly, P(E | T)=A_{i-1}. Plug these into (1), we have:

    A_i=p \ A_{i+1}+q \ A_{i-1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

Further derivation can be made on equation (2).

    A_i=p \ A_i+ q \ A_i=p \ A_{i+1}+q \ A_{i-1}

    p \ A_{i+1}-p \ A_i=q \ A_i - q \ A_{i-1}

    \displaystyle A_{i+1}-A_{i}=\frac{q}{p} \ (A_i-A_{i-1}),\ \ \ \ i=1,2,3,\cdots,n-1

Plugging in the values 1,2,3,\cdots,n-1 for i produces the following n-1 equations.

    \displaystyle \begin{aligned} &A_{2}-A_{1}=\frac{q}{p} \ (A_1-A_{0})=\frac{q}{p} \ A_1 \\&A_{3}-A_{2}=\frac{q}{p} \ (A_2-A_{1})=\bigg[ \frac{q}{p} \biggr]^2 \ A_1 \\&\ \ \ \cdots \\&A_{i}-A_{i-1}=\frac{q}{p} \ (A_{i-1}-A_{i-2})=\bigg[ \frac{q}{p} \biggr]^{i-1} \ A_1 \\&\ \ \ \cdots \\&A_{n}-A_{n-1}=\frac{q}{p} \ (A_{n-1}-A_{n-2})=\bigg[ \frac{q}{p} \biggr]^{n-1} \ A_1 \end{aligned}

Adding the first i-1 equations produces the following equations.

    \displaystyle A_i-A_1=A_1 \ \biggl[\frac{q}{p}+\bigg( \frac{q}{p} \biggr)^{2}+\cdots+\bigg( \frac{q}{p} \biggr)^{i-1} \biggr]

    \displaystyle A_i=A_1 \ \biggl[1+\frac{q}{p}+\bigg( \frac{q}{p} \biggr)^{2}+\cdots+\bigg( \frac{q}{p} \biggr)^{i-1} \biggr] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)

There are two cases to consider. Either \frac{q}{p}=1 or \frac{q}{p} \ne 1. The first case means p=\frac{1}{2} and the second case means p \ne \frac{1}{2}. The following is an expression for A_i for the two cases.

    \displaystyle  A_i = \left\{ \begin{array}{ll}           \displaystyle  i \ A_1 &\ \ \ \ \ \ \displaystyle \frac{q}{p}=1 \\            \text{ } & \text{ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4) \\          \displaystyle  \frac{ \displaystyle 1- \biggl(\frac{q}{p} \biggr)^i}{\displaystyle 1-\frac{q}{p}} A_1 &\ \ \ \ \ \ \displaystyle \frac{q}{p} \ne 1            \end{array} \right.

The above two-case expression for A_i is applicable for i=2,3,\cdots,n. Plugging n into i with A_n=1 gives the following expression for A_1:

    \displaystyle  A_1 = \left\{ \begin{array}{ll}           \displaystyle  \frac{1}{n} &\ \ \ \ \ \ p=\frac{1}{2} \\            \text{ } & \text{ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5) \\          \displaystyle  \frac{\displaystyle 1-\frac{q}{p}}{1-\biggl( \displaystyle\frac{q}{p} \biggr)^n} &\ \ \ \ \ \ p \ne \frac{1}{2}            \end{array} \right.

Plugging A_1 from (5) into (4) gives the following:

    \displaystyle  A_i = \left\{ \begin{array}{ll}           \displaystyle  \frac{i}{n} &\ \ \ \ \ \ p=\frac{1}{2} \\            \text{ } & \text{ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6) \\          \displaystyle  \frac{1-\biggl( \displaystyle \frac{q}{p} \biggr)^i}{1-\biggl( \displaystyle \frac{q}{p} \biggr)^n} &\ \ \ \ \ \ p \ne \frac{1}{2}            \end{array} \right.

The expression for A_i in (6) is applicable for i=1,2,\cdots,n. For the case of p=\frac{1}{2}, the probability of player A owning all the coin is the ratio of the initial amount of coins of player A to the total number of coins, identical to what is stated in the previous section. Recall that B_i is the probability that player B will end up owning all the coins when initially player A has i coins and player B has n-i coins. By symmetry, the following gives the expression for B_i.

    \displaystyle  B_i = \left\{ \begin{array}{ll}           \displaystyle  \frac{n-i}{n} &\ \ \ \ \ \ q=\frac{1}{2} \\            \text{ } & \text{ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (7) \\          \displaystyle  \frac{1- \biggl(\displaystyle \frac{p}{q} \biggr)^{n-i}}{\displaystyle 1- \biggl(\frac{p}{q} \biggr)^n} &\ \ \ \ \ \ q \ne \frac{1}{2}            \end{array} \right.

For the case p=\frac{1}{2}, which is equivalent to q=\frac{1}{2}, it is clear that A_i+B_i=1. For the case p \ne \frac{1}{2}, the following shows that A_i+B_i=1.

    \displaystyle \begin{aligned} A_i+B_i&=\frac{1-\biggl( \displaystyle \frac{q}{p} \biggr)^i}{1-\biggl( \displaystyle \frac{q}{p} \biggr)^n}+\frac{1- \biggl(\displaystyle \frac{p}{q} \biggr)^{n-i}}{\displaystyle 1- \biggl(\frac{p}{q} \biggr)^n} \\&\text{ } \\&=\frac{p^n-p^n\biggl( \displaystyle \frac{q}{p} \biggr)^i}{p^n-q^n}+\frac{q^n- q^n \biggl(\displaystyle \frac{p}{q} \biggr)^{n-i}}{\displaystyle q^n- p^n}\\&\text{ } \\&=\frac{p^n-p^{n-i} \ q^i}{p^n-q^n}-\frac{q^n- q^i \ p^{n-i}}{\displaystyle p^n- q^n} \\&\text{ } \\&=\frac{p^n-q^n}{p^n-q^n}=1  \end{aligned}

The fact that A_i+B_i=1 shows that the gambling game discussed here has to end after certain number of plays and that it will not continue indefinitely.

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Further Discussion

The formulas derived here are quite profound, not to mention impactful and informative for anyone who gambles. More calculations and discussion can be found here in a companion blog on statistics.

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\copyright 2017 – Dan Ma

The birthday problem

Consider this random experiment. You ask people (one at a time) of their birthdays (month and day only). The process continues until there is a repeat in the series of birthdays, in other words, until two people you’ve surveyed share the same birthday. How many people you have to ask before finding a repeat? What is the probability that you will have to ask k people before finding a repeat? What is the median number of people you have to ask? In this post, we discuss this random variable and how this random variable relates to the birthday problem.

In the problem at hand, we ignore leap year and assume that each of the 365 days in the year is equally likely to be the birthday for a randomly chosen person. The birthday problem is typically the following question. How many people do we need to choose in order to have a 50% or better chance of having a common birthday among the selected individuals?

The random experiment stated at the beginning can be restated as follows. Suppose that balls are randomly thrown (one at a time) into n cells (e.g. boxes). The random process continues until a ball is thrown into a cell that already has one ball (i.e. until a repeat occurs). Let X_n be the number of balls that are required to obtain a repeat. Some of the problems we discuss include the mean (the average number of balls that are to be thrown to get a repeat) and the probability function. We will also show how this random variable is linked to the birthday problem when n=365.

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The Birthday Problem

First, we start with the birthday problem. The key is to derive the probability that in a group of k randomly selected people, there are at least two who share the same birthday. It is easier to do the complement – the probability of having different birthdays among the group of k people. We call this probability p_k.

    \displaystyle \begin{aligned} p_k&=\frac{365}{365} \ \frac{364}{365} \ \frac{363}{365} \cdots \frac{365-(k-1)}{365} \\&=\frac{364}{365} \ \frac{363}{365} \cdots \frac{365-(k-1)}{365} \\&=\biggl[1-\frac{1}{365} \biggr] \ \biggl[1-\frac{2}{365} \biggr] \cdots \biggl[1-\frac{k-1}{365} \biggr] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \end{aligned}

where k=2,3,4,\cdots,365. The reasoning for the first line is that there are 365 choices to be picked in the first selection. Each subsequent random selection has to avoid the previous birthday, thus 364 choices for the second person and only 365-(k-1) choices for the kth person.

To answer the birthday problem, just plug in values of k to compute p_k and 1-p_k until reaching the smallest k such that p_k<0.5 and 1-p_k>0.5. The calculation should be done using software (Excel for example). The smallest k is 23 since

    p_{23}= 0.492702766

    1-p_{23}= 0.507297234

In a random group of 23 people, there is a less than 50% chance of having distinct birthdays, and thus a more than 50% chance of having at least one identical birthday. This may be a surprising result. Without the benefit of formula (1), some people may think that it will take a larger sample to obtain a repeat.

The benefit of (1) extends beyond the birthday problem. Let’s consider the case for n, i.e. randomly pick numbers from the set \left\{1,2,3,\cdots,n-1,n \right\} with replacement until a number is chosen twice (until a repeat occurs). Similarly, let p_{n,k} be the probability that in k draws all chosen numbers are distinct. The probability p_{n,k} is obtained by replacing 365 with n.

    \displaystyle \begin{aligned} p_{n,k}&=\frac{n}{n} \ \frac{n-1}{n} \ \frac{n-2}{n} \cdots \frac{n-(k-1)}{n} \\&=\frac{n-1}{n} \ \frac{n-2}{n} \cdots \frac{n-(k-1)}{n} \\&=\biggl[1-\frac{1}{n} \biggr] \ \biggl[1-\frac{2}{n} \biggr] \cdots \biggl[1-\frac{k-1}{n} \biggr] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \end{aligned}

Formula (2) will be useful in the next section.

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The Random Variable

We now look into the random variable discussed at the beginning, either the one for picking people at random until a repeated birthday or throwing balls into cells until one cell has two balls. To illustrate the idea, let’s look at an example.

Example 1
Roll a fair die until obtaining a repeated face value. Let X_6 be the number of rolls to obtain the repeated value. Find the probability function P[X_6=k] where k=2,3,4,5,6,7.

Note that P[X_6=k] is the probability that it will take k rolls to get a repeated die value.

    \displaystyle P(X_6=2)=\frac{6}{6} \times \frac{1}{6}=\frac{1}{6}

    \displaystyle P(X_6=3)=\frac{6}{6} \times \frac{5}{6} \times \frac{2}{6}=\frac{10}{6^2}

    \displaystyle P(X_6=4)=\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6}=\frac{60}{6^3}

    \displaystyle P(X_6=5)=\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{4}{6}=\frac{240}{6^4}

    \displaystyle P(X_6=6)=\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{5}{6}=\frac{600}{6^5}

    \displaystyle P(X_6=7)=\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{1}{6} \times \frac{6}{6}=\frac{720}{6^6}

To get a repeat in 2 rolls, there are 6 choices for the first roll and the second has only one choice – the value of the first roll. To get a repeat in 3 rolls, there are 6 choices for the first roll, 5 choices for the second roll and the third roll must be out of the 2 previous two distinct values. The idea is that the first k-1 rolls are distinct and the last roll must be one of the previous values. \square

The reasoning process leads nicely to the general case. In the general case, let’s consider the occupancy interpretation. In throwing balls into n cells, let X_n be defined as above, i.e. the number of balls that are required to obtain a repeat. The following gives the probability P[X_n=k].

    \displaystyle \begin{aligned} P[X_n=k]&=\frac{n}{n} \times \frac{n-1}{n} \times \cdots \times \frac{n-(k-2)}{n} \times \frac{k-1}{n} \\&=\frac{n-1}{n} \times \cdots \times \frac{n-(k-2)}{n} \times \frac{k-1}{n} \\&=\frac{(n-1) \times (n-2) \times \cdots \times (n-(k-2)) \times (k-1)}{n^{k-1}} \\&=\biggl[1-\frac{1}{n} \biggr] \times \biggl[1-\frac{2}{n} \biggr] \times \cdots \times \biggl[1-\frac{k-2}{n}\biggr] \times \frac{k-1}{n}  \ \ \ \ \ \ \ (3) \end{aligned}

where k=2,3,\cdots,n+1.

The reasoning is similar to Example 1. To get a repeat in throwing k balls, the first k-1 balls must be go into different cells while the last ball would go into one of the k-1 occupied cells. For the first k-1 balls to go into different cells, there are n (n-1) \cdots (n-(k-2)) ways. There are k-1 cells for the last ball to land. Thus the product of these two quantities is in the numerator of (3).

Once the probability function (3) is obtained, the mean E[X_n] can be derived accordingly. For the case of n=365, E[X_{365}]=24.62 (calculated by programming the probability function in Excel). On average it will be required to sample about 25 people to obtain a repeated birthday.

Another interesting quantity is P[X_n>k]. This is the probability that it will take throwing more than k balls to get a repeat. Mathematically this can be obtained by first calculating P[X_n \le k] by summing the individual probabilities via (3). This is a workable approach using software. There is another way that is more informative. For the event X_n>k to happen, the first k throws must be in different cells (no repeat). The event X_n>k is identical to the event that there is no repeat in the first k throws of balls. This is how the random variable X_n is linked to the birthday problem since the probability P[X_n>k] should agree with the probability p_{n,k} in (2).

    \displaystyle P[X_n>k]=\biggl[1-\frac{1}{n} \biggr] \ \biggl[1-\frac{2}{n} \biggr] \cdots \biggl[1-\frac{k-1}{n} \biggr] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)

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Back to The Birthday Problem

Consider the case for X_{365}. What is the median of X_{365}? That would be the median number of people surveyed to obtain a pair of identical birthday. The median of X_{365} would be the least k such that P[X_{365} \le k] is at least 0.5. Note that P[X_{365}>k] is identical to p_k in (1). The above calculation shows that p_{23}=0.4927 and 1-p_{23}=0.5073. Thus the median of X_{365} is 23. Thus when performing the random sampling of surveying birthday, about half of the times you can expect to survey 23 or fewer than 23 people.

The birthday problem is equivalently about finding the median of the random variable X_{365}. A little more broadly, the birthday problem is connected to the percentiles of the variable X_{n}. In contrast, the mean of X_{365} is E[X_{365}]=24.62. The following lists several percentiles for the random variable X_{365}.

    \begin{array}{ccccccc}    k &  \text{ } & P[X_{365}>k]  & \text{ } & P[X_{365} \le k] & \text{ } & \text{Percentile} \\    \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   14 &   \text{ } & 0.77690 &  & 0.22310 &  \\   15 &   \text{ } & 0.74710 &  & 0.25290 & \text{ } & \text{25th Percentile} \\   16 &   \text{ } & 0.71640 &  & 0.28360 &  \\  \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   22 &   \text{ } & 0.52430 &  & 0.47570 &  \\   23 &   \text{ } & 0.49270 &  & 0.50730 & \text{ } & \text{50th Percentile}  \\   24 &   \text{ } & 0.46166 &  & 0.53834 &  \\  \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   31 &   \text{ } & 0.26955 &  & 0.73045 &  \\   32 &   \text{ } & 0.24665 &  & 0.75335 & \text{ } & \text{75th Percentile} \\         33 &   \text{ } & 0.22503 &  & 0.77497 &  \\  \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   40 &   \text{ } & 0.10877 &  & 0.89123 &  \\   41 &   \text{ } & 0.09685 &  & 0.90315 & \text{ } & \text{90th Percentile} \\   42 &   \text{ } & 0.08597 &  & 0.91403 &  \\  \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   46 &   \text{ } & 0.05175 &  & 0.94825 &  \\   47 &   \text{ } & 0.04523 &  & 0.95477 & \text{ } & \text{95th Percentile} \\    48 &  \text{ } & 0.03940 &  & 0.96060 &  \\  \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\    56 &  \text{ } & 0.01167 &  & 0.98833 &  \\  57 &  \text{ } & 0.00988 &  & 0.99012 & \text{ } & \text{99th Percentile} \\  58 &  \text{ } & 0.00834 &  & 0.99166 &  \\      \end{array}

It is clear that in a group of 366 people, it is certain that there will be at least one repeated birthday (again ignoring leap year). This is due to the pigeon hole principle. As the percentiles in the above table shows, you do not need to survey anywhere close to 366 to get a repeat. The median is 23 as discussed. The 75th percentile of X_{365} is 32.

The preceding calculation shows that you do not need a large group to have a repeated birthday. About 50% of the times, you will survey 23 or fewer people, about 75% of the time, 32 or fewer people, About 99% of the time, you will survey 57 or fewer people, much fewer than 365 or 366. So with around 50 in a random group, there is a near certainty of finding a shared birthday. In a random group of 100 people, there should be an almost absolute certainty that there is a shared birthday.

For a further demonstration, we simulated the random variable X_{365} 10,000 times. The range of the simulated values is 2 to 78. Thus the odds for 100 people to survey is smaller than 1 in 10,000. To get a simulated value of 100, we will have to simulate more than 10,000 values of X_{365}. The median of the 10,000 simulated results is 23. The following table summarizes the results.

    \begin{array}{rrrrrr}    \text{Interval} &  \text{ } & \text{Count}  & \text{ } & \text{Proportion} & \\    \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   \text{2 to 9} &   \text{ } &953 &  & 0.09530 &  \\   \text{10 to 19} &   \text{ } & 2870 &  & 0.28700 &  \\   \text{20 to 29} &   \text{ } & 3055 &  & 0.30550 &  \\     \text{30 to 39} &   \text{ } & 1922 &  & 0.19220 &  \\   \text{40 to 49} &   \text{ } & 886 &  & 0.08860 &   \\   \text{50 to 59} &   \text{ } & 253 &  & 0.02530 &  \\     \text{60 to 69} &   \text{ } & 48 &  & 0.00480 &  \\   \text{70 to 78} &   \text{ } & 13 &  & 0.00130 &  \\              \end{array}

Not shown in the table is that 33 of the simulated results are the value of 2. Thus it is possible to ask two people and they both have the same birthday. But the odds of that happening is 33 out of 10,000 according to this particular set of simulations (probability 0.0033). The theoretical probability of 2 is 1/365 = 0.002739726. There are 2 instances of 78 in the 10,000 simulated values. Thus the odds are 2 out of 10,000 with probability 0.0002. The theoretical probability is 0.000037 using (3).

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\copyright \ 2017 \text{ by Dan Ma}

How long does it take to collect all coupons?

This post discusses the coupon collector problem, a classical problem in probability.

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The Coupon Collector Problem

The problem is usually stated as a coupon collector trying to collect the entire set of coupons. For example, each time the coupon collector buys a product (e.g. a box of breakfast cereal), he receives a coupon, which is a prize that can be a toy or a baseball card or other interesting item. Suppose that there are n different types of coupons (prizes) and that the coupon collector wishes to collect the entire set. How many units of the product does the coupon collector have to buy in order to collect the entire set?

A simplified discussion of the coupon collector problem is found in another blog post. This post is more detailed discussion.

This blog post in another blog discusses the coupon collector problem from a simulation perspective.

As shown below, if there are 5 different coupons, it takes 12 purchases on average to get all coupons. If there are 10 different types of coupons, on average it would take 30 purchases. If there are 50 different types of coupons, it would take on average 225 purchases collect the entire set. The first few coupons are obtained fairly quickly. As more and more coupons are accumulated, it is harder to get the remaining coupons. For example, for the 50-coupon case, after the 49 coupons are obtained, it takes on average 50 purchases to get the last coupon.

Suppose that the coupon collector does not want to collect the entire set and only wishes to collect r distinct coupons where r \le n. It turns out that this special case only requires a minor tweak to the case of collecting the entire set. Our strategy then is to focus on the main case. The special case will be discussed at the end of the post.

We first consider the main case that the coupon collector wishes to collect the entire set. The problem can be cast as a random sampling from the population S=\left\{1,2,3,\cdots,n \right\}. Selecting a number at random from S with replacement is equivalent to the coupon collector receiving a coupon. Let X_n be the minimum number of selections such that each number in S is picked at least once. In this post we discuss the probability function of X_n as well as its mean and variance.

Another interpretation of the problem is that it can be viewed as an occupancy problem, which involves throwing balls into n cells at random. The random quantity of interest is the number of balls that are required to be thrown such that each cell has at least one ball. Clearly this formulation is identical to the coupon interpretation and the random sampling interpretation. The angle of occupancy problem is useful since we can leverage the formulas developed in this previous post. A description of the occupancy problem is given here.

Regardless of the interpretation, the goal is obtain information on the random variable X_n, the minimum number of random selections from S=\left\{1,2,3,\cdots,n \right\} in order to have the complete set of n distinct values represented in the sample.

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Mean and Variance

The mean and variance of X_n are easier to obtain. So that is where we begin. The key is to break up X_n into a sum as follows:

    X_n=C_{1}+C_{2}+C_{3}+\cdots+C_{i-1}+C_{i}+\cdots+C_{n} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (0)

where C_{i} is the additional selections from S=\left\{1,2,3,\cdots,n \right\} to get a number that is distinct from the i-1 distinct numbers that have been chosen. For example, C_3 is the number of random selections to get a number that is distinct from the two distinct numbers obtained up to that point.

Note that each C_i involves repeated sampling until some criterion is reached, thus resembling a geometric random variable. Indeed they are. As the sampling continues and as more distinct values are obtained, it is not as easy to obtain a new number. After i-1 distinct numbers have been obtained, the probability of drawing a new distinct number is p=\frac{n-(i-1)}{n}. As geometric random variables, each C_i has the following mean and variance.

    \displaystyle E[C_i]=\frac{1}{p}=\frac{n}{n-(i-1)}

    \displaystyle Var[C_i]=\frac{1-p}{p^2}=\frac{n(i-1)}{[n-(i-1)]^2}

where i=1,2,3,\cdots,n. Note that the random variables C_i are independent. The value of C_i does not depend on how many trials it takes to draw the previous i-1 distinct numbers. The following gives the mean and variance of X_n.

    \displaystyle E[X_n]=\sum \limits_{i=1}^n E[C_i]=\sum \limits_{i=1}^n \frac{n}{n-(i-1)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

    \displaystyle Var[X_n]=\sum \limits_{i=1}^n Var[C_i]=\sum \limits_{i=1}^n \frac{n(i-1)}{[n-(i-1)]^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

The expectation E[X_n] can be rearranged as follows to give more information.

    \displaystyle E[X_n]=n \ \biggl[\frac{1}{n}+ \cdots + \frac{1}{3}+ \frac{1}{2} + 1\biggr]=n \ H_n  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)

The quantity H_n is the partial sum of the harmonic series. Note that H_n \rightarrow \infty as n \rightarrow \infty. Thus E[X_n] \rightarrow \infty as n \rightarrow \infty. The quantity H_n can be interpreted as the average number of units of product that are required to purchase per coupon. The following table lists out the expected values for selected values of n.

Table 1

    \begin{array}{cccccc}    \text{Number of} &  \text{ } & \text{Expected Number of Trials}  & \text{ } & \text{Expected Total Number} &  \\  \text{Coupons} &  \text{ } & \text{per Coupon}  & \text{ } & \text{of Trials (rounded up)} &  \\    n &  \text{ } & E[H_n]  & \text{ } & E[X_n] &  \\  \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   1 &   \text{ } & 1.0000 &  & 1 &  \\   2 &   \text{ } & 1.5000 &  & 3 &  \\   3 &   \text{ } & 1.8333 &  & 6 &  \\   4 &   \text{ } & 2.0833 &  & 9 &  \\   5 &   \text{ } & 2.2833 &  & 12 &  \\   6 &   \text{ } & 2.4500 &  & 15 &  \\   7 &   \text{ } & 2.5929 &  & 19 &  \\   8 &   \text{ } & 2.7179 &  & 22 &  \\         9 &   \text{ } & 2.8290 &  & 26 &  \\   10 &   \text{ } & 2.9290 &  & 30 &  \\   20 &   \text{ } & 3.5977 &  & 72 &  \\   30 &   \text{ } & 3.9950 &  & 120 &  \\   40 &   \text{ } & 4.2785 &  & 172 &  \\   50 &   \text{ } & 4.4992 &  & 225 &  \\    60 &  \text{ } & 4.6799 &  & 281 &  \\    70 &  \text{ } & 4.8328 &  & 339 &  \\  80 &  \text{ } & 4.9655 &  & 398 &  \\  90 &  \text{ } & 5.0826 &  & 458 &  \\  100 &  \text{ } & 5.1874 &  & 519 &  \\    \end{array}

Table 1 gives an estimate on how long to expect to collect the entire set of coupons for selected coupon sizes. The third column gives the expected total number of purchases to obtain the entire coupon set. The second column gives an estimate of how many purchases on average to obtain one coupon. For the 50-coupon case, it takes on average about 4.5 purchases to obtain one coupon. However, it does not tell the whole story. To get the 50th coupon, it takes on average 50 trials. Note that E[C_{50}]=50 in the 50-coupon case in formula (1). In a simulation of the 50-coupon problem, it took 54 trials to obtain the 50th coupon. To get the 49th coupon, it takes on average 50/2 = 25 trials.

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The Occupancy Problem

We now view the coupon collector problem as an occupancy problem in order to leverage a formula from a previous post. Suppose that we randomly throw k balls into n cells. Let Y_{k,n} be the number of empty cells as a result of randomly assigning k balls into n cells. The following gives the probabilities P(Y_{k,n}=j) where j=0,1,2,\cdots,n-1.

    \displaystyle P(Y_{k,n}=0)=\sum \limits_{i=0}^{n} (-1)^{i} \binom{n}{i} \biggl(1-\frac{i}{n}\biggr)^{k} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)

    \displaystyle P(Y_{k,n}=j)=\binom{n}{j} \sum \limits_{i=0}^{n-j} (-1)^{i} \binom{n-j}{i} \biggl(1-\frac{j+i}{n}\biggr)^{k} \ \ \ \ \ \ \ \ \ \ \ \ \ (5)

    where j=1,2, \cdots, n-1.

The notation \binom{n}{j} is the binomial coefficient, which is the number of ways to choose j objects out of n objects where order does not matter. The calculation is defined by \displaystyle \binom{n}{j}=\frac{n!}{j! (n-j)!}.

The formula (5) gives the probability of having j empty cells. In throwing k balls into n cells, the probability of having w occupied cells is then P[Y_{k,n}=n-w].

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The Probability Function

We now discuss the probability function of the random variable X_n, namely P[X_n=k] for k=n,n+1,n+2,\cdots. The event X_n=k means that all n cells are occupied after throwing k balls with the first k-1 balls landing in n-1 cells. Putting it in another way, there are zero empty cells after throwing k balls and there is 1 empty cell after throwing the first k-1 balls. This can be stated using the notation in the preceding section on the occupancy problem as follows:

    P(X_n=k]=P[Y_{k,n}=0 \text{ and } Y_{k-1,n}=1]  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)

Consider the following derivation.

    \displaystyle \begin{aligned} P(X_n=k]&=P[Y_{k,n}=0 \text{ and } Y_{k-1,n}=1] \\&=P[Y_{k,n}=0 \ \lvert \ Y_{k-1,n}=1] \times P[Y_{k-1,n}=1] \\&=\frac{1}{n} \times \binom{n}{1} \sum \limits_{i=0}^{n-1} (-1)^i \binom{n-1}{i} \biggl[ 1-\frac{1+i}{n} \biggr]^{k-1} \\&=\sum \limits_{i=0}^{n-1} (-1)^i \binom{n-1}{i} \biggl[ 1-\frac{1+i}{n} \biggr]^{k-1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (7) \end{aligned}

    where k=n,n+1,n+2,\cdots.

Rather than memorizing the probability function in (7), a better way is to focus on the thought process that is inherent in (6).

One comment about the calculation for (7). The summation for P[X_n=k] has n terms. A given probability may involve multiple values of k, e.g.

    P[X_6>15]=1-P[6 \le X_6 \le 14].

Unless the number of values for k is very small, the calculation should be done using software. Microsoft Excel is an excellent way to perform the calculation. The calculations for the examples below are programmed in Excel.

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Examples

Example 1
Suppose that a fair die is rolled until all 6 faces have appeared. Find the mean number of rolls and the variance of the number of rolls. What is the probability that it will take at least 12 rolls? What is the probability that it will take more than 15 rolls?

Using the notation developed above, the random variable of interest is X_6. Its mean and variance are:

    \displaystyle E[X_6]=6 \ \biggl[ 1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6} \biggr]=6 \times 2.45 = 14.7 \approx 15

    \displaystyle Var[X_6]=\sum \limits_{i=1}^6 \frac{6(i-1)}{[6-(i-1)]^2}=38.99

The following is the probability function for X_6.

    \displaystyle P[X_6=k]=\sum \limits_{i=0}^{5} (-1)^i \binom{5}{i} \biggl[ 1-\frac{1+i}{6} \biggr]^{k-1}

    where k=6,7,8,\cdots

For each k, the quantity P[X_6=k] requires 6 calculations. Performing the calculations in Excel, the desired probabilities are:

    P[X_6 \ge 12]=1-P[6 \le X_6 \le 11]=1-0.356206419=0.643793581

    P[X_6 > 15]=1-P[6 \le X_6 \le 15]=1-0.644212739=0.355787261

Even though the average number of trials is 15, there is still a significant probability that it will take more than 15 trials. This is because the variance is quite large. \square

Example 2
An Internet startup is rapidly hiring new employees. What is the expected number of new employees until all birth months are represented? Assume that the 12 birth months are equally likely. What is the probability that the company will have to hire more than 25 employees? If the company currently has more than 25 employees with less than 12 birth months, what is the probability that it will have to hire more than 35 employees to have all 12 birth months represented in the company?

The random variable of interest is X_{12}. The following shows the mean and probability function.

    \displaystyle E[X_{12}]=12 \ \biggl[ 1 + \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{12} \biggr]=37.23852814 \approx 38

    \displaystyle P[X_{12}=k]=\sum \limits_{i=0}^{11} (-1)^i \binom{11}{i} \biggl[ 1-\frac{1+i}{12} \biggr]^{k-1}

    where k=12,13,14,\cdots

Performing the calculation in Excel, we obtain the following probabilities.

    P[X_{12} > 25]=1-P[12 \le X_{12} \le 25]=1-0.181898592=0.818101408

    P[X_{12} > 35]=1-P[12 \le X_{12} \le 35]=1-0.531821149=0.468178851

    \displaystyle P[X_{12} > 35 \ \lvert \ X_{12} > 25]=\frac{P[X_{12} > 35]}{P[X_{12} > 25]}=\frac{0.468178851}{0.818101408}=0.572274838

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A Special Case

We now consider the special case that the coupon collector only wishes to collect r distinct coupons where r \le n. Of course, n is the total number of distinct coupon types. Let X_{n,r} be the minimum number of purchases such that r distinct coupons have been obtained. In the random sampling interpretation, X_{n,r} would be the minimum sample size such that r distinct elements have been chosen from the sample space S=\left\{1,2,3,\cdots,n \right\}. The mean and variance of X_{n,r} follow from the same idea. Each X_{n,r} is the independent sum of geometric random variables as in (0).

    X_{n,r}=C_{1}+C_{2}+C_{3}+\cdots+C_{i-1}+C_{i}+\cdots+C_{r} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)

    where r \le n.

Thus E[X_{n,r}] and Var[X_{n,r}] would be like (1) and (2) except that the summation is through r instead of n.

For the probability function of X_{n,r}, we only need to tweak the thought process expressed in (6) slightly. For the event X_{n,r}=k to happen, exactly r cells are occupied after throwing k balls with the first k-1 balls landing in r-1 cells. In other words, there are exactly n-r empty cells after throwing k balls and there are exactly n-r+1 empty cells after throwing k-1 balls. The following expresses this condition in terms of the occupancy problem, i.e. similar to (6).

    P[X_{n,r}=k]=P[Y_{k,n}=n-r \text{ and } Y_{k-1,n}=n-r+1]  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (9)

Here’s the important components that need to go into P[X_{n,r}=k] with the first one coming from the occupancy formula (5).

    \displaystyle P[Y_{k-1,n}=n-r+1]=\binom{n}{n-r+1} \sum \limits_{i=0}^{r-1} (-1)^i \binom{r-1}{i} \biggl[ 1-\frac{n-r+1+i}{n} \biggr]^{k-1}

    \displaystyle P[Y_{k,n}=n-r \ \lvert \ Y_{k-1,n}=n-r+1]=\frac{n-r+1}{n}

Multiply the above two probabilities together produces the desired probability P[X_{n,r}=k] for k=r,r+1,r+2,\cdots.

    \displaystyle P[X_{n,r}=k]=\binom{n-1}{r-1} \sum \limits_{i=0}^{r-1} (-1)^i \binom{r-1}{i} \biggl[ 1-\frac{n-r+1+i}{n} \biggr]^{k-1} \ \ \ \ \ \ \ \ (10)

Note that when n=r (collecting the entire set of coupons), formula (10) would be identical to (7). The following example demonstrates the calculation.

Example 3
Consider the 6-coupon case discussed in Example 1. Suppose that the coupon collector is interested in collecting r=4 coupons. What is the expected number of purchases to get 4 coupons? What is the probability that it will take more than 6 purchases to get 4 coupons? What is the probability that it will take more than 8 purchases to get 4 coupons? Compare these results with Example 1.

The random variable of interest is X_{6,4}. The mean is:

    \displaystyle E[X_{6,4}]=6 \biggl[ \frac{1}{6}+\frac{1}{5}+\frac{1}{4}+\frac{1}{3} \biggr]=5.7 \approx 6

Note that it is much faster to obtain 4 coupons than the entire set of six. The following gives the probability function for X_{6,4}.

    \displaystyle P[X_{6,4}=k]=\binom{5}{3} \sum \limits_{i=0}^{3} (-1)^i \binom{3}{i} \biggl[ 1-\frac{3+i}{6} \biggr]^{k-1}

    where k=4,5,6,\cdots

Performing the calculations in Excel gives the following probabilities.

    P[X_{6,4} > 6]=1-P[12 \le X_{12} \le 25]=1-0.74845679=0.25154321

    P[X_{6,4} > 8]=1-P[12 \le X_{12} \le 35]=1-0.928712277=0.071287723

The first probability shows there is still a good chance that it will take more then the mean number of trials to get 4 coupons. The wait time is much less than in Example 1 since the probability of wait time more than 8 is fairly small. \square

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Moment Generating Function

One distributional quantity that is easy to obtain is the moment generating function (MGF) for the random variable X_n (the case of collecting the entire set of coupons) as well as for the random variable X_{n,r} (the partial case). Since both of these random variables are the independent sum of geometric random variables, their MGFs would simply be the product of the individual geometric MGFs. The following gives the results.

    \displaystyle M_{X_n}(t)=\prod \limits_{i=1}^n \frac{[n-(i-1)] e^t}{n-(i-1) e^t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (11)

    \displaystyle M_{X_{n,r}}(t)=\prod \limits_{i=1}^r \frac{[n-(i-1)] e^t}{n-(i-1) e^t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (12)

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\copyright \ 2017 \text{ by Dan Ma}

Looking for a Match

I recently gave an exam in my statistics course, which turned out to be an excellent example for the matching problem, a classic problem in probability. There were 66 students taking the test. I wrote down the names of the students in the order of turning in the exam. The following table shows the positions of the students in alphabetical order and in the order they turned in the exam.

For example, the first student in the alphabetical order was the 12th student who turned in the exam. The first student who turned in the exam was the 37th student in the alphabetical order. It turned out that there was a student who had the same position in both orders. Such a student is called a match (see the following table).

This example captures the essence of a classic problem in probability called the matching problem. There are many other colorful descriptions of the matching problem. All of them involve the pairing of two orderings on a set of objects. We have a set of objects or items that are ordered in some natural way. Then we scramble the items randomly (or according to some ordering that is unpredictable in advance). Whenever an item has the same position in both the original order and in the scrambled order, it is called a match.

In our exam example, one version of the matching problem asks: what is the probability that there is at least one student that is a match?

In fact, in matching two orderings, such as the one described here, a match happens more often than not. The probability of having at least one match is roughly 0.63. Specifically, when n is the number of students taking the exam, the probability of finding at least one match approaches 1-e^{-1}=0.632121 as n \rightarrow \infty. The derivation of this probability is based on the inclusion-exclusion principle and is discussed in the blog post called The Matching Problem.

Even though the probability of having at least one match is a function of n (the number of items), the probability converges to 1-e^{-1} pretty rapidly. Thus for all practical purposes, we can say that the probability of having at least one match is 0.63 (roughly two-third), whenever the number of items involved in the random scrambling is reasonably large (as in the 66 students taking the exam).

Instead of finding the probability of having at least one match, we can also ask for the probability of having exactly k matches, where k is any integer from 0 to n. Let X_n be the number of matches when we match the “exam turning in” order with the alphabetical order for n students. The probability function P(X_n=k) is derived in the blog post called More About the Matching Problem.

The blog post More About the Matching Problem also points out that P(X_n=k) is approximated by the Poisson distribution with parameter \alpha=1. Thus we have:

\displaystyle (1) \ \ \ \ \ P(X_n=k) \approx \frac{e^{-1}}{k!} \ \ \ \ \ \ \ \ \ \ k=0,1,2,\cdots,n

The following are the first 4 probabilities of P(X_n=k).

\displaystyle (2) \ \ \ \ \ P(X_n=0) \approx \frac{e^{-1}}{0!}=0.3679

\displaystyle (3) \ \ \ \ \ P(X_n=1) \approx \frac{e^{-1}}{1!}=0.3679

\displaystyle (4) \ \ \ \ \ P(X_n=2) \approx \frac{e^{-1}}{2!}=0.1839

\displaystyle (5) \ \ \ \ \ P(X_n=3) \approx \frac{e^{-1}}{3!}=0.0313

In the random experiment of matching two orderings on the same set of objects, about 37% of the time, there is no match and about 37% of the time there is exactly one match. Having no matches and having exactly one match are the mostly scenarios (occurring about 74% of the time). Having 2 matches is possible, but only happens about 18% of the time. It is rare to have 3 ore more matches.

Another interesting observation is that if a match occurs, it is mostly likely that there is only one match (such as the example discussed here).

There are many colorful descriptions of the matching problem. The possibilities are limitless. One example is that of n married couples going to a ball room dancing class. The instructor randomly assign the ladies to the gentlemen as dancing partners. A match occurs if a wife is assigned as
the dancing partner of her husband.

A previous blog post (Tis the Season for Gift Exchange) presents an example involving gift exchange. Each person attending a party brings a gift for gift exchange. The gifts are put in a pile and each person randomly selects a gift from the pile. A match occurs if a person selects his or her own gift.

In another blog post (A lazy professor who lets students do their own grading), a professor randomly returns quizzes to the students for grading. A match occurs if a students is assigned his or her own quiz.

A lazy professor who lets students do their own grading

After reading the example discussed here, any professor or instructor who is contemplating randomly distributing quizzes back to the students for grading perhaps should reconsider the idea! This is one of several posts discussing the matching problem. Go to the end of this post to find links to these previous posts.

Consider the following problem. Suppose that a certain professor is lazy and lets his students grade their own quizzes. After he collects the quizzess from his n students, he randomly assigns the quizzes back to these n students for grading. If a student is assigned his or her own quiz, we say that it is a match. We have the following questions:

  • What is the probability that each of the n students is a match?
  • What is the probability that none of the n students is a match?
  • What is the probability that exactly k of the n students are matches?
  • What is the probability that at least one of the n students is a match?

The above problem is called the matching problem, which is a classic problem in probability. In this post we solve the last question indicated above. Though the answer is in terms of the total number of quizzes n, it turns out that the answer is independent of n and is approximately \frac{2}{3}. Thus if the professor assigns the quizzes randomly, it will be very unusual that there is no match.

The last question above is usually stated in other matching situations. One is that there are n couples (say, each couple consists of a husband and a wife) in a class for ballroom dancing. Suppose that the dance instructor randomly matches the men to the ladies. When a husband is assigned his own wife, we say that it is a match. What is the probability that there is at least one couple that is a match?

The key to answering this question is the theorem stated in Feller (page 99 in chapter 4 of [1]). We state the theorem and make use of it in the solution of the last question above. A sketch of the proof will be given at the end. For ideas on the solutions to the first three questions above, see this previous post.

The union of n events
For any n events E_1,E_2, \cdots,E_n that are defined on the same sample space, we have the following formula:

(1) \ \ \ \ \ P[E_1 \cup E_2 \cup \cdots \cup E_n]=\sum \limits_{m=1}^{n} (-1)^{m+1} \thinspace S_m where

S_1=\sum \limits_{r=1}^{n}P[E_r],

S_2=\sum \limits_{j<k}P[E_j \cap E_k],

S_m=\sum P[E_{i(1)} \cap E_{i(2)} \cap \cdots \cap E_{i(m)}]

Note that in the general term S_m, the sum is taken over all increasing sequence i(\cdot), i.e. 1 \le i(1) < i(2) < \cdots < i(m) \le n. For n=2,3, we have the following familiar formulas:

\displaystyle P[E_1 \cup E_2]=P[E_1]+P[E_2]-P[E_1 \cap E_2]

\displaystyle \begin{aligned}P[E_1 \cup E_2 \cup E_3]=& \ \ \ \ P[E_1]+P[E_2]+P[E_3]\\&-P[E_1 \cap E_2]-P[E_1 \cap E_3]-P[E_2 \cap E_3]\\&+P[E_1 \cap E_2 \cap E_3] \end{aligned}

The Matching Problem

Suppose that the n students are labeled 1,2, \cdots, n. Let E_i be the even that the i^{th} student is assigned his or her own quiz by the professor. Then P[E_1 \cup E_2 \cup \cdots \cup E_n] is the probability that there is at least one correct match.

Note that P[E_i]=\frac{(n-1)!}{n!}=\frac{1}{n}. This is the case since we let the i^{th} student be fixed and we permute the other n-1 students. Likewise, P[E_i \cap E_j]=\frac{(n-2)!}{n!}, since we fix the i^{th} and j^{th} students and let the other (n-2)! students permute. In general, whenever i(1),i(2),\cdots,i(m) are m distinct integers and are increasing, we have:

\displaystyle P[E_{i(1)} \cap E_{i(2)} \cdots \cap E_{i(m)}]=\frac{(n-m)!}{n!}

We now apply the formula (1). First we show that for each m where 1 \le m \le n, S_m=\frac{1}{m!}. Since there are \binom{n}{m} many ways to have m matches out of n students, we have:

\displaystyle \begin{aligned}S_m&=\sum P[E_{i(1)} \cap E_{i(2)} \cdots \cap E_{i(m)}]\\&=\binom{n}{m} \frac{(n-m)!}{n!}\\&=\frac{1}{m!} \end{aligned}

Applying the formula for the union of n events, we have:

\displaystyle P[E_1 \cup E_2 \cup \cdots \cup E_n]=1-\frac{1}{2!}+\frac{1}{3!}-\cdots+(-1)^{n+1}\frac{1}{n!}

\displaystyle 1-P[E_1 \cup E_2 \cup \cdots \cup E_n]=1-1+\frac{1}{2!}-\frac{1}{3!}+\cdots+(-1)^{n}\frac{1}{n!}

Note that the left-hand side of the above equality is the first n+1 terms in the expansion of e^{-1}. Thus we have:

\displaystyle \begin{aligned}\lim_{n \rightarrow \infty}P[E_1 \cup E_2 \cup \cdots \cup E_n]&=\lim_{n \rightarrow \infty}\biggl(1-\frac{1}{2!}+\frac{1}{3!}-\cdots+(-1)^{n+1}\frac{1}{n!}\biggr)\\&=1-e^{-1}\\&=0.6321205588 \end{aligned}

The above limit converges quite rapidly. Let P_n=P[E_1 \cup E_2 \cup \cdots \cup E_n]. The following table lists out the first several terms of this limit.

\displaystyle \begin{pmatrix} n&\text{   }&P_n \\{2}&\text{   }&0.50000 \\{3}&\text{   }&0.66667 \\{4}&\text{   }&0.62500 \\{5}&\text{   }&0.63333 \\{6}&\text{   }&0.63194 \\{7}&\text{   }&0.63214 \\{8}&\text{   }&0.63212\end{pmatrix}

Sketch of Proof for the Formula
The key idea is that any sample point in the union E_1 \cup E_2 \cdots \cup E_n is counted in exactly one time in the right hand side of (1). Suppose that a sample point is in exactly t of the events E_1,E_2,\cdots,E_n. Then the following shows the number of times the sample point is counted in each expression:

\displaystyle \sum \limits_{m=1}^{n}P[E_m] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t \text{ times}

\displaystyle \sum \limits_{a<b}P[E_a \cap E_b] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \binom{t}{2} \text{ times}

\displaystyle \sum \limits_{a<b<c}P[E_a \cap E_b \cap E_c] \ \ \ \ \ \ \ \ \ \binom{t}{3} \text{ times and so on}

Thus the sample point in question will be counted exactly H times in the right hand side of the formula.

\displaystyle H=\binom{t}{1}-\binom{t}{2}+\binom{t}{3}- \cdots + (-1)^{t+1}\binom{t}{t}

The following is the derivation that H=1.

\displaystyle 0=(1-1)^t=\sum \limits_{a=0}^{t} \binom{t}{a}(-1)^{a}(1)^{t-a}=\binom{t}{0}-\binom{t}{1}+\binom{t}{2}+ \cdots +(-1)^t \binom{t}{t}

\displaystyle 1=\binom{t}{0}=\binom{t}{1}-\binom{t}{2}- \cdots +(-1)^{t+1} \binom{t}{t}=H

Reference

  1. Feller, W., An Introduction to Probability Theory and its Applications, Vol. I, 3rd ed., John Wiley & Sons, Inc., New York, 1968

Previous Posts on The Matching Problem
The Matching Problem

More About the Matching Problem

Tis the Season for Gift Exchange

Tis the Season for Gift Exchange

Suppose that there are 10 people in a holiday party with each person bearing a gift. The gifts are put in a pile and each person randomly selects one gift from the pile. In order not to bias the random selection of gifts, suppose that the partygoers select gifts by picking slips of papers with numbers identifying the gifts. What is the probability that there is at least one partygoer who ends up selecting his or her own gift? In this example, selecting one’s gift is called a match. What is the probability that there is at least one match if there are more people in the party, say 50 or 100 people? What is the behavior of this probability if the size of the party increases without bound?

The above example is a description of a classic problem in probability called the matching problem. There are many colorful descriptions of the problem. One such description is that there are n married couples in a ballroom dancing class. The instructor pairs the ladies with the gentlemen at random. A match occurs if a married couple is paired as dancing partners.

Whatever the description, the matching problem involves two lists of items that are paired according to a particular order (the original order). When the items in the first list are paired with the items in the second list according to another random ordering, a match occurs if an item in the first list and an item in the second list are paired both in the original order and in the new order. The matching problem discussed here is: what is the probability that there is at least one match? Seen in this light, both examples described above are equivalent mathematically.

We now continue with the discussion of the random gift selection example. Suppose that there are n people in the party. Let E_i be the event that the i^{th} person selects his or her own gift. The event E=E_1 \cup E_2 \cup \cdots \cup E_n is the event that at least one person selects his or her own gift (i.e. there is at least one match). The probability P(E)=P(E_1 \cup E_2 \cup \cdots \cup E_n) is the solution of the matching problem as described in the beginning. The following is the probability P(E)=P(E_1 \cup E_2 \cup \cdots \cup E_n).

\displaystyle (1) \ \ \ \ \ \ P[E_1 \cup E_2 \cup \cdots \cup E_n]=1-\frac{1}{2!}+\frac{1}{3!}-\cdots+(-1)^{n+1} \frac{1}{n!}

The answer in (1) is obtained by using an idea called the inclusion-exclusion principle. We will get to that in just a minute. First let’s look at the results of (1) for a few iterations.

\displaystyle (2) \ \ \ \ \ \  \begin{bmatrix} \text{n}&\text{ }& P[E_1 \cup E_2 \cup \cdots \cup E_n]  \\\text{ }&\text{ }&\text{ } \\ 3&\text{ }&0.666667  \\ 4&\text{ }&0.625000  \\ 5&\text{ }&0.633333  \\ 6&\text{ }&0.631944  \\ 7&\text{ }&0.632143  \\ 8&\text{ }&0.632118 \\ 9&\text{ }&0.632121 \\ 10&\text{ }&0.632121 \\ 11&\text{ }&0.632121 \\ 12&\text{ }&0.632121 \end{bmatrix}

In a party with random gift exchange, it appears that regardless of the size of the party, there is a very good chance that someone will end up picking his or her own gift! A match will happen about 63% of the time.

It turns out that the answers in (1) are related to the Taylor’s series expansion of e^{-1}. We show that the probability in (1) will always converge to 1-e^{-1}=0.632121. Note that the Taylor’s series expansion of e^{-1} is:

\displaystyle (3) \ \ \ \ \ \ e^{-1}=\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots+(-1)^n \frac{1}{n!}+\cdots

Consequently, the Taylor’s series expansion of 1-e^{-1} is:

\displaystyle \begin{aligned} (4) \ \ \ \ \ \   1-e^{-1}&=1-\biggl(1-1+\frac{1}{2!}-\frac{1}{3!}+\cdots+(-1)^n \frac{1}{n!}+\cdots \biggr) \\&=1-\frac{1}{2!}+\frac{1}{3!}-\cdots+(-1)^{n+1} \frac{1}{n!}+\cdots  \end{aligned}

Note that the sum of the first n terms of the series in (4) is the probability in (1). As the number of partygoers n increases, the probability P[E_1 \cup E_2 \cup \cdots \cup E_n] will be closer and closer to 1-e^{-1}=0.6321. We have:

\displaystyle (5) \ \ \ \ \ \ \lim \limits_{n \rightarrow \infty} P[E_1 \cup E_2 \cup \cdots \cup E_n]=1 - e^{-1}

\displaystyle (6) \ \ \ \ \ \ \lim \limits_{n \rightarrow \infty} P[(E_1 \cup E_2 \cup \cdots \cup E_n)^c]=e^{-1}

The equation (5) says that it does not matter how many people are in the random gift exchange, the answer to the matching problem is always 1-e^{-1} in the limit. The equation (6) says that the probability of having no matches approaches e^{-1}. There is a better than even chance (0.63 to be more precise) that there is at least one match. So in a random gift exchange as described at the beginning, it is much easier to see a match than not to see one.

The Inclusion-Exclusion Principle

We now want to give some indication why (1) provides the answers. The inclusion-exclusion principle is a formula for finding the probability of the union of events. For the union of two events and the union of three events, we have:

\displaystyle (7) \ \ \ \ \ \ P[E_1 \cup E_2]=P[E_1]+P[E_1]-P[E_1 \cap E_2]

\displaystyle \begin{aligned} (8) \ \ \ \ \ \   P[E_1 \cup E_2 \cup E_3]&=P[E_1]+P[E_1]+P[E_3] \\& \ \ \ \ -P[E_1 \cap E_2]-P[E_1 \cap E_3]-P[E_2 \cap E_3] \\& \ \ \ \ +P[E_1 \cap E_2 \cap E_3]  \end{aligned}

To find the probability of the union of n events, we first add up the probabilities of the individual events E_i. Because the resulting sum overshoots, we need to subtract the probabilities of the intersections E_i \cap E_j. Because the subtractions remove too much, we need to add back the probabilities of the intersections of three individual events E_i \cap E_j \cap E_k. The process of inclusion and exclusion continues until we reach the step of adding/removing of the intersection E_1 \cap \cdots \cap E_n. The following is the statement of the inclusion-exclusion principle.

\displaystyle (9) \ \ \ \ \ \ P[E_1 \cup E_2 \cdots \cup E_n]=S_1-S_2+S_3- \ \ \cdots \ \ +(-1)^{n+1}S_n

In (9), S_1 is the sum of all probabilities P[E_i], S_2 is the sum of all possible probabilities of the intersection of two events E_i and E_j and S_3 is the sum of all possible probabilities of the intersection of three events and so on.

We now apply the inclusion-exclusion principle to derive equation (1). The event E_i is the event that the i^{th} person gets his or her own gift while the other n-1 people are free to select gifts. The probability of this event is \displaystyle P[E_i]=\frac{(n-1)!}{n!}. There are \displaystyle \binom{n}{1}=\frac{n!}{1! (n-1)!} many ways of fixing 1 gift. So \displaystyle S_1=\binom{n}{1} \times \frac{(n-1)!}{n!}=1.

Now consider the intersection of two events. The event E_i \cap E_j is the event that the i^{th} person and the j^{th} person get their own gifts while the other (n-2) people are free to select gifts. The probability of this event is \displaystyle P[E_i \cap E_j]=\frac{(n-2)!}{n!}. There are \displaystyle \binom{n}{2}=\frac{n!}{2! (n-2)!} many ways of fixing 2 gifts. So \displaystyle S_2=\binom{n}{2} \times \frac{(n-2)!}{n!}=\frac{1}{2!}.

By the same reasoning, we derive that \displaystyle S_3=\frac{1}{3!} and \displaystyle S_4=\frac{1}{4!} and so on. Then plugging \displaystyle S_i=\frac{1}{i!} into (9), we obtain the answers in (1).

The matching problem had been discussed previously in the following posts.

The matching problem

Mote about the matching problem