John and Mark found 16 dollars in a paper bag. Rather than...

John and Mark found 16 dollars in a paper bag. Rather than splitting the cash they decided to flip a coin for it. They decided that the one who first wins 10 tosses gets all the money. After 15 tosses of the coin John has won eight times and Mark seven times. On the 16th flip the coin rolled away and was lost, so they decided to divide the 16 dollars based on their respective chances of winning if they started up again. Clearly, John should get more than Mark, but exactly how much should each receive? Note: This is a special case of a general problem called the "problem of points" first solved successfully by Pascal.

Key Concepts

Problem of Points

This is a classical problem in probability theory that deals with how to fairly divide stakes in a game of chance when the game is interrupted before its conclusion. The solution involves calculating the probability of each player winning if the game were to continue, and then allocating the stakes accordingly.

Fair Division Based on Expected Value

Fair division in this context means splitting any reward or stakes according to the players' respective probabilities of winning. This ensures that each player's share reflects their chance of success had the game continued, embodying the principle of expected value in decision making.

Combinatorial Probability

Combinatorial probability involves counting the number of possible outcomes in a random process and using these counts to determine probabilities for various events. In games of chance, such as coin tossing, the outcomes can be calculated using combinations, which is crucial for determining the likelihood of each player reaching their required number of wins.

Conditional Probability

This is the calculation of the probability of an event occurring, given that another event has already occurred. In the context of the interrupted game, the calculation of each player's chance of winning is based on the current state of play, making conditional probability a key concept in determining the fair division of the stakes.

Transcript

00:03 I told that two siblings matt and liz both play roulette and that matt has brought a $10 to a casino and liz has brought $15, that there are different tables that they each made a dollar wagers on red on consecutive spins, where matt has made 10 spins, and liz has made 15.

00:32 We're given the random variable x, which represents the number.

00:36 Of times mat 1 and y, which represents the number of times liz 1.

00:41 In part a, we're asked to determine a reasonable probability model for x.

00:59 We have, as suggested by the hint, that successive spins of a roulette wheel are independent.

01:12 Think of the spins as trials, this means that the trials are independent, and moreover, we have that the probability of landing out in red is equal to 1838ths, which is a constant.

01:39 So in each trial there is a constant probability of success.

01:46 And you can either have success or failure.

01:50 So the conditions of a binomial experiment are clearly met, and so it would be reasonable to say that the random variable x, is binomially distributed with parameters, well the number of trials we had that matt made 10 trials and probability of success 1838 on each trial.

02:37 In part b we're asked to find a reasonable probability model for y.

02:47 Following the same reasoning as a, we again have that the successive spins on the relay wheel are independent, the probability of landing on red is 1838th, which is constant.

03:12 And if we see the spins as trials and landing on red as a success, we have that successive trials are independent, that we can either have a success or a failure in each trial, and the probability of success on each trial is the same.

03:32 So by the same reasoning as in part a, the conditions of a binomial experiment are met, and therefore it makes sense to say that y is binomially distributed with sample size, 15, because liz makes 15 spins, and probability of success on each trial, 1838.

04:11 In part c, we're asked to find a reasonable probability model for x plus y, which represents the total number of times that matt and liz win that day.

04:31 We have the x plus y is the combined number of times that matt and liz win, know that they play a total of 10 plus 15 or 25 games.

04:52 All the games are independent.

05:01 Now the reason that they're independent is because matt and liz are sitting at different tables.

05:15 So they're independent, and we also have that their games are all identical.

05:26 So these trials are independent and they have identical probabilities of success and failure.

05:41 In particular we have that p is equal to 1838 for each game.

05:54 So the conditions of a binomial experiment are net again, and therefore it makes sense to say that x plus y is binomially distributed with sample size of 25 and probably ability of success, 1838.

06:34 In part b, we're asked to use moment generating functions, along with our answers from parts a and b, prove that the answer in part c is in fact correct, and not just reasonable.

06:54 So we have the moment -generating function of a binomial distribution with parameters n and p.

07:15 Recall that this is m m of t is equal to 1 minus p plus p e to the t all to the n...

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