# Geometric distribution problem

• October 21st 2008, 09:20 PM
kid funky fried
Geometric distribution problem
I am am trouble with the following homework problem.
(Note:This is exactly how the problem is stated)

Two players each put one dollar into a pot. They then decide to throw a pair of dice alternately. The first one who throws a sum of 5 wins the pot. How much should the player who starts add to the pot to make this a fair game?

I think it is a geometric distribution.
Which is defined as p(x)= p(1-p)^x and the mean is (1-p)/p.
I also know the probability of 2 dice summing to 5 is 4/36 or approx. .1111.
I can determine the probability that player A wins when P{X=1}, P{X=3}...
Similarly, for player B P{X=2}, P{X=4}
The professor also gave the following clues:
P{A wins}=? P{B wins} =?
E[A's reward]=P{A wins}*1
E[B's reward]=P{B wins}*(1+$x) But I am lost. Any help would be greatly appreciated. • October 21st 2008, 09:42 PM mr fantastic Quote: Originally Posted by kid funky fried I am am trouble with the following homework problem. (Note:This is exactly how the problem is stated) Two players each put one dollar into a pot. They then decide to throw a pair of dice alternately. The first one who throws a sum of 5 wins the pot. How much should the player who starts add to the pot to make this a fair game? I think it is a geometric distribution. Which is defined as p(x)= p(1-p)^x and the mean is (1-p)/p. I also know the probability of 2 dice summing to 5 is 4/36 or approx. .1111. I can determine the probability that player A wins when P{X=1}, P{X=3}... Similarly, for player B P{X=2}, P{X=4} The professor also gave the following clues: P{A wins}=? P{B wins} =? E[A's reward]=P{A wins}*1 E[B's reward]=P{B wins}*(1+$x)

But I am lost.
Any help would be greatly appreciated.

Calculate Pr(A wins). Call it p.

Pr(B wins) = 1 - p.

For a fair game, the expected winnings of each player should be zero.

If A wins s/he wins $1. If A loses s/he 'wins' -$(1 + x).

Therefore p(1) - (1-p)(1 + x) = 0.

Solve for x.
• October 21st 2008, 10:17 PM
kid funky fried
Quote:

Originally Posted by mr fantastic
Calculate Pr(A wins). Call it p.

Pr(B wins) = 1 - p.

For a fair game, the expected winnings of each player should be zero.

If A wins s/he wins $1. If A loses s/he 'wins' -$(1 + x).

Therefore p(1) - (1-p)(1 + x) = 0.

Solve for x.

I haven't done the calculation but symmetry tells me that p is probably equal to 2/3 ......

Is the probability of player A winning 4/36???
How do i determine this?

p(1).. is this 4/36 * (32/36)^1 ???
• October 22nd 2008, 12:25 AM
mr fantastic
Quote:

Originally Posted by kid funky fried
Is the probability of player A winning 4/36???
How do i determine this?

p(1).. is this 4/36 * (32/36)^1 ???

No. That would mean the proability of B winning was 32/36. But obviously the person who starts first must have the advantage (because they can win on the very first throw, before their opponent has even thrown). In fact, 4/36 = 1/9 is the probability of either player winning on a single throw.

There are many different way of doing the calculation.

Let the probability of A winning be p and the probability of B winning be q.

Let the probability of success in a single throw be r.

$p = r + (1-r)^2 r + (1 - r)^4 r + .... = r (1 + [1-r]^2 + [1-r]^4 + ....)$

$q = (1-r) r + (1 - r)^3 r + (1-r)^5 r + .... = r (1-r) (1 + [1-r]^2 + [1-r]^4 + ....) = r (1 - r) p$.

Therefore:

p + q = 1 .... (1)

q = r (1 - r) p .... (2)

Solve equations (1) and (2) simultaneously:

$p = \frac{1}{2 - r}$ and $q = \frac{1- r}{2 - r}$.

In your case $r = \frac{4}{36} = \frac{1}{9}$.

Therefore $p = \frac{1}{2 - \frac{1}{9}} = \frac{9}{17}$, showing that A has (as expected) a slight advantage over B by starting first.

Note that as r --> 1, p --> 1 and q --> 0 (as expected).
• October 22nd 2008, 04:32 PM
kid funky fried
Thanks Mr Fantastic. I do not think I would have gotten the answer without your help. But, on the positive side, I do understand now how you got the answer.