statistic problem

Question

1. Two persons, A and B, are to play a serious of independent games, with A’s winning probability in each game = p and B’s winning probability = q = 1 – p. The game ends when the total number of winnings of one of the players is 2 greater than the number of winnings of the other player, and that player with more winnings wins the whole game. (For example, suppose the result of first 6 games is [A wins, B wins, B wins, A wins, A wins, A wins]. Then, A wins 4 games and B wins 2 games. So, A wins the whole game.)

Show that the probability that A wins the whole game is .



2. Two players A and B plays a series of independent games, with A’s probability of winning for each game = p and B’s probability of winning = 1 – p. When A wins, B pays A $1, and when B wins, A pays B $1. A starts with $M, and B starts with an infinite amount of money (i.e. B will never ruin). The whole game will end only when A runs out of money (i.e. “ruins”). (We do not specify A’s maximum amount of money for the game to end; or equivalently, we assume that the end will continue whenever A has a positive amount of money.)

1 if t = q/p ( 1 (i.e. p ( ½);
Show that A’s probability of getting ruined = (not A’s probability of winning) t M if t = q/p < 1 (i.e. p > ½).


[Hint: You need to consider 3 different cases: t > 1; t = 1; and t < 1.

Answer 1

For the first question I think the answer should be:

 

P(A) win whole game = p^2 / (p^2 + (1-p)^2)

 

Did you get the right answer yet?

 

My intuition

 

LET:

 

A represents A winning single game

B represents B winning single game

 

Then combinations for winning whole game for both are:

 

 

AA

BB

AB AA

AB BB

BA AA

BA BB

ABAB AA

ABAB BB

ABBA AA

ABBA BB

BABA AA

BABA BB

BAAB AA

BAAB BB

....

 

Notice that given a certain history length the chance of winning for AA or BB is the same as the original first AA or BB.  In other words the problem is the same as always playing 2 rounds and then discarding if there is no winner after 2 rounds, and starting over until we get AA or BB. In this case after playing 2 rounds the chance of getting AA is p^2 and the chance of getting  BB is (1-p)^2. So I am interested in how much more likely BB is compared to AA and so the result for the winner has to be standardized and so divide by P(AA) +P(BB). And so P( A winning) = P(AA) /(P(AA) + P(BB)) = p^2 / (p^2 + (1-p)^2).