dtweah wrote:Tom has $2 and Bill has $1. A die is rolled and if it comes up 1 or 2 then Tom wins and Bill gives him $1; if it comes up 3,4,5 or 6 then Bill wins and Tom gives him $1. This is repeated until Tom or Bill has no money and the other is the victor. What is the probability that Tom is the victor?

(a) 1/2
(b) 2/5
(c) 3/5
(d) 4/7
(e) 3/7

Prob of Tom to win his throw = 2/6 = 1/3 and for Bill it is 4/6 = 2/3

Tom can win either in his first throw or in his second throw or in his third throw and so on......

Probability that Tom wins in the first throw = 2/6 = 1/3

Now if Tom has to win in his second win, Tom must loose his first throw and Bill also must loose in his first throw and then Tom wins his second throw.

Hence the probability = 2/3 x 1/3 x 1/3 = (2/9) x 1/3


Similarly if Tom has to win in his third throw, both Tom and Bill must loose their first 2 throws

Hence the probability = 2/3 x 1/3 x 2/3 x 1/3 x 1/3 = (2/9)^2 x 1/3


So you can observe that the chance of Tom winning is

1/3 + (2/9)x1/3 + (2/9)^2 x 1/3 + (2/9)^3 x 1/3 +................

This is an infinite GP with common ratio 2/9 .

So the sum = (1/3)/(1-2/9) (Sum to infinite GP = a/(1-r))

=1/3 x 9/7 = 3/7

This is a very fascinating problem and the solutions offered by Ian and Sura are great. I solved it through the route of conditional probability by defining three events and conditioning on them: A= event of 1,2 occurring. F=Event that game ends on first draw and W=event that Tom wins. Don't have time to write that solution now. The OA gives two methods and I find the first method terribly ingenious. In my solution I tinkered with the second method using a tree diagram but always ended up with one tree going to infinity and didn't happen on the idea of a geometric sequence, so I switched to conditional probability. I will post the conditional solution later. All these approaces need to be studied well.

(e) Method I: Let p be the probability that Tom is the victor. If Tom wins the first roll then he is the victor. If Tom loses the first roll and wins the second roll then the game is returned to the original position. It follows that p = 1/3 + (2/3)(1/3)p from which p = 3/7.

Method II: Let W,L respectively denote win and loss for Tom. Then Tom wins with the sequences W, (LW)W, (LW)(LW)W, (LW)(LW)(LW)W,… . Addition of the corresponding probabilities gives a geometric series with initial term 1/3 and ratio (2/3)(1/3) = 2/9. The sum of the series is (1/3)/[1 – (2/9)] = 3/7.