Question 125652: A urn has 9 white balls and 11 black balls. A ball is drawn and then replaced. If you draw a white ball you win 5 cents, if you draw a black ball you lose 5 cents. You have a dollar to gamble with, your opponent has 50 cents. IF you play until one of you loses his money, what is the probability you will lose your dollar?
Answer by kev82(151) (Show Source):
You can put this solution on YOUR website! This is a one dimensional random walk with absorbing barriers, more commonly known as the gambler's ruin problem.
At this point note that there are 3 outcomes I win, you win, game goes on indefinitely, so P(I win)+P(you win) does not necessarily add to 1.
Let's think of this in terms of 5 cents. to make it easier. Initially I have 1 dollar(20x5cents) and you have half of that (10x5cents). The game is over when one of us has nothing and the other has 30x5 cents.
Let i be the number of 5 cents someone has, be my winning probability from that position and be your winning probability. Let be the probability I win a game 9/20 and be the probability you win a game 11/20.
Now, the probability I win with i coins is the probability I win this game, and that I win with i+1 coins, or that I lose this game and win with i-1 coins. (Not these two events are mutually exclusive) So
If we have 0 coins we can't win, sp and if we have 30 coins we have won so . Our task is to find .
I can write so I can write . Substituting this in and rearranging gives
Now let me define . This leads to the recurrence This has the well known solution . The proble is that we don't know . Let's move on though and consider
Notice that it is a collapsing sum and actually evaluates to . Doing the actual sum gives a standard geometric series which I'm sure you can evaluate. This is great because we know so using this we can calculate . Given this we can either use the p recurrence, or the sum of the z recurrence (we now know ) so we can calculate . I get it as
Which I calculate to be about 13.2%
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