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Can you please help me with this question:

In a gambling game, each turn a player throws 2 fair dice. If the sum of numbers on the dice is 2 or 7, the player wins a dollar. If the sum is 3 or 8, the player loses a dollar. The player starts to play with 10 dollars and stops the game if he loses all his money or if he earns 5 dollars. What's the probability for the player to lose all the money and what's the probability to finish the game as a winner? If there some 3rd possibility to finish the game? If yes, what's its probability?

Thanks a lot!

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I tried to make the phrasing and grammar a bit better, as well as making the title more descriptive. –  Zev Chonoles Dec 20 '12 at 22:45
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2 Answers 2

up vote 2 down vote accepted

[edit: Apparently I misread the question. The player starts out with 10 dollars and not five.]

Given that "rolling a 2 or 7" and "rolling a 3 or 8" have the same probability (both occur with probability 7/36), the problem of the probability of a player earning a certain amount of money before losing a different amount of money is the same as the problem of the Gambler's Ruin.

What's different when considering individual rounds is that there's a possibility of a tie. But because a tie leaves everything unchanged, the Gambler's Ruin still applies simply because we can simply consider only events that do change the state of each player's money.

Therefore, the probability that the player makes \$5 before losing \$10 is the same probability as flipping coins against somebody with $5, or 2/3. And the probability of the opposite event is 1/3.

The third outcome, that the game goes on forever, has a probability that vanishes to zero.

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Perfect! Thank you –  Tina Dec 26 '12 at 11:15
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The game is fair as $P(2 or 7)=P(3 or 8)=\frac 7{36}$, so if we ignore the non-paying rolls, it just flipping a coin. As the game is fair, $\frac 23$ of the time he will earn $5$ and $\frac 13$ of the time lose $10$.

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Thanks a lot, now it's clear.... –  Tina Dec 26 '12 at 11:15
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