Re: probability problem
- From: "jmorriss@xxxxxxxxxxx" <jmorriss@xxxxxxxxxxx>
- Date: 7 Dec 2005 18:14:37 -0800
Greenl...@xxxxxxxxxxxxxx wrote:
> How do you do a question of the type, 2 players are throwing dice,
> taking turns, and the first to get a 6 wins the game.
> What are the chances that the first person to throw the dice
> will win, and of the second person?
>
> I get an answer of 7/12 for the first person, but an answer
> of 5/12 + (1/12) * (5/6)^n for the second person. It makes
> sense if n tends to infinity but if not (and in a real game it wouldn't)
> then how would you interpret the result, since then the chances
> that either of the two players will win is greater than 1. I think there
> was a mistake or something.
>
> Thanks in advance.
Not clear what the problem really is... How many dice? Get a six
showing, or a total of six?
In any case, consider the game as a series of <pairs> of throws...
A throws, then B
A has a probability of p of winning on that throw... if he fails
(Probability [1-p]) then B has a probability of p of winning on that
throw, for a probability of B winning of p*(1-p)...
The same ratio applies for the next pair, if no one wins...
.
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- probability problem
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