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anonymous
 4 years ago
Two dice are rolled in a game. I win if the sum of the outcomes of the two dice is 7 or 11 and lose if the sum is 2, 3 or 12. I keep rolling until one of these sums occurs. Using conditional probabilities, what is the probability of winning this game?
anonymous
 4 years ago
Two dice are rolled in a game. I win if the sum of the outcomes of the two dice is 7 or 11 and lose if the sum is 2, 3 or 12. I keep rolling until one of these sums occurs. Using conditional probabilities, what is the probability of winning this game?

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P0sitr0n
 4 years ago
Best ResponseYou've already chosen the best response.0winning is getting 2 / 12 1/6

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yes, it sounds like a game of craps.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm... why is the winning getting 2/12 and eventually 1/6?

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1I guess it is a little different

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Say I let W be winning, then the chance of getting the winning rolls would be \[\frac{6}{36}+\frac{2}{36} = \frac{8}{36}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02/3 is the the chance of getting the winning rolls or the eventual probability of winning the game?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Something that I find weird about this question is that isn't rolling the winning outcome the probability of winning the game?

beginnersmind
 4 years ago
Best ResponseYou've already chosen the best response.0There are 4 ways of getting a losing roll and 8 of getting a winning one. The rest of the time (24/36) you roll again. There's a (24/36)*(8/36) chance that you win on your second roll. There's a (24/36)*(24/36)*(8/36) that you win on your third roll. And so on.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac{6+2}{6+2+1+2+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then I would sum it all to infinity?

beginnersmind
 4 years ago
Best ResponseYou've already chosen the best response.0You could some over infinity. But there's an easier way. Note that in every single round the probability of winning is exactly twice as much as the probability of losing. Eventually the game has to end, and at every point you are twice as likely to win than to lose. Hence the chance of winning is 2/3 and losing is 1/3.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Zarkon, 6 is for getting 7, 2 is for getting 11. And I could just take the winning ways over the losing ways?

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1numer of ways to win 6+2 number of ways to end the game...6+2+1+2+1

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1(number of ways to win)/(number of ways to end the game)

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1the game ends when you win or lose so the game ends or a roll of 2,3,7,11 or 12 there are 1+2+6+2+1 ways to do this

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1the proability that you roll forever is zero...so we really don't need to consider thoses values.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1the game essentially starts over if we don't roll a 2,3,7,11 or 12.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yeah! the game starts over if we don't roll a 2,3,7,11 or 12. And is totally memoryless of the previous rolls and that's why ending a game is just considering the win and losing ways! ohh... thanks!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But how come according to the question, I need to use conditional probability to solve it?

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1you can write it out formally using conditional probability...but this problem is not hard enough to put in the effort. :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What would I condition on if I had to use conditional probability?

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1you could condition on the round of the game...you would get something like beginnersmind first wrote

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh I see....thank you so much for all the help Zarkon! :)

beginnersmind
 4 years ago
Best ResponseYou've already chosen the best response.0If you absolutelutely need to work conditional probability into it you could take \[S_n=\sum_{0}^{\infty}(24/36)^{n}*(8/36)\] The nth term in the series represents the probability that you win on exactly the nth roll. If I didn't mess up the value of the sum should be 2/3.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1the only thing you need to do is fix your notation \[S_n=\sum_{k=0}^{n}(24/36)^{k}*(8/36)\] then \[S_n\to\frac{2}{3}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh...yea! I just tried to calculate it and it really works too. I tried it this way: \[S_\infty =\frac{1}{1\frac{24}{26}}\cdot\frac{8}{36} = \frac{2}{3} \] interesting. Thanks! :)
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