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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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P0sitr0nBest ResponseYou've already chosen the best response.0
winning is getting 2 / 12 1/6
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
oh yes, it sounds like a game of craps.
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
hmm... why is the winning getting 2/12 and eventually 1/6?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
I guess it is a little different
 one year ago

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

xEnOnnBest ResponseYou've already chosen the best response.0
2/3 is the the chance of getting the winning rolls or the eventual probability of winning the game?
 one year ago

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

beginnersmindBest ResponseYou've already chosen the best response.0
There 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.
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
\[\frac{6+2}{6+2+1+2+1}\]
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
Then I would sum it all to infinity?
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.0
You 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.
 one year ago

xEnOnnBest 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?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
numer of ways to win 6+2 number of ways to end the game...6+2+1+2+1
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
(number of ways to win)/(number of ways to end the game)
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
the 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
 one year ago

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

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

xEnOnnBest ResponseYou've already chosen the best response.0
oh 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!
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
But how come according to the question, I need to use conditional probability to solve it?
 one year ago

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

xEnOnnBest ResponseYou've already chosen the best response.0
What would I condition on if I had to use conditional probability?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
you could condition on the round of the game...you would get something like beginnersmind first wrote
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
oh I see....thank you so much for all the help Zarkon! :)
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.0
If 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.
 one year ago

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

xEnOnnBest ResponseYou've already chosen the best response.0
ohh...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! :)
 one year ago
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