Here's the question you clicked on:
xEnOnn
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?
winning is getting 2 / 12 1/6
oh yes, it sounds like a game of craps.
hmm... why is the winning getting 2/12 and eventually 1/6?
I guess it is a little different
Say I let W be winning, then the chance of getting the winning rolls would be \[\frac{6}{36}+\frac{2}{36} = \frac{8}{36}\]
2/3 is the the chance of getting the winning rolls or the eventual probability of winning the game?
Something that I find weird about this question is that isn't rolling the winning outcome the probability of winning the game?
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.
\[\frac{6+2}{6+2+1+2+1}\]
Then I would sum it all to infinity?
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.
@Zarkon, 6 is for getting 7, 2 is for getting 11. And I could just take the winning ways over the losing ways?
numer of ways to win 6+2 number of ways to end the game...6+2+1+2+1
(number of ways to win)/(number of ways to end the game)
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
the proability that you roll forever is zero...so we really don't need to consider thoses values.
the game essentially starts over if we don't roll a 2,3,7,11 or 12.
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!
But how come according to the question, I need to use conditional probability to solve it?
you can write it out formally using conditional probability...but this problem is not hard enough to put in the effort. :)
What would I condition on if I had to use conditional probability?
you could condition on the round of the game...you would get something like beginnersmind first wrote
oh I see....thank you so much for all the help Zarkon! :)
If you absolutelutely need to work conditional probability into it you could take \[S_n=\sum_{0}^{\infty}(24/36)^{n}*(8/36)\] The n-th term in the series represents the probability that you win on exactly the n-th roll. If I didn't mess up the value of the sum should be 2/3.
the only thing you need to do is fix your notation \[S_n=\sum_{k=0}^{n}(24/36)^{k}*(8/36)\] then \[S_n\to\frac{2}{3}\]
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! :)