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

- schrodinger

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- Zarkon

craps

- P0sitr0n

winning is getting 2 / 12
1/6

- anonymous

oh yes, it sounds like a game of craps.

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## More answers

- anonymous

hmm... why is the winning getting 2/12 and eventually 1/6?

- Zarkon

I guess it is a little different

- anonymous

Say I let W be winning, then the chance of getting the winning rolls would be
\[\frac{6}{36}+\frac{2}{36} = \frac{8}{36}\]

- Zarkon

2/3

- anonymous

2/3 is the the chance of getting the winning rolls or the eventual probability of winning the game?

- anonymous

Something that I find weird about this question is that isn't rolling the winning outcome the probability of winning the game?

- beginnersmind

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.

- Zarkon

\[\frac{6+2}{6+2+1+2+1}\]

- anonymous

Then I would sum it all to infinity?

- beginnersmind

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.

- beginnersmind

some=sum

- anonymous

@Zarkon, 6 is for getting 7, 2 is for getting 11. And I could just take the winning ways over the losing ways?

- Zarkon

numer of ways to win 6+2
number of ways to end the game...6+2+1+2+1

- Zarkon

(number of ways to win)/(number of ways to end the game)

- Zarkon

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

- Zarkon

the proability that you roll forever is zero...so we really don't need to consider thoses values.

- Zarkon

the game essentially starts over if we don't roll a 2,3,7,11 or 12.

- anonymous

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!

- anonymous

But how come according to the question, I need to use conditional probability to solve it?

- Zarkon

you can write it out formally using conditional probability...but this problem is not hard enough to put in the effort. :)

- anonymous

What would I condition on if I had to use conditional probability?

- Zarkon

you could condition on the round of the game...you would get something like beginnersmind first wrote

- anonymous

oh I see....thank you so much for all the help Zarkon! :)

- beginnersmind

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.

- Zarkon

the only thing you need to do is fix your notation
\[S_n=\sum_{k=0}^{n}(24/36)^{k}*(8/36)\]
then
\[S_n\to\frac{2}{3}\]

- anonymous

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! :)

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