Question

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?

Answer 1

craps

Answer 2

winning is getting 2 / 12
1/6

Answer 3

oh yes, it sounds like a game of craps.

Answer 4

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

Answer 5

I guess it is a little different

Answer 6

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

Answer 7

2/3

Answer 8

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

Answer 9

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

Answer 10

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.

Answer 11

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

Answer 12

Then I would sum it all to infinity?

Answer 13

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.

Answer 14

some=sum

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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!

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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.

Answer 28

the only thing you need to do is fix your notation

\[S_n=\sum_{k=0}^{n}(24/36)^{k}*(8/36)\]

then

\[S_n\to\frac{2}{3}\]

Answer 29

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! :)