anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Zarkon
  • Zarkon
craps
P0sitr0n
  • P0sitr0n
winning is getting 2 / 12 1/6
anonymous
  • anonymous
oh yes, it sounds like a game of craps.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
hmm... why is the winning getting 2/12 and eventually 1/6?
Zarkon
  • Zarkon
I guess it is a little different
anonymous
  • 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
  • Zarkon
2/3
anonymous
  • anonymous
2/3 is the the chance of getting the winning rolls or the eventual probability of winning the game?
anonymous
  • anonymous
Something that I find weird about this question is that isn't rolling the winning outcome the probability of winning the game?
beginnersmind
  • 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
  • Zarkon
\[\frac{6+2}{6+2+1+2+1}\]
anonymous
  • anonymous
Then I would sum it all to infinity?
beginnersmind
  • 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
  • beginnersmind
some=sum
anonymous
  • 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
  • Zarkon
numer of ways to win 6+2 number of ways to end the game...6+2+1+2+1
Zarkon
  • Zarkon
(number of ways to win)/(number of ways to end the game)
Zarkon
  • 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
  • Zarkon
the proability that you roll forever is zero...so we really don't need to consider thoses values.
Zarkon
  • Zarkon
the game essentially starts over if we don't roll a 2,3,7,11 or 12.
anonymous
  • 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
  • anonymous
But how come according to the question, I need to use conditional probability to solve it?
Zarkon
  • Zarkon
you can write it out formally using conditional probability...but this problem is not hard enough to put in the effort. :)
anonymous
  • anonymous
What would I condition on if I had to use conditional probability?
Zarkon
  • Zarkon
you could condition on the round of the game...you would get something like beginnersmind first wrote
anonymous
  • anonymous
oh I see....thank you so much for all the help Zarkon! :)
beginnersmind
  • 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
  • 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
  • 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! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.