Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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?

  • one year ago
  • one year ago

  • This Question is Closed
  1. Zarkon
    Best Response
    You've already chosen the best response.
    Medals 1

    craps

    • one year ago
  2. P0sitr0n
    Best Response
    You've already chosen the best response.
    Medals 0

    winning is getting 2 / 12 1/6

    • one year ago
  3. xEnOnn
    Best Response
    You've already chosen the best response.
    Medals 0

    oh yes, it sounds like a game of craps.

    • one year ago
  4. xEnOnn
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  5. Zarkon
    Best Response
    You've already chosen the best response.
    Medals 1

    I guess it is a little different

    • one year ago
  6. xEnOnn
    Best Response
    You've already chosen the best response.
    Medals 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
  7. Zarkon
    Best Response
    You've already chosen the best response.
    Medals 1

    2/3

    • one year ago
  8. xEnOnn
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  9. xEnOnn
    Best Response
    You've already chosen the best response.
    Medals 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
  10. beginnersmind
    Best Response
    You've already chosen the best response.
    Medals 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
  11. Zarkon
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  12. xEnOnn
    Best Response
    You've already chosen the best response.
    Medals 0

    Then I would sum it all to infinity?

    • one year ago
  13. beginnersmind
    Best Response
    You've already chosen the best response.
    Medals 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
  14. beginnersmind
    Best Response
    You've already chosen the best response.
    Medals 0

    some=sum

    • one year ago
  15. xEnOnn
    Best Response
    You've already chosen the best response.
    Medals 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
  16. Zarkon
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  17. Zarkon
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  18. Zarkon
    Best Response
    You've already chosen the best response.
    Medals 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
  19. Zarkon
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  20. Zarkon
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  21. xEnOnn
    Best Response
    You've already chosen the best response.
    Medals 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
  22. xEnOnn
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  23. Zarkon
    Best Response
    You've already chosen the best response.
    Medals 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
  24. xEnOnn
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  25. Zarkon
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  26. xEnOnn
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  27. beginnersmind
    Best Response
    You've already chosen the best response.
    Medals 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 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.

    • one year ago
  28. Zarkon
    Best Response
    You've already chosen the best response.
    Medals 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
  29. xEnOnn
    Best Response
    You've already chosen the best response.
    Medals 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
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.