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xEnOnn Group Title

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?

  • 2 years ago
  • 2 years ago

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  1. Zarkon Group Title
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    craps

    • 2 years ago
  2. P0sitr0n Group Title
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    winning is getting 2 / 12 1/6

    • 2 years ago
  3. xEnOnn Group Title
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    oh yes, it sounds like a game of craps.

    • 2 years ago
  4. xEnOnn Group Title
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    hmm... why is the winning getting 2/12 and eventually 1/6?

    • 2 years ago
  5. Zarkon Group Title
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    I guess it is a little different

    • 2 years ago
  6. xEnOnn Group Title
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    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 years ago
  7. Zarkon Group Title
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    2/3

    • 2 years ago
  8. xEnOnn Group Title
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    2/3 is the the chance of getting the winning rolls or the eventual probability of winning the game?

    • 2 years ago
  9. xEnOnn Group Title
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    Something that I find weird about this question is that isn't rolling the winning outcome the probability of winning the game?

    • 2 years ago
  10. beginnersmind Group Title
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    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.

    • 2 years ago
  11. Zarkon Group Title
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    \[\frac{6+2}{6+2+1+2+1}\]

    • 2 years ago
  12. xEnOnn Group Title
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    Then I would sum it all to infinity?

    • 2 years ago
  13. beginnersmind Group Title
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    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.

    • 2 years ago
  14. beginnersmind Group Title
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    some=sum

    • 2 years ago
  15. xEnOnn Group Title
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    @Zarkon, 6 is for getting 7, 2 is for getting 11. And I could just take the winning ways over the losing ways?

    • 2 years ago
  16. Zarkon Group Title
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    numer of ways to win 6+2 number of ways to end the game...6+2+1+2+1

    • 2 years ago
  17. Zarkon Group Title
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    (number of ways to win)/(number of ways to end the game)

    • 2 years ago
  18. Zarkon Group Title
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    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

    • 2 years ago
  19. Zarkon Group Title
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    the proability that you roll forever is zero...so we really don't need to consider thoses values.

    • 2 years ago
  20. Zarkon Group Title
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    the game essentially starts over if we don't roll a 2,3,7,11 or 12.

    • 2 years ago
  21. xEnOnn Group Title
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    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!

    • 2 years ago
  22. xEnOnn Group Title
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    But how come according to the question, I need to use conditional probability to solve it?

    • 2 years ago
  23. Zarkon Group Title
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    you can write it out formally using conditional probability...but this problem is not hard enough to put in the effort. :)

    • 2 years ago
  24. xEnOnn Group Title
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    What would I condition on if I had to use conditional probability?

    • 2 years ago
  25. Zarkon Group Title
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    you could condition on the round of the game...you would get something like beginnersmind first wrote

    • 2 years ago
  26. xEnOnn Group Title
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    oh I see....thank you so much for all the help Zarkon! :)

    • 2 years ago
  27. beginnersmind Group Title
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    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.

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

    • 2 years ago
  29. xEnOnn Group Title
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    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! :)

    • 2 years ago
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