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badreferences Group Title

Probability question challenge. Two players roll an \(n\)-sided dice. The player who fails to get better rolls than previous ones loses. How likely is it for the first player to win?

  • 2 years ago
  • 2 years ago

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  1. badreferences Group Title
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    @UnkleRhaukus @TuringTest @KingGeorge @experimentX @estudier

    • 2 years ago
  2. TuringTest Group Title
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    I am terrible at probability, but everyone always asks me about it.

    • 2 years ago
  3. TuringTest Group Title
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    I'll be curious to hear what others say though

    • 2 years ago
  4. badreferences Group Title
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    So am I. Took me like 3 days to solve this. But I do have the solution.

    • 2 years ago
  5. Libniz Group Title
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    I think this is geometric distribution?

    • 2 years ago
  6. badreferences Group Title
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    Well, no I don't think so.

    • 2 years ago
  7. badreferences Group Title
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    @Mariantheduke You might be interested. This problem came from the physics dept.

    • 2 years ago
  8. badreferences Group Title
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    @mukushla

    • 2 years ago
  9. badreferences Group Title
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    I do not think this solution works.

    • 2 years ago
  10. sauravshakya Group Title
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    1/n * (n)/n + 1/n *(n-1)/n + 1/n * (n-2)/n+...

    • 2 years ago
  11. sauravshakya Group Title
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    1/n^2 (n+n-1+n-2+n-3+...n-n)

    • 2 years ago
  12. badreferences Group Title
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    on the right track, but not it.

    • 2 years ago
  13. sauravshakya Group Title
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    Did I missed something?

    • 2 years ago
  14. badreferences Group Title
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    The answer is\[\left(1-\frac1n\right)^{n-r}\]where \(r\) is the roll that has just occurred, and \(n\) the number of sides the dice has.

    • 2 years ago
  15. badreferences Group Title
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    @saifoo.khan This would interest you, methinks.

    • 2 years ago
  16. badreferences Group Title
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    I'm sorry, that's the probability for the first player to lose. Wait a second.

    • 2 years ago
  17. badreferences Group Title
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    The probability that the first player wins is\[1-\left(1-\frac1n\right)^n\]where \(r=0\).

    • 2 years ago
  18. sauravshakya Group Title
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    I got |dw:1349578761373:dw|

    • 2 years ago
  19. badreferences Group Title
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    I have it in good confidence that my answer is correct. I don't know what assumption you made that was wrong, though.

    • 2 years ago
  20. sauravshakya Group Title
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    What is r ?

    • 2 years ago
  21. badreferences Group Title
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    The roll just made.

    • 2 years ago
  22. badreferences Group Title
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    Well, goodnight, Sir. I will see you another day perhaps.

    • 2 years ago
  23. sauravshakya Group Title
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    WELL GOOD NIGHT

    • 2 years ago
  24. saifoo.khan Group Title
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    @badreferences : No man! I can't do probability. :S

    • 2 years ago
  25. sauravshakya Group Title
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    when n=1 and n=2 your and my answer gives the same result Now, when n=3 according to u the answer is 0.7037 and according to me the answer is 0.66666 Now, when n=3 we can do it in a simple way. 1/3 *1 + 1/3 *2/3 + 1/3 *1/3 = 0.66666

    • 2 years ago
  26. gregohb Group Title
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    If you know r, then its a conditional probability. the actual question was what was the overall probability of the first player winning (regardless of his roll)

    • 2 years ago
  27. badreferences Group Title
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    @gregohb That sets \(r=0\).

    • 2 years ago
  28. UnkleRhaukus Group Title
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    each player rolles the same dice with expectation value \(\langle roll\rangle\) the players are expected to roll the same number there will be variation in the result , hence player one will win half of these and play two will win the other half , occasionally the players will roll the same number and in this event play one wins, so if they play for a long time player one is more likely to win

    • 2 years ago
  29. gregohb Group Title
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    There is something wrong with the original problem. If they roll one each, and stop, and first guys loses. Thats one thing. Or if they keep rolling until one loses, what if they both lose simultaneously. Or what if they keep rolling over and over again and each one wins some and loses some? Then do they keep count? Its not clear. The prob of one guy losing on his own from one roll to the next is (1/n)*(0 + 1/n + 2/n + ... (n-1)/n) = (n-1)*n/(2n^2) = (n-1)/(2n) = a The probablity of the other guy to not lose is 1-a. So For 1st guy to lose and second guy to win (since these are indepedendent events) is a*(1-a). Now if there is a series of rolls and first must continue to lose and 2nd guy continue to win, or whatever, then that needs to be expanded. but i am not clear what the problem is actually asking precisely.

    • 2 years ago
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