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Another die rolling game

Posted by Chris on August 2, 2014 – 2:46 pm

Sue and Bob take turns rolling a fair 6-sided die. Once either person rolls a 6 the game is over. Sue rolls first, if she doesn’t roll a 6, Bob rolls the die, if he doesn’t roll a 6, Sue rolls again. They continue taking turns until one of them rolls a 6.

If Bob rolls a 6 before Sue, what is the probability that he did it on his second roll?


This post is under “Logic, MathsChallenge” and has 10 respond so far.
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  1. 1. Wizard of Oz Said:

    5/6 * 5/6 * 5/6 * 1/6 = 125/1296

  2. 2. Chris Said:

    Hi Wiz. That’s the probability that Bob wins on his second roll. That’s not was is being asked for, but your result is a useful stepping stone.

  3. 3. jan jansen Said:

    prob, x, of bob winning is
    x=5/36+(25/36)*(5/36)+…+(25/36)^n*(5/36)
    x-(25/36)*x=5/36
    x=5/11

    prob, y, of winning on his second roll is
    y=(25/36)*(5/36)

    So the probability we are looking for is y/x or approximately 21.2%.

    By the way i did use brute force in the other question.

  4. 4. Chris Said:

    Hi Jan. Thanks for the confession re the other problem. I’m both relieved and disappointed that no-one has found a nice way to do that one.

    Let S and s denote the events of Sue rolling a 6 or not, and let B and b denote the events of Bob rolling a 6 or not. In each case that’s for a single roll of the die. So:
    P(S) = P(B) = 1/6 and P(s) = P(b) = 5/6.

    The question implies that we need the ratio of the number of games that Bob wins on his second roll to the total number of games that he wins (on any roll).

    At the outset of a game, the probability that Bob will win on his second roll is:
    P(s)P(b)P(s)P(B) = (5/6)(5/6)(5/6)(1/6) = 125/1296

    Let p denote the probability of Bob winning a game. Here are three ways to calculate p. First the first principles way:

    p = P(s)P(B) + P(s)P(b)P(s)P(B) + P(s)P(b)P(s)P(b)P(s)P(B) + …
    = P(s)P(B) (1 + P(s)P(b) + (P(s)P(b))^2 + (P(s)P(b))^3 + (P(s)P(b))^4 + …)
    Using 1/(1 – x) = 1 + x + x^2 + x^3 + … when |x| < 1, we get
    p = P(s)P(B) / (1 – P(s)P(b)) = (5/6)(1/6) / (1 – (5/6)(5/6)) = 5/11

    Now for a slicker way: Bob wins either because Sue fails to roll 6 and Bob rolls a 6, or Sue fails to roll a 6 and Bob also fails to roll a 6, and in that case, we’re back at square one and the remaining probability for Bob to win is p. So:
    p = P(s)p(B) + P(s)P(b) p => p(1 – P(s)P(b)) = P(s)P(B) =>
    p = P(s)P(B) / (1 – P(s)P(b)) = 5/11 as before.

    The slickest way: every trial ends with either S or sB – any previous rolls are irrelevant. So p = P(sB) / ( P(sB) + P(S)) = P(s)P(B) / ( P(s)P(B) + P(S))
    = (5/6)(1/6) / ((5/6)(1/6) + (1/6)) = 5/11

    Finally, the probability of Bob winning on his second roll, given that he wins, is:
    (125/1296) / (5/11) = 275/1296 ≈ 0.212…

  5. 5. Chris Said:

    Here’s another slick way to calculate the probability of Bob winning.

    After Sue fails to roll a 6 (the probability of that being 5/6), then by symmetry, the probability of Bob now winning must be the same as Sue’s initial probability of winning, and that’s simply 1 – p. So p = (5/6)(1 – p) => (1 + 5/6)p = 5/6
    => p = 5/11.

  6. 6. Chris Said:

    If Bob rolls first, what is the probability now? You might be surprised at the result.

  7. 7. Chris Said:

    I’ll redo it Jan’s way, but with more steps:
    p = P(s)P(B) + P(s)P(b)P(s)P(B) + P(s)P(b)P(s)P(b)P(s)P(B) + …
    = P(s)P(B) + P(s)P(b)(P)s)P(B) + P(s)P(b)P(s)P(B) + …)
    = P(s)P(B) + P(s)P(b) p
    => (1 – P(s)P(b))p = P(s)P(B)
    => p = 5/11

    Obviously, Jan has effectively proved the geometric series identity stuff.

  8. 8. Karl Sharman Said:

    Coming to this question a little late – been on holiday!

    But, I’ll just throw the cat amongst the pigeons and say it would be a probability of 1/6, as the dice has no memory, and past results are not a guide to future performance.
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    I’ve left a big gap for Chris to wonder if I am serious… Bob only wins if Sue fails to roll 6, so I went along the lines of Chris’s “slicker” solution in Post 4, which should have got me to the “slickest way” by extension, but I gave up after getting the solution!

  9. 9. Chris Said:

    Hi Karl. I won’t reveal my source, but some of the responses that people were giving there did include 1/6 as the answer. I wouldn’t be surprised if someone really said 1. 5/36 and 125/1296 were popular answers there.

    The most common mistake is to have decided that Sue’s rolls could be completely ignored because she never rolls a 6 in the games that Bob wins.

    One particular poster was amazing, his lack of understanding of probability and infinite processes beggars belief.

  10. 10. Chris Said:

    OK, nobody’s tried my post 6 challenge. The probability is the same, regardless of Sue or Bob going first.

    If Bob goes first, then the probability of him winning is 6/11. The probability (at the outset) of him winning on his second roll is now P(b)B(s)P(B) = 25/108 = 150/1296. So the relative probability is (150/1296)/(6/11) = 275/1296 as before.

    At least one poster on the source site used that as evidence that Sue’s involvement was irrelevant.

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