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Probability : Rolling Dice, Flipping Coins, etc.

  1. May 13, 2012 #1
    The easiest way I can explain this is russian roulette. You have a six-sided die, and I need to find out the probability of the other person rolling a 6 before me, or dying in russian roulette with a 6-chamber revolver.

    I know it's different depending on who goes first, so for my example, I went first. This is what I came up with.

    Total[(5/6)^x×1/6 where x = {1, 3, 5}]

    This gives me the sum of the probability of winning on the first, second, and third round. The problem is, I want the overall probability of winning. The number of rounds may vary by a lot. You could roll a 6 on the first roll, or you may have 20 rounds before either of you roll it.

    I just don't understand how I would come up with an answer to this.

    Any help is appreciated.
  2. jcsd
  3. May 13, 2012 #2


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    Rolling dice until someone gets a six and playing russian roulette with a six chamber revolver are not equivalent games. For example in russian roulette the game is over by the sixth turn, this is not the case with the dice rolling problem.

    If you sum up the probability your opponent rolls a 6 on his first turn, or his second turn, or his third turn, etc. you get

    [tex] \sum_{n=0}^{\infty} \left( \frac{5}{6} \right)^{2n+1} \frac{1}{6} = \frac{5}{36} \sum_{n=0}^{\infty} \left( \frac{25}{36} \right)^2 [/tex]

    which is an infinite geometric series. Do you know how to calculate those?
  4. May 13, 2012 #3


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    Hey IceXaos and welcome to the forums.

    The one I need to clarify is what one 'roll' is. I will assume that you 'take turns' like you do in Russian roulette by throwing a dice-roll each where each paired roll is independent from the others and every roll is independent of every other roll.

    This means that both of you will have exactly the same distribution for 'shooting yourself' and because of this you only need one distribution.

    The distribution you need is known as the geometric distribution. From the wiki site:


    In your case, p = 1/6 and 1 - p = 5/6.

    The actual distribution allows for infinitely many 'trials' between the both of you since there is a chance (although it gets very very small) that you could say do this 100 times or a 1000 times and still not find a loaded chamber.

    But you can get an idea of the 'average' number of times it would take by looking at the mean of the distribution which is 1/(1/6) which is 6 tries (from wiki-pedia).

    If you wanted to calculate the probability you use the PDF which is P(X = k) = (1-p)^(k-1)xp. If you want to calculate say P(X > 3), then you need to calculate say P(X = 4) + P(X = 5) + P(X = 6) and so on. For this you would be better off using the fact that P(X > 3) = 1 - P(X = 1) + P(X = 2) + P(X = 3) (assuming we have at least 1 trial).

    So to calculate the actual probability for some number of trials, you need to calculate P(X < n) for the probability that the game lasted n times for one player.
  5. May 13, 2012 #4
    No, I don't understand how to calculate anything infinite.

    Also, the russian roulette idea, I was talking if you spun the revolver each turn. I forgot about the way of playing with a single spin and all taking a turn. Anyways ... off topic.
  6. May 13, 2012 #5


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    I think what he means is that they empty the chamber after each trigger pull, put in a new bullet and do a 'random' spin. In this case, you have the situation that is analogous to the dice problem.
  7. May 13, 2012 #6
    Just a little beyond me, so please try not to get annoyed.

    I know there is a 5/36 chance of winning with one trial. So, what I think your equation is saying is ..

    (win) * sum of (dont lose from trial 2 on ..)

    Not that I understand how to calculate this, and wiki page is confusing me a bit. Could you calculate it and show me by chance?

    /e I must be wrong because the exponent would be (n*2)-1 if that were the case I believe.
  8. May 13, 2012 #7


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    Woops there was a typo. The correct equation should be

    [tex] \sum_{n=0}^{\infty} \left( \frac{5}{6} \right)^{2n+1} \frac{1}{6} = \frac{5}{36} \sum_{n=0}^{\infty} \left( \frac{25}{36} \right)^n [/tex]

    The left hand side is exactly what you wrote. n=0 is the probability that your opponent loses on his first roll. n=1 is the probability your opponent loses on his second roll. n=2 is the probability he loses on his third roll. Etc.

    The right hand side is algebraic manipulation... pull out a factor of 1/6 and a factor of 5/6 and each term in the sum is (5/6)2n = (25/36)n.

    Writing this out without any sigma notation the probability of winning is
    [tex] \frac{5}{36} (1 + \frac{25}{36} + \left( \frac{25}{36} \right)^2 + .... ) [/tex]
    In general when summing ANY series of the form
    [tex] 1+x+x^2+x^3+x^4+...[/tex]
    as long as the absolute value of x is smaller than one this is equal to [itex] 1/(1-x)[/itex] (a handwavy proof is not difficult if you want it presented). In our case x=25/36, so our probability is
    [tex] \frac{5}{36} (1 + \frac{25}{36} + \left( \frac{25}{36} \right)^2 + .... ) = \frac{5}{36} \frac{1}{1-25/36} = \frac{5}{36} \frac{36}{11} = \frac{5}{11} [/tex]
  9. May 13, 2012 #8


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    Basically this is the geometric distribution above but you are considering both people pulling the trigger which will mean that instead a 6 denominator its a 36 since you have two trigger pulls for each 'turn' (I didn't specify this properly).

    The idea is that you are calculating when a 6 happens and this means that every event until a 6 is not a six. You use the fact that A and B are independent and that P(A OR B) = P(A) + P(B) - P(A AND B) but P(A AND B) = P(A)P(B).

    The easiest way though is to look at the geometric series in terms of a binary situation where you either 'stop' or keep going and using P(A AND B) = P(A)P(B), then you can derive the probability at a given N using this using induction.

    Again look at the geometric distribution and consider what the probability of getting shot is (consider each 'turn' and what happens for both players in each 'turn'), what happens in a 'win' and what happens when 'nothing happens' and that will give you the answer.
  10. May 13, 2012 #9


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    There's a much easier way.
    Let p be the probability that the first to roll is the first to roll a 6.
    The first roller either rolls a 6 straight away or doesn't. If they don't, their probability of getting the first 6, now as second roller, is 1-p.
    So p = 1/6 + (5/6)(1-p)
    Whence p = 6/11.
  11. May 13, 2012 #10
    Sorry, it took a few minutes but now I think I understand. Here, I'll attempt it, changing the game a little. Say, the first to roll a 1 or 6.

    sum of ((1/3)^(2n+1))(1/3) = (2/9) * sum of (?)^n

    I don't understand how you got 25/36, or where it came from.

    If this is correct, I like it, but I'm pretty sure the answer is 5/11. I just need guided on how to get to it.
  12. May 13, 2012 #11


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    6/11 is the probability that the first roller gets the 6 first. This must be more than 1/2. 5/11 is the probability that the second roller gets the 6 first.
  13. May 13, 2012 #12
    Oh, then that makes sense.

    Now, I can see how both are done, but there's one thing I can't do without a calculator at the moment and that's dealing with the variable in the exponent. I know that's what I'm missing to understand what he was telling me
  14. May 13, 2012 #13
    Yay, cool reasoning!
  15. May 15, 2012 #14
    Wanted to say thank you. I fully understand what you've showed now and can use it for a few things I have goin'. (:
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