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Probabilities-Russian roulette

  1. Jan 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello,
    A player places a single round in a revolver leaving 5 empty emplacements.

    a) What is the probability to stay alive after playing N times
    b) What is the probability to stay alive after playing N-1 times and die the next shot ?
    c) How many times can a player participate on average ?

    2. Relevant equations



    3. The attempt at a solution
    a) (5/6)N
    b)(5/6)(N-1)1/6
    c) Really don't know how to solve this we saw basic law of probabilities to introduce the course of thermodynamics, the answer is given it is 6

    Thanks !
     
  2. jcsd
  3. Jan 20, 2014 #2

    LCKurtz

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    I think for (b) you mean ##\left(\frac 5 6\right )^{N-1}\left( \frac 1 6\right )## don't you? In that case you have (a) and (b) correct. From part (b) you have that if ##T## = time of death then ##P(T = n) = \left(\frac 5 6\right )^{n-1}\left( \frac 1 6\right )##. The average time of death is just the expected value of ##T##. So you have two problems: What is the formula for ##E(T)## and can you calculate it? Can you take it from there? If not, come back with what you try.
     
  4. Jan 20, 2014 #3
    What do you mean by E(T) ?

    Thanks
     
  5. Jan 21, 2014 #4

    haruspex

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    The expected value of T. (=average value.)
     
  6. Jan 21, 2014 #5
    The formula to calculate the average is
    E(T)=∑P(T)*T (sum n=0 to N)
    =∑(5/6)N-1*N/6
    How can i get to E(T)=6 from here ?
    Solved !!! :smile:
     
    Last edited: Jan 21, 2014
  7. Jan 21, 2014 #6

    haruspex

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    Good.
    There is a way to get the answer without summing a series.
    Suppose the expected value is E. After pulling the trigger once, there's a 1 in 6 chance it's all over. Otherwise, the expected number of rounds remaining is still E:
    E = 1 + (1/6)*0 + (5/6)*E
     
  8. Jan 21, 2014 #7

    PeroK

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    That's very clever!
     
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