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Probability of losing if a 5 comes up (dice problem)

  1. Aug 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A player goes on rolling one six sided normal die (unbiased) infinite times. But if he gets a 5 then he loses and stops rolling anymore. However until and unless he gets a 5, he continues to roll the die.What's the probability of losing??


    2. Relevant equations
    Look, I know that the probability of a 5 turning up is 1/6. But in an infinite number of rolls this can be huge (if the 5 doesn't turn up)!:smile: I mean, we need the probability of losing and not a 5 turning up. So maybe I'll get an infinite G.P.?? You lose when you get a 5. I am really messed up with this problem so, please would anyone help me?? I will be really pleased.
     
  2. jcsd
  3. Aug 26, 2012 #2
    Not sure if it is the right way of reasoning on the problem, but I would say it's still 1/6.

    I mean, if you consider the frequentist approach to probability, that 1/6 is not an inner part of the dice, but it's something that arises after a number of rolls n that tends to infinity. Hence, this should be the answer to your question.

    Btw, I advice you to wait for more qualified posters. :smile:
     
  4. Aug 26, 2012 #3

    Ray Vickson

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    What is the probability that the first toss is not 5? What is the probability that the second toss is not 5? In general, what is the probability that all of the first n tosses are not 5?

    RGV
     
  5. Aug 27, 2012 #4

    HallsofIvy

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    If the game is that 'he rolls a single die until he gets a 5 and then he loses', the probability of losing is 1! You say, "we need the probability of losing and not a 5 turning up" but, according to what you say the only way to lose is to roll a 5.
    And, apparently, he is not allowed to stop rolling until he gets a five- there is no way to get a 5 except by losing!

    As far as the numbers are concerned, you say "I know that the probability of a 5 turning up is 1/6. But in an infinite number of rolls this can be huge". No, it is not. The only way he can not lose on a given roll is to roll anything other than a 5 and the probability of that is 5/6. The probability of rolling a string of 'not fives' n times is [itex](5/6)^n[/itex]. Multiplying a number less than 1 by itself repeatedly makes it smaller not larger. The limit, as n goes to infinity, of [itex](5/6)^n[/itex] is 0.
     
  6. Aug 27, 2012 #5

    Ray Vickson

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    In my post I had asked the OP to (essentially) perform the computation you did here, guided by a short sequence of simple questions to answer. I had been hoping the OP would do that, but he/she did not respond with answers to my questions, so I don't know whether or not he/she ever got the point.

    RGV
     
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