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Number of rolls of a die to get a particular #

  1. Jun 24, 2010 #1
    How many rolls of a single die gives you an even chance of rolling, say, a six? Does this problem, as stated, even have a unique answer? Three rolls gives you a less than even chance, while four rolls gives a greater than even chance. It is tempting to solve

    (5/6)^x = .5 and get

    x = ln 2 / ln (6/5) ~ 3.8

    But since the rule that the probability of several independent events is just the product of their individual probabilities only applies for an integral number of events, this answer isn't really justified. All it tells you is that the answer, if there is one, is between 3 and 4.
     
  2. jcsd
  3. Jun 25, 2010 #2

    mathman

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    You have answered the question for yourself. Three is not enough and four is too many, so there is no way to get a probability of exactly one half.
     
  4. Jun 25, 2010 #3
    Right, no single trial is going to give even chances. But in a repeated trial situation, there may be ways to make the chances even. One way would be to use some other randomizer (like random number generator) to decide whether you get 3 or 4 rolls on any particular trial. Actually I guess doing it this way, with the number of rolls being a random variable, really can make a single trial have even chances. But that's getting a little fancy for practical situations. I asked this question because I heard a story about a legendary professional gambler, Titanic Thompson, using this proposition to beat someone out of a fair amount of cash. He allowed the person to alternate rolling three and four rolls, for an average of 3.5 rolls per trial. The person telling it explained that the correct line is 3.8. I wondered how he got that answer. I think the answer depends on the structure of the bet. Do you let the person roll the die three times some fraction of the time and four rolls the other, or one roll and six rolls, two and ten, or whatever? It turns out that when you randomly get 3 or 4 rolls in such a way that chances are even, the average number of rolls is indeed close to 3.8. I haven't tried it yet, but it may turn out that in all these situations, the average number of rolls is close to 3.8. But I really doubt they are all exactly the same. I also thought it was a neat coincidence that naively solving (5/6)^x = .5 gives this answer, because I don't see how that would apply here.
     
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