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Rolling 3 dices at different speeds

  1. Jul 25, 2014 #1
    I have 3 fair dices. The probability of 2 of them lying in the same number without the 3rd doing so is given by [tex]\frac{N (N-1)}{N^3}[/tex], with N=6 in a regular dice.

    What if I roll the 3rd dice twice as fast (i.e. 2 times for every time I roll the other dices)? Or three times as fast? Simple simulation results show that the above formula still holds, the question is: how to prove it?

    Thank you for your help.
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  3. Jul 25, 2014 #2


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    Your question is confusing. How are the extra rolls of the third die being treated in relation to the first two? Are they seperate accounts?
  4. Jul 26, 2014 #3
    They are all independent. However, every time I roll the first two dice, I roll the 3rd dice twice or 3 times.

    I then want the count the number of times the first two dice land on the same number without the 3rd dice doing so. Does that make it more clear?
  5. Jul 26, 2014 #4


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    That was not the point of mathman's question. If you roll die "A" two or three times as fast as "B" and "C" how are you counting its number? Are you adding its number, on every roll, to the numbers on "B" and "C" while their numbers stay the same? Or are you only looking at the number on "A" that comes up when "B" and "C" are rolled?
  6. Jul 26, 2014 #5
    Say I roll dices B and C and they land on the same number X. I then roll dice A say two times. If none of them falls on the number X, then I count 2 times. Does that answer your question?
  7. Jul 26, 2014 #6


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    The original question asked for a proof. As far as I understand what you are trying to do, the proof is no different than the original proof. The probability that the second die has the same number as the first is 1/n. The probability that the third has a different number is (n-1)/n, each time you roll.

    (all rolls are assumed independent)
  8. Jul 28, 2014 #7
    What you say is correct, but how does one incorporate the "each time you roll" factor for the 3rd dice on the formulas?
  9. Jul 28, 2014 #8


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    By the way, "dices" means "chops into small pieces". As far a small cubes with dots on them is concerned "dice" is already the plural of "die".
  10. Jul 28, 2014 #9


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    Since the rolls of the third die are all independent of each other, pretend they are different dies. Label the 5 rolls, R1, R2, R3, R4, R5. Prove the equation for the 5 rolls, taking into account any constraints on each roll. So the labels are on the rolls, not on the dice.
  11. Jul 28, 2014 #10


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    OK it's easy we have die 1 die 2 and first roll of die 3: you can figure out how often the thing happens that you're interested in.

    For die 1 and die 2 and second roll of die 3 it's obviously, in the long run, equally often.

    So in the two sequences combined it is equally often.

    What you simulated is, I suppose, the long run relative frequency of what you were interested in. Each sub experiment comes in two parts, within each pair the two parts have the same distribution, but they are dependent of one another. Dependence within pairs does not spoil the long run relative frequency.
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