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nouser
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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.
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.