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.(adsbygoogle = window.adsbygoogle || []).push({});

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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# Rolling 3 dices at different speeds

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