I am working on a Dice game involving players rolling dice to match a shared die.

Example: 6 6-sided dice are thrown, one Black and the rest white. Players score more points for each white die that comes up matching the Black die.

How do you calculate the probability that one or more white dice with match the the black die?

Is there a definable formula for N white dice, if more or less white dice are thrown?

At what point does it statistically guarantee (within a reasonable margin) that at least one white die will match the black die?


So you throw $N+1$ dice with $S$ sides, $N$ of those dice being white and one black. For each white dice, the probability that it shows the same side as the black die is $\frac{1}{S}$. All the white dice are independent, so the probability of $k$ white dice matching the black die is the probability of exactly $k$ arbitrary independent events out of a total of $N$, each occuring with probability $\frac{1}{S}$, happening.

The probability of $k$ specific events happening is $\left(\frac{1}{S}\right)^k\left(1-\frac{1}{S}\right)^{N-k}$. To get the probability for arbitrary $k$ events, we have to multiply that with the number of possible ways to pick $k$ events from $N$, which is $\binom{N}{k}$. That yields the overall probability for $k$ out of $N$ white dice with $S$ sides matching the black die \begin{eqnarray} P(k) &=& \binom{N}{k}\left(\frac{1}{S}\right)^k\left(1-\frac{1}{S}\right)^{N-k} \\ &=& \binom{N}{k}\frac{(S-1)^{N-k}}{S^N} \text{.} \end{eqnarray}

For just one white die, i.e. $N=1$, the probabilities of it matching ($k=1)$ respectively not matching ($k=0$) the black die are then \begin{eqnarray} P(1) &=& \frac{1}{S} \\ P(0) &=& \frac{S-1}{S} \end{eqnarray} which seems right.

For arbitrary many white dice, the probability of none of them matching the black die is $$ P(0) = \left(\frac{S-1}{S}\right)^N \text{,} $$ which means that for the probability of no match to be less than $\sigma$, you need to throw more than $$ \frac{\log \sigma}{\log \left(\frac{S-1}{S}\right)} $$ white dice.


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