Probability of getting the largest/smallest number of $k$ dice

Question

$k$ dice are thrown.

I'm trying to compute the probabilities of the following two events:

$A$ - the largest number that occurred is $i$.

$B$ - the smallest number that occurred is $i$.

So far I know that the $|W|= 6^k$.

For $A$, I think that this is the answer:

$$\sum_{j=1}^k \frac{(i-1)^{k-j}\binom{k}{j}}{6^k}.$$

The reason is:

First, I'm choosing what die will get $i$, then for the rest of the dice, each one of them has $i-1$ options and dividing all with the $|W|$. In addition, I add all of the options of the amount of the dice that will get $i$ (from $1$ to $k$).

For $B$:

$$\sum_{j=1}^k \frac{(6-i)^{k-j}\binom{k}{j}}{6^k}.$$

Thank you for the help.

Answer 1

The probability that the largest is equal to $i$ is the difference between the probability for all being at most $i$, and that for all being less than $i$.

$$\mathsf P(\max{[X_j]}_{j=1}^k=i)~=~ \mathsf P(\bigcap_{j=1}^k \{X_j\leq i\})-\mathsf P(\bigcap_{j=1}^k \{X_j < i\})$$

Which may be evaluated using independence and identicality of distributions for the dice.

The probability that the least is equal to $i$ is likewise the difference between the probability for all being at least $i$, and that for all being greater than $i$.

$$\mathsf P(\min{[X_j]}_{j=1}^k=i)~=~ \mathsf P(\bigcap_{j=1}^k \{X_j\geq i\})-\mathsf P(\bigcap_{j=1}^k \{X_j > i\})$$

Which may again be evaluated using independence and identicality of distributions for the dice.

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$$\sum_{j=1}^k \frac{(i-1)^{k-j}\binom{k}{j}}{6^k}$$

This is the probability that from $1$ to $k$ dice show equal to $i$ and the rest show less than $i$.   Which is okay but inelegant.   It would be tricky to find a closed form from this; but it is not wrong.

PS: To evaluate for $i=1$ you have to use the convention that $0^0=1$.

Answer 2

Your answer, for case A, is correct, as it translates to:

$$ \begin{array}{l} N(i,k) = \sum\limits_{1\, \le \,j\, \le \,k} {\left( \begin{array}{c} k \\ j \\ \end{array} \right)\left( {i - 1} \right)^{\,k - j} } = \sum\limits_{0\, \le \,j\, \le \,k} {\left( \begin{array}{c} k \\ j \\ \end{array} \right)\left( {i - 1} \right)^{\,k - j} } - \left( {i - 1} \right)^{\,k} = \\ = i^{\,k} - \left( {i - 1} \right)^{\,k} \\ \end{array} $$

That is: the number of ways to get a max $= i$ equals the number of ways to get $\leqslant i$ minus the number of ways to get $< i$.

For case B, you go of course for a complementary computation.