# Probablity in dice game [closed]

In a game of dice, five dice are rolled simultaneously. Find the probability that the roll produced at least four of a kind.

Please give me the solution.

• Calculate the probabilities of five of a kind and four of a kind and add them. – 655321 Jan 21 '15 at 7:20
• Presumably this means $5$ of a kind or $4$ of a kind. For $5$ of a kind, easy I think, $\frac{6}{6^5}$. For $4$ of a kind, the kind we have $4$ of can be chosen in $6$ ways. For each way, the other kind can be chosen in $5$ ways. Once these are chosen, the kind we have $1$ of can be placed in any one of $5$ places, for a probability of $\frac{(6)(5)(5)}{6^5}$. Add the two probabilities. – André Nicolas Jan 21 '15 at 7:25

Imagine instead rolling the dice one at a time, and recording the result as a word of length $5$ over the alphabet $\{1,2,3,4,5,6\}$. There are $6^5$ such words, all equally likely. We now count the favourables.

There are $6$ "five of a kind" words. So the probability of five of a kind is $\frac{6}{6^5}$.

Now we find the probability of "four of a kind." The kind we have four of can be chosen in $6$ ways. For each choice, the kind we have one of can be chosen in $5$ ways. For each of the $(6)(5)$ choices so far, there are $5$ places where the kind we have one of can be put, for a total of $(6)(5)(5)$. So the probability of four of a kind is $\frac{(6)(5)(5)}{6^5}$.

It follows that the probability of at least four of a kind is $\frac{6}{6^5}+\frac{(6)(5)(5)}{6^5}$.

Probability of five of a kind:

First dice can be anything, $P_1 = 1$, but the next four dices all have to land on the same number as the first one: $P_2 = P_3 = P_4 = P_5 = 1/6$. The probability of five of a kind is then $$\prod_{i=1}^5 P_i = \frac{1}{6^4}.$$

Probability of four of a kind:

First dice can be anything, $P_1 = 1$, but the next three dices all have to land on the same number as the first one: $P_2 = P_3 = P_4 = 1/6$. The last dice has to land on a different number, so $P_5 = 5/6$. The probability of four of a kind is then $$\prod_{i=1}^5 P_i = \frac{5}{6^4}.$$

The probability of having at least four of a kind is then the probability of having five or four of a kind $$\frac{1}{6^4} + \frac{5}{6^4} = \frac{1}{6^3} = \frac{1}{216}.$$

The total cases are $6^{5}$ as there are $6$ states for each dice
Now, we have to find out the number of ways we can get $4$ same numbers with no restriction on the last die and then subtract that from the total cases. We first choose the number for the four dice by ${6\choose1}=6$ ways. The remaining die can have any of $6$ numbers.
So, by principle of inclusion/exclusion we get $6^{5}-6\times6=7740$ favourable cases (cases with atleast 4 distinct numbers on the dice). Therefore, the probability is $\frac{7740}{6^{5}}=\frac{215}{216}$ as the final answer.