# Poker Dice (Combinational Probability)

I've seen similar questions to this asked, and have searched the internet. I haven't found the exact questions here and the internet doesn't have the correct answers.

Play poker using 5 fair dice rather than a deco of cards. Roll the five dice onto the table. The possible has are: five-of-a-kind, four-of-a-kind, a full house (3 dice with a number and common, and the other two dice with another number in common, i.e. 3,3,3,2,2), three-of-a-kind, two pair, one pair, a straight (5 dice in a sequence, such as 1,2,3,4,5 or 2,3,4,5,6), and nothing.

I know that five-of-a-kind is $\binom{6}{1}$*$({1\over 6})^5$ = .00077

four-of-a-kind I'd think i'd be $\binom{6}{1}$*$({1\over 6})^4$ * $\binom{5}{1}$*${1\over 5}$

yet this gives me the wrong answer, my logic is the same on the full house and three-of-a-kind, and I don't know what I'm doing wrong.

for two pair it'd think it'd be $\binom{6}{1}$*$({1\over 6})^2$*$\binom{5}{1}$*$({1\over 5})^2$*$\binom{4}{1}$*${1\over 4}$but this gives .0333, and the correct answer is .06173.

my reasoning for a single pair is also the same.

Obviously nothing would be 1-the summation of all other probabilities.

btw, here are the answers given in the back of the book.

five-of-a-kind = .00077 four-of-a-kind = .01929 full house = .03086 three-of-a-kind = .03858 two pair = .06173 one pair = .15432 a straight = .23148 nothing = .46296

• I wrote a blog article about this a few years ago. It ignores straights, but they are easy to account for at the end. – MJD Sep 6 '15 at 7:47

In 4-of-a-kind, e.g. the last factor 1/5 is wrong.It should be 1/6. Also, you are not taking permutations of each type into account In fact, simplify the problem by taking a common denominator of $6^5$ for all types, and just count the numerator (number of occurrences)

I shall work out a few so that you get the drift

5 of a kind: ${6\choose 1}\cdot\frac{5!}{5!} = 6$

4-1 of a kind: ${6\choose1}{5\choose 1}\cdot\frac{5!}{4!} = 150$

3-2 of a kind (full house): ${6\choose1}{5\choose 1}\cdot\frac{5!}{3!2!}= 300$

Note for further guidance that for 2 pairs (2-2-1) the selection would be ${6\choose2}{4\choose1}$, not ${6\choose1}{5\choose1}{4\choose 1}$ because choosing 4 & 5, say for the two pairs is the same as choosing 5 & 4

You're making three mistakes.

One is that you're varying the factor $\left(\frac16\right)^5=\frac1{6^5}$, which has the number $6^5$ of possible rolls in the denominator; it should be fixed, since independent of what you're considering as favourable outcomes in the various cases, the total number of possible outcomes is always $6^5$.

The second mistake is that you're not accounting for the different orders in which the rolls can occur. Thus, for four of a kind, there are $\binom51$ different choices for the position of the single roll, so the probability is

$$\frac{\binom61\binom51\binom51}{6^5}=\frac{25}{1296}\approx0.01929\;,$$

and for full house there are $\binom52$ different choices for the positions of the pair, so the probability is

$$\frac{\binom61\binom51\binom52}{6^5}=\frac{25}{648}\approx0.03858\;,$$

which is what you quote for three of a kind; the third mistake is that you mixed up the quoted probabilities and assigned them to the wrong hands.