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
Thank you for your help in advance.
Best Answer
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