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.
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.