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You play poker by rolling 5 ordinary die. Numbers on the die correspond to ranks on the cards. For example, two pairs means you rolled the numbers $aabbc$ in any order, where $a,b,c$ are different numbers. Calculate:

a) P(full house), ie the pattern $aabbb$

Attempt: By counting, we have $ \frac{5!}{2! 3!}$ possible rearrangements. Dividing this by the sample space, $|s| = 6^5$ gives approx $0.04,$ I believe this is the right answer

b) P(one pair), ie the pattern $aabcd$

Attempt: So there are $\frac{5!}{2!}$ possible rearrangements. Divide by $|s|$ to get a number > 1. I am not sure where I went wrong here. My reasoning for the $5!/2!$ was that for any one combination of 5 numbers, you can order them 5! ways. However, 2 of the numbers are the same, so we divide by 2!.

c) P({$aabbc$}) = $\frac{5!}{2!.2!}$ all divided by $6^5$ gives 0.46, exactly twice the suggested answer.

I also wonder now why the sample space would not be $(6^5 . 5!)$?

d) How many rolls you need for a full house on average? Attempt: So to get a full house, you need $aabbb$. Roll a die. Roll a die again.The expected number of rolls needed to get a matching number on the first roll is 6. I tried to extend this arguement to find how many rolls you need to roll a different number other than the one already got, but didn't get very far. I suppose this method also neglects the fact that we don't care about the order of the full house. Could I have a hint on this one (not the answer)?

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Poker dice has some probabilities – Henry Dec 4 '12 at 21:46
For the dots as multiplication, use \cdot as in $6\cdot5\cdot4$. – FrenzY DT. Dec 5 '12 at 16:43
up vote 4 down vote accepted

One pair: There are $6$ ways to choose what we will have two of. For each such way, there are $\dbinom{5}{2}$ ways to decide where the numbers shall be in the list.

For every way of getting this far, there are $\dbinom{5}{3}$ ways to decide on the collection of singletons, and then $3!$ ways to list them. Multiply.

Or else more easily, once we have dealt with the pair, there are $(5)(4)(3)$ ways to fill the gaps.

Two pairs: This one is tricky, it is quite common to get an answer twice as big as the truth. The two kinds that we will have pairs of can be chosen in $\dbinom{6}{2}$ ways. For every such choice, the lonely number can be chosen in $4$ ways.

Now the cards of the higher chosen kind can be placed in $\dbinom{5}{2}$ ways, and for each such choice, the lonely number can be placed in $3$ ways.

(Instead you may have reasoned like this: The first kind we have a pair in can be chosen in $6$ ways, and then the second kind we have a pair in can be chosen in $5$ ways. However, there is no "first kind" or second kind, so if we use that sort of reasoning we need to then divide by $2$.)

Waiting time: Let the probability of a full house be $p$. This was calculated by you earlier. Now it is a standard fact that for a geometric distribution where the probability of success is $p$, the mean waiting time if $\dfrac{1}{p}$.

There are various ways to show this. One is to sum the infinite series $$p+2(1-p)p+3(1-p)^2p+4(1-p)^3p+\cdots.$$ Another is to let the mean waiting time be $w$. On the first toss, we either have success (probability $p$) or failure. We used up one toss. If success, it's over, need no more tosses. If failure, on average we need w. Thus $$w=1+(1-p)w.$$ Solve for $w$.

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Where exactly did I go wrong with my answers for b) and c)? I used the same reasoning for these questions as I did in a) and in a), I got the correct answer. Thanks. – CAF Dec 4 '12 at 20:57
Also, how is it clear that we have a geometric distribution? – CAF Dec 4 '12 at 23:07
By the definition of geometric distribution. Suppose we have probability $p\ne 0$ of "success" on any trial, and repeat experiment until we get the first success. Let $X$ be the number of trials. Then $X$ has geometric distribution. – André Nicolas Dec 4 '12 at 23:55
So you are defining the probability of success $p$ as getting 5 rolls which give a full house and failure if the 5 rolls are not a full house. Then each trial is rolling the five die? I know the meaning of geometric distribution, i'am just trying to apply it here. Can you tell me where I went wrong in part b and c? Thanks . – CAF Dec 5 '12 at 10:08
For (b) the problem is the $5.\cdot 4\cdot 3$. Maybe you thought thusly: Pick a number other than the one we have $2$ of ($5$ choices). Then pick another number ($4$ choices), then another ($3$ choices). But for example picking $6$, then $1$, then $2$ is the same as picking $2$ then $6$ then $1$. So you are overcounting by a factor of $3!$. The issue with two pairs is similar, a pair of $2$'s and a pair of $6$'s is same as pair of $6$'s, pair of $2$'s. There was no problem with full house, since three $5$'s and two $4$'s is not same as three $4$'s and two $5$'s. – André Nicolas Dec 5 '12 at 16:27

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