Probability question about poker. Doubt about the order of events.
Poker dice is played by simultaneously rolling 5 dice. In how many ways can I have a) a pair; b) a double pair?
To the first question I did: $\binom52\cdot\binom61\cdot\binom51\cdot\binom41\cdot\binom31=3600$ ways.
Of 5 dices, I select 2 to form a pair, and to this pair, i'll select 1 number from 6; After this, I'll have 3 dices left, and 5 numbers left to pick. So, to each dice I'll select 5*4*3 numbers, respectively.
To the second question I did: $\binom52\cdot\binom32\cdot\binom61\cdot\binom51\cdot\binom41=3600$ ways
From 5 dices, I pick 2, and then from 3 left, I pick 2. Then, for one pair I'll pick one number from 6, and to the other pair I'll pick one from 5 numbers. Lastly, i'll pick 1 from 4 numbers to the single dice.
Why did I get it wrong if I'm doing the same technique? I'm distributing one number to 1 pair of dices, or to each dice. Shouldn't it get right since I'm doing the exact same thing on both cases?
Thanks
$\endgroup$ 52 Answers
$\begingroup$Your first one is right.
Other easy ones are no pair, which is $6\cdot 5\cdot 4 \cdot 3\cdot 2=720$, triplets which is $\binom{5}{2}\cdot 6\cdot 5\cdot 4 =1200$, quads which is $5\cdot 6\cdot 5 = 150$, full house which is $\binom{5}{2}\cdot 6\cdot 5=300$ and of course five a a kind, which is just $6$.
Your second is wrong because you double counted hands with pairs on $(A=B),(C=D)$ along with $(C=D),(A=B)$ -- those are the same ways of picking which dice form the two pairs, but you are counting them separately. One way to look at it is that you pick one die to not be paired (five ways) then pick how the remaining four dice are paired up -- three ways, not six. Then you assign a number to the pair involving the leftmost paired dice, a number to the other pair , and a number to the loner. This gives $5\cdot 3\cdot 6\cdot 5\cdot4 = 1800$.
You can then add the ways to get $6^5=7776$, as must be the case.
$\endgroup$ 3 $\begingroup$You are selecting values and positions, so order matters, but only once. You can select the values (unorderly) then positions (orderly), or you can select positions (unordered) then values (orderly).
So you might select two values, then places into which you put the highest value, places into which you put the lowest, and a value for the lone into the remaining place.
$$\binom 62~\binom 5 2\binom 32~\binom 4 1$$
Or you might select four places and divide them into two pairs (indistinguishable), then select values for the leftmost and rightmost pair, and a value for the remaining lone .
$$\binom 54\binom 42\frac 1{2!}~\binom 61\binom 51~\binom 41$$
Same thing.
Where as in the first answer you correctly selected places for the pair and a value for it, then values for each of the remaining three lones. You could also have selected three values for the lones then places for each value.
$$\binom 61\binom 52\binom 51\binom 41\binom 31=\binom 61\binom 52\cdot\binom 53\binom 31\binom 21\binom 11$$
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