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For example, the probability of being dealt a hand with 3 spades and 2 hearts from a standard deck of cards. I would think the geometric probability is: $\dfrac{13*12*11*13*12}{52*51*50*49*48}$ because I usually don't use the binomial coefficient with geometric probabilities, but the answer is incorrect unless I multiply by the binomial coefficient (which is a factor of 10 here).

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    $\begingroup$ K is needed only in binomial probability, in geometric probability you do not need it. Your question is too general $\endgroup$ – Jimmy R. Mar 25 '14 at 22:37
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    $\begingroup$ Card hands are unordered collections. For the purpose of solving your problem, may, if you wish, think in terms of ordered collections since the probability that an unordered collection has a certain number of spades and hearts equals the probability that an ordered collection has that number of spades and hearts. I hope this is conceptually clear; it can also be understood computationally by observing that ordered collections of five cards are in $5!$-to-$1$ correspondence with unordered collections of five cards. At the same time, ordered collections of five cards containing $3$ ... $\endgroup$ – Will Orrick Mar 30 '14 at 20:42
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    $\begingroup$ ... spades and $2$ hearts are in $5!$-to-$1$ correspondence with unordered collections of five cards containing $3$ spades and $2$ hearts. Thinking of probabilities as ratios, one sees that there is a cancellation between numerator and denominator of the factor $5!$ when one takes the ordered approach. I gather that you are taking the ordered approach since your denominator, $52\cdot51\cdot50\cdot49\cdot48,$ is the number of ordered collections of five cards, and not the number of card hands. That means that your numerators should also be interpreted as a number of ... $\endgroup$ – Will Orrick Mar 30 '14 at 22:00
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    $\begingroup$ ... ordered collections. The numerator $13\cdot12\cdot11\cdot13\cdot12$ can be interpreted as the number of ordered collections in which the first three cards are spades and the last two are hearts. This, of course, leaves out many ordered collections, namely all those in which the spades and hearts occur in some other order. The factor $10=\frac{5!}{3!\,2!}$ is needed to account for the different orderings of spades and hearts. This is exactly the same ordering factor that is included in the binomial formula in the situation where there are three successes and two failures for ... $\endgroup$ – Will Orrick Mar 30 '14 at 22:07
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    $\begingroup$ ... a total of five trials. This counting factor is not always a binomial coefficient, however. If you wanted to know the number of hands with two spades, one heart, and two diamonds, the counting factor would be a multinomial coefficient, $\frac{5!}{2!\,1!\,2!}$. There are also problems where no such factor would occur: for example, you might really want to know the probability that if five cards were dealt from the top of the deck first three would be hearts and the last two would be hearts. I should point out that I usually wouldn't usually use the ordered approach in your .. $\endgroup$ – Will Orrick Mar 30 '14 at 22:16
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If $X$ is a binomial random variable with parameters $n, p$ then it's probability mass function is given by $$P(X=k)=\dbinom{n}{k}p^k(1-p)^{n-k}$$ for all $0\le k \le n$. The binomial coefficient $$\dbinom{n}{k}$$ counts the possible ways to get $k$ successful results in the total of the $n$ trials.

For example if $n=10$ and $k=2$, then you can obtain the two successes in trials $1$ and $2$, or in trials $1$ and $3$ or in trials $7$ and $10$ and so on. The binomial coefficient counts exactly all these possibilities and therefore it can't be omitted. It's value is equal to $1$ when $k=1$ or when $k=n$ - as you pointed out - and thus these are the only cases in which it can be omitted.


In the geometric distribution you do not need a binomial coefficient since, there the order of the events is determined! Specifically, you have $k-1$ fails and in the $k-th$ trial you have a success.

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  • $\begingroup$ I updated my question with a specific problem I found that requires the binomial coefficient with the geometric probability. $\endgroup$ – mr eyeglasses Mar 26 '14 at 3:16

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