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I have been working on my probability. I have "possibly" memorized how to solve but, I really like to understand it with my heart. But.. I am kinda stuck.

I am trying to find out the possible outcomes of four of kind (unordered) Here is my intuition:

$$\frac{52 \cdot 3 \cdot 2 \cdot 1 \cdot 48}{5!}$$

I do not see why this is wrong.

Please help me out!

N. F. Taussig
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    This is the vague"est" qustion that i've ever read. Can you explain a little bit what you mean by "outcomes of four of kind"? – Shervin Sorouri Jul 12 '18 at 08:25
  • What are the numbers 58, 3, etc...? – Shervin Sorouri Jul 12 '18 at 08:26
  • I assume that you have a deck of cards and you want the probability of selecting 4 different cards? – Shervin Sorouri Jul 12 '18 at 08:29
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    You are dividing by 5! to allow for the fact that the multiplications counts the number of ways to pick the 5 cards in any order, and so count every possible combination only once. However, that is not what the product is counting. It is actually picking any 4 same-value cards in any order, followed by the odd card out. So you should only divide by 4! instead of 5!. – Jaap Scherphuis Jul 12 '18 at 08:47

3 Answers3

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I suppose that you are playing poker with a deck of 52 cards and you're asking what is the probability of dealing a hand $$ xxxxy\qquad\qquad\qquad\qquad(\star) $$ where $x$ is just any card from $2$ to $A$ and $y$ is any (and $y\neq x$ since there are only four $x$ in the deck).

Since the order of dealing is unimportant, you're just counting how many subsets with $5$ elements of the form ($\star$) you can form out of $52$ cards.

To form any such set you have to choose $x$, for which you have $13$ choices, and any of these choice can be coupled with any $y$ for which you have $52-4=48$ possibilities.

Thus you have $13\cdot48=624$ subsets of the form ($\star$).

Since there are alltogether $\binom{52}5$ possible hands dealt (the number of all subsets with $5$ elements chosen out of $52$) the probability to be dealt a Four of a Kind in poker is $$ \frac{624}{\binom{52}5}=624\frac{5!\cdot47!}{52!}. $$ I'll leave to you the task to come with the actual number.

Andrea Mori
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Consider drawing 4 cards:

1st card: any card is fine

2nd card: only 3 cards can make 4of a kind: 3/51

3rd card: 2/50

4th card: 1/49

So the answer is $\frac{3\times 2 \times 1}{51 \times 50 \times 49}$

J Tg
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  • This is when drawing only 4 cards. I'm fairly sure the question is about getting a 4-of-a-kind in a poker hand of 5 cards (because there is a 5! in the attempted solution to the question). Also you are calculating a probability, when the question seems to be about the number of ways a 4-of-a-kind can occur. – Jaap Scherphuis Jul 12 '18 at 08:37
  • Thanks for pointing out – J Tg Jul 12 '18 at 08:56
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My interpretation of your question is that you are asking how many 5-card poker hands include 4 of a kind, where the order of the cards in the hand is not considered relevant (so, for example, the hand 2D, 2H, 2S, 2C, 8C is the same as 8C, 2C, 2S, 2H, 2D).

There are $13$ ways to pick the rank of the 4-of-a-kind. Once the rank is chosen, the hand must include all four cards of that rank. This can only be done in $1$ way, since we don't consider the order of the cards significant. Then the remaining card can be any one of the $48$ cards remaining. So in all, there are $$13 \times 1 \times 48$$ hands.

If you wanted to compute the probability of four of a kind, you would need to divide by the number of five-card hands, $\binom{52}{5} = 2,598,960$.

awkward
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