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You are watching: Probability of getting 4 of a kind

One that the an initial things that concerned me was $\dfrac5252 \cdot \dfrac5151 \cdot \dfrac350 \cdot \dfrac249 \cdot \dfrac148$ however this to be of food wrong.

Then, ns realized the the 3rd card can be the same as the first OR 2nd card, so i tried $\dfrac5252 \cdot \dfrac5151 \cdot \dfrac750 \cdot \dfrac249 \cdot \dfrac148$, which is additionally wrong.

Then ns realized that we more than likely need to include up the probabilities of cases where the an initial card is the one that doesn"t match, or the second is the one the doesn"t match, etc. I think this an approach leads come the solution, however I don"t think we"re to plan to deal with it this way. I think the solution looks something choose $\dfracx 52\choose5$, yet I"m not certain what have to be in the numerator.

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edited Jun 8 "16 at 21:28

N. F. Taussig

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asked Jun 8 "16 in ~ 21:09

OviOvi

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First, due to the fact that there space 13 sets of four-of-a-kind, variety of ways of picking one that the 13 ranking is $\beginpmatrix13\\1\endpmatrix$Then, once you have actually chosen the rank, the variety of ways to draw the various other three cards of that rank is $\beginpmatrix4\\4\endpmatrix$Finally, the variety of ways to draw the critical card. I m sorry is to choose from 48 various other cards: $\beginpmatrix48\\1\endpmatrix$The number of possible 5 card hands dealt from a 52 map deck is $\beginpmatrix52\\5\endpmatrix$So the final answer is:$$\frac\beginpmatrix13\\1\endpmatrix\beginpmatrix4\\4\endpmatrix\beginpmatrix48\\1\endpmatrix\beginpmatrix52\\5\endpmatrix$$

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edited Nov 12 "17 in ~ 22:07

bearacuda13

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answered Jun 8 "16 at 21:22

RinaRina

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$\underline\textUseful hint because that future:$

In drawing w/o instead of (hypergeometric distribution)

**when bespeak doesn"t matter**, using combinations provides the answer directly:

$$\texte.g. For your question\;\; \frac\binom131\binom44\binom481\binom525$$

whereas if stimulate **does** matter, multiply probabilities offers the price directly,

$$\texte.g. If the "single" must be second,\;\; \frac5252\cdot\frac4851\cdot\frac350\cdot\frac249\cdot\frac148$$

and if you choose instead to usage the various other process, (which girlfriend *can*)you **must** usage a multiplication/division variable to compensate.

That is why, because order doesn"t matter for your question,

if you usage the multiplication of probabilities approach, you must multiply through $5$

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edited Jun 9 "16 in ~ 4:28

reply Jun 9 "16 at 4:17

true blue aniltrue blue anil

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You have the right to compute the variety of ordered color etc or the variety of unordered combine of 5 cards. The probability will come out the same as lengthy as the denominator is computed the exact same way. Your last sentence is correct because that unordered combinations, therefore $x$ must be the number of unordered hands the contain four of a kind. How many ways to choose the rank you have four of? How many ways to pick the weird card, offered the location of four is already chosen?

Your earlier approach is do the efforts to count ordered means to draw the hand, i m sorry is why the denominator comes the end $52 \cdot 51 \dots 48$ In that case, you deserve to compute the ordered variety of ways to draw the four of a kind first, climate multiply by the $5$ location the weird card have the right to be in.

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answered Jun 8 "16 in ~ 21:20

Ross MillikanRoss Millikan

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Assuming all $52 \choose 5$ five card poker-hands space equally likely, the probability of gift dealt 4 cards that the exact same rank is the ratio $x / 52 \choose 5$, where $x$ is the variety of five card poker-hands that contain 4 cards the the same rank. For instance, a poker hand that contributes to $x$ is $\2,2,2,2,Q\$. The number of ways to select the rank is $13 \choose 1=13$, and also the number of ways to select the odd card is the variety of ways to pick a card from the continuing to be 48 cards, which is $48 \choose 1=48$. So $x=13 \cdot 48$.

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answer Jun 9 "16 in ~ 4:03

Ashwin GanesanAshwin Ganesan

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The most basic explanation might be the following:there are $52\choose4$ possible combinations that 4 cards in a deck of 52. Then, v 5 cards, you have the right to have 13 * 5 possible four the a kind. Division the last by the former.

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