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In a standard deck of cards plus 1 additional joker (53 cards), I am trying to calculate the probability of drawing a Three of a Kind and a Joker in the same 5 card hand.

At the moment I know the probability for drawing a three of a kind is:

${^{13}\mathrm C_1}\times{^4\mathrm C_3}\times{^{12}\mathrm C_2}\times({^4\mathrm C_1})^2 ~/~ {^{53}\mathrm C_5}$

I am just wondering how I would add in the joker

(this is for school and any help would very appreciated)

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  • $\begingroup$ Do you mean Ace Ace Ace Joker King (which would be 4 Aces) or Ace Ace Joker King Queen? $\endgroup$ – Daniel Mathias Mar 17 '20 at 2:51
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One of the non-3-of-a-kind cards must be the Joker, so in your notation $$\dfrac{\,^{13}C_1 \times \,^{4}C_3\times \,^{12}C_1 \times \,^{4}C_1 \times \,^{1}C_1}{\,^{53}C_5}$$

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The probability for obtaining three from four suits in one from thirteen kinds and two other cards, when selecting five from fifty-three cards is actually:$$\def\C#1#2{\mathop{^{#1}\mathrm C_{#2}}}\dfrac{\C{13}{1}\times\C 43\times\left(\C{12}2\times\C42^2+\C{12}1\times\C41\times\C11\right) }{\C{53}5}$$

Note: those two other cards are either: two from four suits in each of two from twelve kinds, or one from four suits in one from twelve kinds and the joker.

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