1
$\begingroup$

I am trying to calculate the probability of getting a full house but instead of drawing only $5$ cards, I am drawing $x$ number of cards from the $52$ card deck. I know for a $5$ card hand, I would take

$$\frac{(13\times 4) \times (12\times 6)}{52 \choose 5} = 0.00144$$

How would I generalize this to $x$ cards?

$\endgroup$
  • 1
    $\begingroup$ Probably you need this? $\displaystyle \frac {13!} {1!1!(13-1-1)!} \times \frac {\displaystyle \binom {4} {3} \binom {4} {2} \binom {52-4-4} {x-3-2}} {\displaystyle \binom {52} {x}}$ The latter fraction corresponding to the probability of obtaining a specific full house, while the former one count the number of specific cases. $\endgroup$ – BGM Jan 26 '18 at 9:52
  • 1
    $\begingroup$ What does a "full house" mean for a hand of $x$ cards? $\endgroup$ – Graham Kemp Jan 26 '18 at 14:24
  • $\begingroup$ It would mean the same thing as with a 5 card hand. So as long as there is a 3 of a kind and at least a pair among those x cards, then it would be considered a full house. The probability would only get higher as x increased. I'm just unsure how to formulate the problem. $\endgroup$ – zrelova Jan 26 '18 at 23:16
  • $\begingroup$ What about 3 kings, 2 jacks, 2 tens? or 4 kings, and 2 jacks, or 3 kings, 3 jacks? It is important to define exactly what you are looking for in order to apply the correct inclusion - exclusion rules. $\endgroup$ – Doug M Feb 3 '18 at 1:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.