The Jokers added to the deck in this case would be useless/junk cards, not wilds¹.

My intuition is that there would be no impact to how we'd normally calculate the possible combinations/frequency for each standard Poker hand, though this would add new mixes to the total number of all possible hands and thus decrease the overall probability of each standard poker hand due to the now increased total combination pool/denominator.

I believe this is because the number of possible combinations for those specific hands still remain the same regardless of how many additional exotic cards you'd add to the deck if the exact definitions for those hands aren't altered, i.e. even if there were a baker's dozen Jokers added, you still could not have a Joker Flush since a standard Flush would still be defined only from the original 4 suits and 13 ranks, which in binomial coefficients (already excluding Straight Flushes) would still be C(4,1) suits to work with and C(13,5) ranks to choose your 5-card hand from.

The only thing that'd change from including Jokers is the addition of new hand combinations that would add to the total number of all possible hands, bringing that binomial coefficient C(52,5) to C(54,5) assuming the usual 2 Jokers that come with most decks. By increasing the total number of all possible hand combinations, we have effectively reduced the overall chance for each standard poker hand by increasing the total denominator, but the actual frequency/numerator for each hand is still the same I believe, or am I missing something since my gut still has some doubts on this?

Bonus Question: For the earlier example of a 'Joker Flush', if we were to count this specific type of Flush as an entirely new, distinct kind of hand, I believe it'd simply have a total frequency of C(13,5) assuming 13 Jokers added to the deck and no straights if Jokers are just rankless junk cards?

Thank you for your time and consideration on my errant curiosity.

¹ In researching this question though, I did find an interesting article about this kind of problem but with Jokers as Wildcards which opens up an entirely different kind of question: Gadbois, S. `Poker with Wild Cards - A Paradox?' http://www2.denizyuret.com/ref/gadbois/gadbois96.pdf

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    $\begingroup$ Your arguments appear predicated on not allowing Jokers to substitute for anything. But that's not usually how they work, and when they are allowed as wildcards, then on a relative basis their inclusion hugely increases the chances of the rarest hands. $\endgroup$
    – whuber
    Commented Feb 26, 2022 at 17:19
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    $\begingroup$ @whuber Yes, the useless/junk Joker version of this problem is exactly the question I'm asking here for a variety of reasons. I did end up stumbling into the Wild Card variant of the question from the article in my footnote and I believe that's a substantially different and much more complicated problem than the question I have here, which I suspect just comes from me not being well-versed and confident enough in the relevant fundamentals. $\endgroup$
    – Lilium
    Commented Feb 26, 2022 at 17:26

1 Answer 1


I think you are mistaken: the The probability of a given poker hand most certainly changes: the chance of drawing any single card is now $\frac{1}{52+\text{No. jokers}}$, and so is less probable than in a joker-free deck. Therefore, the probability of drawing a single hand with a specific category of sets of cards decreases compared to the joker-free deck. Another way of viewing a joker-leavened deck: your chance of a specific good hand decreases, since you now have a non-zero chance of 'polluting' your hand with jokers.

You write:

I believe this is because the number of possible combinations for those specific hands still remain the same regardless of how many additional exotic cards you'd add to the deck if the exact definitions for those hands aren't altered

Where I see the flaw in this logic, is that you are not adding in all the possible combinations that now include a joker. To make a simpler example: if I hand you an urn with one of your favorite red candies in it, and ask you to pick one you win every time with 100% chance of favorite red candy on a single draw. But if I then leaven that urn with ten of your least-favorite black candies (I will assume you dislike licorice*), on a single draw you now have only a $\frac{1}{11}$ chance of drawing that red candy: even though all original 'combinations' of the red candy remain.

* I love black licorice.

  • $\begingroup$ I believe we're talking past each other a bit since I agree with your points and I suspect my question's way too specific, strange, and pedantic to be immediately intuitive. I absolutely agree that leavening the deck with useless Jokers reduces the probability of the standard hands since now we have all these Jokers uselessly joking around. My overly specific curiosity was actually your point at the very end, that "all original 'combinations' of the red candy remain" regardless of adding jokers, which is sorta inherently obvious but my intuition's saying: "but, like, we've added more cards??" $\endgroup$
    – Lilium
    Commented Feb 26, 2022 at 17:36
  • $\begingroup$ My confusion is also more specifically about using binomial coefficients to calculate raw frequencies/possible combinations, rather than the binomial distribution or even the probability of the hands. This is sorta because, going at this in a roundabout way, the probability of a hand can also just be the total # of combinations for that hand, which I believe should (obviously, I think??) still be the same if we keep their exact definitions regardless of adding jokers. The denominator would indeed increase though because of all new possible hand combinations now including useless Joker fluff. $\endgroup$
    – Lilium
    Commented Feb 26, 2022 at 17:44
  • $\begingroup$ To put this in a different way, I also really enjoy black licorice, but for some reason I really don't like the usual chewy string ones but love the salted hard candy ones from Northern Europe. $\endgroup$
    – Lilium
    Commented Feb 26, 2022 at 17:50
  • $\begingroup$ The binomial distribution has nothing to do with this question. The hypergeometric probability of 4 Aces in a randomly dealt 5-card hand from an ordinary 52 card deck is larger than the probability of 4 aces in a randomly dealt 5-card hand from a 54-card deck with 2 jokers. In R, code dhyper(4, 4,48, 5) returns $1.846893e-05$ while dhyper(4, 4,50, 5) returns $1.581023e-05.$ $\endgroup$
    – BruceET
    Commented Feb 26, 2022 at 18:23
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    $\begingroup$ @BruceET Yeah . . . I can simply talk about the Bernoulli distribution. :) Edited my answer. $\endgroup$
    – Alexis
    Commented Feb 26, 2022 at 18:24

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