My *nix world

Poker probability

Have you ever wondered what are the chances that you get a quads (four of a kind) when playing poker? Well, that depends on how many cards the hand has because there are several poker variants, right? These kind of problems are simply called poker probability.

Let's assume a hypothetical poker game where we have 36 cards (numbered from 1 to 9 of diamonds,hearts,clubs,spades).

What is the probability of getting 4 of 1s if a hand contains 9 cards? The order of how the cards are arranged doesn't matter.

 

My answer is roughly 2 in a 1000 chance.

Let's see why:

  1. firstly let's see how many hands are theoretically possible (A)
  2. secondly let's count how many hands that includes 4 of 1s are theoretically possible (B)
  3. the chances are B in a A (or 100*B/A percent)

Note that the order doesn't matter, i.e. a hand with 1,2,3 is considered to be the same as 2,3,1 or 3,1,2, and so on, because we focus on what cards the hand has and not how are they arranged/sorted.

(i) The number of possible permutations with 36 cards taken by 9 is 36P9=\frac{36!}{(36-9)!}

  • anyway, because the order doesn't matter we have to take into account that some of these pairs are "counted twice" (i.e 123 is the same like 312 or 132,etc) so we have to divide that number (36P9) by the number of how 9 empty slots can be arranged, i.e. 9!
  • The real answer would be \frac{36P9}{9!}=\frac{36!}{9!(36-9)!}=36C9

(ii) The number of hands of 9 cards that includes 4 of 1s is determined like this:

  • we have 9 cards and let's suppose that some of them (like the first four of them) are 1s
  • after we have picked those 4 of 1s, how many cards available are in the deck? 36-4=32 cards
  • how many unique permutations can we do with 32 cards taken by 5 (because there are only 5 empty slots) ?
    • The answer is 32P5=\frac{32!}{(32-5)!}; anyway, because the order doesn't matter we have to take into account that some of these pairs are "counted twice" (i.e 123 is the same like 312 or 132,etc) so we have to divide that number (32P5) by the number of how 5 empty slots can be arranged, i.e. 5!
    • The real answer would be \frac{32P5}{5!}=\frac{32!}{5!(32-5)!}=32C5
#1 #2 #3 $3 #5 #6 #7 #8 #9
1 1 1 1 _ _ _ _ _

(iii) So, the probability of getting 4 of 1s from a deck of 36 cards is equal to the \frac{32C5}{36P9}\approx 0.2\% which is like 2 in 1000 chance.

Now, if you think that this article was interesting don't forget to rate it. It shows me that you care and thus I will continue write about these things.

 
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Poker probability

Eugen Mihailescu

Founder/programmer/one-man-show at Cubique Software
Always looking to learn more about *nix world, about the fundamental concepts of math, physics, electronics. I am also passionate about programming, database and systems administration. 16+ yrs experience in software development, designing enterprise systems, IT support and troubleshooting.
Poker probability

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