Brute Force probability of a pair in a flop:
(+/%#)2=;#&.>~.&.>{(3 comb 52){4#1 to 13
0.169412
Skip Cave
Cave Consulting LLC
On Thu, Oct 14, 2021 at 12:10 AM Thomas McGuire <[email protected]>
wrote:
> Devon presented more of his Poker simulations now using Jd (the J
> database) at the most recent NYCJUG meeting (
> https://code.jsoftware.com/wiki/NYCJUG/2021-10-12 <
> https://code.jsoftware.com/wiki/NYCJUG/2021-10-12>)
>
> He came up with an interesting problem of calculating the expected number
> of pairs in the initial flop in a game of Omaha.
>
> 3 cards dealt into the flop. Consider those with only pairs in them, there
> would be 3 ways that the pairs could be dealt out.
>
> X X Y, X Y X, Y X X
>
> Taking the first configuration the number of different hands you could
> make would be:
>
> */52 3 48 NB. this matches part of Devon’s calculation
> 7488
>
> NB. anyone of 52 cards can be dealt, once the card is dealt then only 1 of
> 3 can be dealt to make the pair
> NB. then we have to exclude the remaining 2 cards that match and therefore
> the third card will be anyone of 48
>
> NB. since there are 3 configurations the cards can be dealt in, the other
> 2 would be calculated:
> */52 48 3
> 7488
> */48 52 3
> 7488
>
> NB. this makes
> 3*7488. NB. Different 3 card hands with only pairs in them
> 22464
>
> To calculate the percentage of the total number of hands Devon made the
> calculation using only one of the pairs hand configurations. Then used 3!52
> for combinations of 52 things taken 3 at a time.
>
> This was off by a factor of 2 from his simulation, where he enumerated all
> the possibilities.
>
> I finally realized (it took me 2 days of intermittent thought) that the
> order in which the cards are dealt matter. Not so much in the scoring of a
> hand but it does matter for the number of ways the same hand can be dealt
> out.
>
> NB. So the total number of 3 card hands from 52 cards is:
> */52 51 50
> 132600
>
> NB. which is permutations of 52 things taken 3 at a time
> NB. which like the Combinations function in J has a representation called
> the Stope function:
> 52 ^!._1 (3)
> 132600
>
> NB. Devon’s original calculation was:
> (*/52 3 48)%3!52
> 0.338824
>
> NB. However based on the analysis above it should be:
> (3 * */52 3 48)%52 ^!._1 (3)
> 0.169412
>
> NB. Which is very close to his simulated percentage: 0.169418
>
> Devon was more concerned with scoring a hand rather than the order they
> were dealt in, when dealing with his calculation.
> So the best I could come up with to follow what I think was Devon’s
> thought process was the following:
>
> NB. there are 13 different card values
> NB. there are 2!4 combinations of pairs due to the different suits
> NB. there will be 48 cards left over to make the flop since we exclude 3
> of a kind (that’s a different type of poker hand than a pair)
> NB. So:
> (*/13,(2!4),48)%3!52
> 0.169412
>
> So the question is did I calculate these both correctly or did I just come
> up with 2 methods that match the simulation and just think my logic is
> correct?
>
>
>
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