I finally got around to working on this again. This problem features
prominently in today's NYCJUG (spoiler alert):
https://code.jsoftware.com/wiki/NYCJUG/2022-01-11 .
On Fri, Oct 15, 2021 at 11:48 AM Skip Cave <s...@caveconsulting.com> wrote:
> Brute Force probability of a pair in 5 cards:
>
> (+/%#)4=;#&.>~.&.>{(5 comb 52){4#1 to 13
>
> 0.422569
>
>
> Skip Cave
> Cave Consulting LLC
>
>
> On Fri, Oct 15, 2021 at 9:43 AM Devon McCormick <devon...@gmail.com>
> wrote:
>
> > I have not had a chance to look over Thomas's calculations in detail. To
> > check them, I would extend the logic to some other similar problems to
> see
> > how they do. Come to think of it, I believe I have calculated the chance
> > of a pair in five cards and that's something you could look up as well.
> >
> > The brute force calculations are a good sanity check but fall down for
> > larger problems. As I said in the meeting, I tend to trust my
> simulations
> > more than my calculations.
> >
> > On Fri, Oct 15, 2021 at 4:52 AM Skip Cave <s...@caveconsulting.com>
> wrote:
> >
> > > 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 <tmcguir...@gmail.com>
> > > 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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> > > >
> > > ----------------------------------------------------------------------
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> > >
> >
> >
> > --
> >
> > Devon McCormick, CFA
> >
> > Quantitative Consultant
> > ----------------------------------------------------------------------
> > For information about J forums see http://www.jsoftware.com/forums.htm
> >
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
>
--
Devon McCormick, CFA
Quantitative Consultant
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
