Re: Another poker probability problem

I think I've come up with an answer, albeit through a combination of
brute force and calculations. I think it's correct.

As mentioned earlier with ace being both high and low there are ten
possible straight flushes. A2345, 23456, 34567 all the way through
TJQKA.

Within each straight flush there are ten distinct combinations of three
cards one can form (5 choose 3). If we look at three such examples
side-by-side-by-side we see (table best viewed with a monospace font):

A2345 23456 34567
----- ----- -----
A23 [234] [345]
A24 [235] [346]
A25 236 347
A34 [245] [356]
A35 246 357
A45 256 367
234 [345] [456]
235 346 457
245 356 467
345 456 567

Taking the A2345 as the first straight flush we can easily see that any
combination of cards in successive straight flushes that involve three
cards found in previous straight flushes are duplicated (denoted by
brackets). One could also say "Only those combinations utilizing the
one card not available in the previous straight flush are not
duplictaed" This pattern holds true all the way through TJQKA. These
can easily be eliminated from consideration.

Counting up all such enumerated hands and eliminating the duplicates we
arrive at 64 different combinations of the three cards to a straight
flush (seems like a result one should have been able to arrive at by
some other calculation rather than brute force...).

Now we examine the possibilities for the fourth card. Looking at the
64 different combinations we can identify three classes of
three-card-combinations:

(1) Three-card-straight-flush (sub-defined by how much room on either
end of the ranks there is to fill that straight flush; e.g. a
three-card-straight-flush starting or ending with an Ace has less ways
to make a straight flush ("outs") than a three card straight flush of
6-7-8)

(2) Three cards which completely define only one possible straight
flush (e.g. A-3-5 or 6-8-T), and

(3) Three cards which allow only two possible straight flushes (e.g.
5-7-8 can only be used to make 45678 or 56789)

Understanding that there will be special cases as you get to the
extremes of the ranks - A23 is a three card straight that can be used
to fill only one straight flush, JKA are three cards which allow only
one possilbe straight flush) - we can still quickly see that the
possibilities for the fourth card (which does not further our progress
towards a straight flush) is either 47, 46 or 45, cards depending.

Counting up the cases I found 36 instances out of the 64 possible three
card combinations above where 47 of the remaining cards would qualify,
20 instances where only 46 of the remaining cards would qualify and 8
instances where only 45 of the remaining cards would qualify. We have
to multiply the whole thing by 4 because our analysis above covers only
one suit and there are four suits in the deck:

47 x 36 = 1692
46 x 20 = 920
45 x 8 = 360
----
2972
x 4
----
11,888

I get 11,888 four-card combinations of exactly three cards to a
straight flush (as an aside, there are only 164 four-card combinations
which yield four cards to a straight flush, so 11,888 + 164 = 12,052
combinations where you're three-or-better to a straight flush)

Some of these combinations will yield four-card combinations which are
"three to a straight flush, four to a flush"; others will yield "three
to a straight flush, one pair" but given that my strategy is geared
towards whether or not I'd continue the hand, having the flush as a
backup would obviously indicate I would continue. Slight modifications
to the math above can break out those scenarios if desirable.

Continuing along this line of analysis (and in response to some of the
earlier comments about how figuring this out was the wrong way to go
about things), I calculated the number of combinations where you have
three cards to a straight - but with no pair kicking.

The base number of 64 different combinations of three cards to a
straight is identical, as is the breakdown of how those hands are
classified. As we are now talking straights and not straight flushes
and I am also excluding pairs the number of qualifying fourth cards is
significantly lower, however the multipliers grow because each of the
64 three card combinations can happen any of 60 different ways - four
suits for each of three cards = 4 x 4 x 4 = 64 minus the four
combinations where all three cards are of the same suit (we don't want
to double-count the straight-flush possibilities) = 60 possible
mixed-suit combinations of the first three cards.

Each three-card combination, however, is multiplied by a smaller
multiplier: 32, 28 or 24 in place of the original 47, 46 and 45.

Doing the math:

32 x 36 = 1152
28 x 20 = 560
24 x 8 = 192
----
1904
x 60
----
114,240

I get 114,240 four card combinations which contain exactly three to a
straight and no pair.

As there are only 270,725 possible four card combinations I'm concerned
that this seems a little high (one wouldn't think that over 42% of the
time you'd have three to a straight in only four cards... but the math
seems to yield that conclusion), but upon looking at the three classes
of "three cards to a straight" that there are I understand that a good
poker player would only chase a very small subset of those - only those
where you have three cards in succession with some room on either end,
table action depending of course.

As another aside we can mention the number of ways you'd have four to a
straight. Any four ranked cards can be arranged with the four
different suits 4 x 4 x 4 x 4 = 256 different ways. Subtract the four
ways in which all four cards are of the same suit (to eliminate the 164
combinations discussed earlier where we're four to a straight flush)
and you get 252 as a multiplier x 41 possible four-to-a-straight
combinations = 10,332 possible four-to-a-straight (but not a straight
flush) possibilities.

As with before, slight modifications to the math can yield the number
of combinations within that 10,332 where you'd have an open-ended
straight draw or "four to a straight, three to a flush"

If I've made any errors I'd appreciate a second set of eyes pointing
them out.

Thanks,
-Chuck

.

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