David Sklansky's 2-7 lowball math wrong and so is pokenum's

[SumZero]

Here is the pokenum's answer:

two dimes simulation

which suggests that non-flush 2345 is an even bigger dog than Sklansky says winning just 845/1892 times not 885/1892. But it turns out that the pokenum hand evaluation doesn't know that A2345 beats A2367 and thus is off. But the 2345 catching an Ace is supposed to account for 52/1892 wins. So the answer should be (845+52)/1892 = 897/1892. So why the difference?

Sklansky's easy math is wrong (after getting the harder counting right):
The numbers given for when 2345 wins when hitting a 4 and 5 are obviously wrong.

3/44 * 6/43 = 18/1892 not the 12/1892 given. And hence he is low in his count by 6 twice. That makes up the 12 difference and the true answer is 897/1892.

So the somewhat counter intuitive 2345 being worse than 2367 is true, but it is a slightly smaller dog than the article makes out.

[Luv2DriveTT]

I can verify SumZero's calculations. David made an error when the Hero wins when pairing the 4 or the 5.

Pairing the 4: There are 3 cards that hero can draw to win if the villain draws one of the two remaining 6s or 7s (6 cards). 3x6=18, not 12 as is shown in the numerator of David's calculation.

Pairing the 5: There are 3 cards that hero can draw to win if the villain draws one of the two remaining 6s or 7s (6 cards). 3x6=18, not 12 as is shown in the numerator of David's calculation.

Therefore the total chances of 5234x winning against 2367x is 995/1892, which makes 5234x a 995-to-897 underdog.

Math is not my strong point, please someone correct me if my calculations are shown to be in error. On a related note, I am very happy David chose to use this subject as I have been struggling with Triple Draw 2-7 calculations for some time now. I hope either David or another poster can revisit this subject with three or two draws remaining... that would be fascinating.

PS: Personally I didn't find this conclusion surprising at all, any thinking who has spent enough hours playing this game should know the outcome of this draw instinctively... but its still great none the less to finally see the math of the problem worked out. What did surprise me was the size of the underdog status that 2345x held.

TT [img]/images/graemlins/club.gif[/img]

[Gabe]

yeah, it didn't seem counterintuitive to me either.

I think a better example would have been 5432 vs. 8753.

(I screwed up my math on it somewhere. I get 8753 as a 1038-854 favorite, but according to twodimes, after you add 52, it should be 996-896. Is adding 52 to twodimes right?)

Another good one would be 5432 vs. 9832.

Or how about 5432 vs. the ugly but blocking 9874.

The trick to 2to7 td is to know how important it is to have a 7 and a 2, or an 8 and a 3 or 2.

[SumZero]

[ QUOTE ]
yeah, it didn't seem counterintuitive to me either.

[/ QUOTE ]

Your intuition is better than mine then. I knew I'd rather have 2ww7 than 2345. But it is surprising to me that 2367 beats 2345 and that even 9874 beats 2345 with one draw to go.

[ QUOTE ]
I think a better example would have been 5432 vs. 8753.

(I screwed up my math on it somewhere. I get 8753 as a 1038-854 favorite, but according to twodimes, after you add 52, it should be 996-896. Is adding 52 to twodimes right?)

[/ QUOTE ]

The 52 came about because that was the number of ways that 2345 could catch an A, win according to real 2-7 rules, but lose because of a straight in pokenum hand evaluation. For different draws the number that you need to add can be different - it isn't always adding 52. In general you need to make this correction any time either hand is drawing to cards that could hit A2345.

In your particular example 2345 could hit an A 4/44 different ways. Once it does, it will beat 3578 only if 3578 pairs or also hits an ace. 2 3's left, 2 5's left, 3 7's left, 3 8's left and 3 A's left. That makes 13 out of 43 cards left that will have A2345 win. So 4/44 * 13/43 which is 52/1892. Two dimes reports 838 wins for 2345 vs. 1054 losses. But that counts the 52/1892 wins as losses. So the real answer is 890 wins for 2345 and 1002 losses.

[ QUOTE ]

Another good one would be 5432 vs. 9832.

[/ QUOTE ]

twodimes gives us 5432 wins 866 vs. 1026 losses. But it has 4/44 * (2+2+3+3+3)/43 which is again 52/1892 misidentified losses. So really 5432 is only a 918 vs 974 underdog.

[ QUOTE ]

Or how about 5432 vs. the ugly but blocking 9874.

[/ QUOTE ]

twodimes gives 884 wins for 5432 vs. 1008 losses. But there are 4/44 * (2+3+3+3+3)/43 which is now 56/1892 (basically b/c only one card overlaps there is an extra card to pair with). So the real answer is 2345 is near coinflip dog winning 940 vs. lossing 952 times.

[ QUOTE ]
The trick to 2to7 td is to know how important it is to have a 7 and a 2, or an 8 and a 3 or 2.

[/ QUOTE ]

That, or how important it is to play against really, really, really bad opponents. [img]/images/graemlins/smile.gif[/img]

[capybara]

Sklansky's second paragraph should be enough to give careful readers a hint that 7632 is the favorite. The error in the "common sense" reasoning is the swift conclusion that 5432 is the favorite.

To recap, the reasoning is as follows: there are three cases where 7632 almost always loses: (a) catching a 7 (b) catching a 6 (c) catching the same card as 5432.

There are three cases where 5432 almost always loses: (A) catching a 5 (B) catching a 4 (C) catching a 6

Intuitively, (a) and (A) should roughly cancel, as (b) and (B). But (c) and (C) are not similar events.

The chance of catching a 6 is 3/44, while the chance of getting the same card is about (28/44)x(3/44), since this event is only likely/relevant if the card is one of AKQJT98.

Thus, (C) is more likely than (c) so 5432 should be an underdog.