2-7 theory: strange and disturbing

[MarkGritter]

Ordinarily, with a hand like 2c 8h 9s Td Ts, you would never consider dropping T8. You'd draw one. In some cases you might decide that a T wouldn't win and you'd draw to 289 or 28. But 29T--- never?

Yet there are circumstances where, with perfect knowledge of your opponents' hands, discarding the 8 is the correct answer.

Suppose you're playing face-up TD so you never make a mistake in the Sklansky sense. You don't know what order cards will come off the deck, but you can deduce what's left in the stub.

It's down to the last draw. Your opponent has a pat J. You have the 289TT hand. But, you know the deck (which has just six cards left--- it's been a wild hand) has 3 6's and 3 8's in it.

If you draw to 289T you will lose 50% of the time with a pair of 8s, and win the other 50% of the time.

But, if you draw to 29T, you will win 60% of the time. The two cards you draw are one of 15 combinations.

66: 3 combinations
88: 3 combinations
68: 9 combinations.

So you lose just 6 out of 15 times, or 40%.

Unfortunately putting more cards in the stub doesn't necessarily improve matters. Let's add 4 3's and 4 7's to the remaining deck (14 total cards.)

Drawing to 289T: you lose 3/14 of the time by pairing the 8 = 21.4%.

Drawing to 29T: out of 14C2 = 91 possible draws, you lose with

33: 6 combinations
66: 3 combinations
77: 6 combinations
88: 3 combinations
total of 18 out of 91 = 19.8%

(Before you get too smug about rough draws, consider 23588, or 23577, with a deck containing just three fives and three sixes.)

[2461Badugi]

[ QUOTE ]
(which has just six cards left--- it's been a wild hand)

[/ QUOTE ]

Is this the home game version where you can buy an extra draw by matching the pot?

Actually, that could be kind of fun.

[DeathDonkey]

Hi Mark,

Isn't it extremely unlikely for this situation to occur in practice? It would be rare for there to ever be more nines left in the deck than eights, because people tend to hold eights. Or in the more practical example you gave at the end, people tend to hold fives more than sixes (though usually they keep both) so it would be tough for me to think of a time when the deck would be five-rich.

I'm just trying to think of how I could ever apply this idea of yours to an actual hand and I'm pretty sure the answer is we would never have perfect enough information to risk tossing the second key card hoping to catch it back.

-DeathDonkey

[MarkGritter]

There is no question that either of these would be extremely unlikely situations, and that in practice a player would not have sufficient information to make the perfect play.

The problem is that, without a theory that explains or limits the degenerate cases, how can we be sure how extensive such surprising cases are? We can hand-wave all day long, but without a solid mathematical backing all our rules of thumb are suspect.

I.e, Just because the only example I can come with is extremely unlikely, does not imply that all likely cases are straightforward. For that we need some mathematical description of under what cases the 'sensible' draws are actually correct.

[*TT*]

[ QUOTE ]
Isn't it extremely unlikely for this situation to occur in practice?

[/ QUOTE ]

In practice I'd say the scenario is impossible because the hero never has enough information to make such a decision, but in theory Mark is correct. I recall being part of a similar discussion here ages ago, I assume Mark was involved in it as well (because he usually is) - I wish I remembered the name of the thread so we could review it to see how far we have all come in our understanding of the game.

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

[MarkGritter]

Now that I'm convinced that the hand evaluation is correct (screw you, pokenum!) I have to believe that the suit of a single card--- which is getting discarded anyway--- can affect the correct draws for both players, and cause a 6.5% switch in equity. Bizarre.

Best draws:

0: 2c3h7dKs/Kd 0.483159
1: 2h5h9hTh/Kc 0.516841

Swap Kc -> Kh, new best draws:

0: 2c3h7d/KsKd 0.54756
1: 2h5h9h/ThKh 0.45244