A theoretical draw question

[MarkGritter]

Suppose you're playing single-draw lowball (or the last draw in TD) with perfect information about which cards your opponent(s) have in their hands and have discarded. So, it is impossible to make a "mistake" in the Sklansky sense.

Does position still matter? Is there any case in which the order in which draws are announced will influence strategy? Why or why not?

Here's a way of stating it in the two player case: A seeks to increase his EV, B seeks to decrease it. Let DD,DP,PD,PP denote the four cases where A and B, respectively, either Draws one or stands Pat. I am interested in whether there is a case where

max(min(DD,DP), min(PD,PP)) != min(max(DD,PD), max(DP,PP))

I'd also be interested in any answers for more complicated situations, i.e., draw 1 or 2, multiple players, etc.

My guess is that position does not matter with just two players, but it might with three.

[Snoogins47]

The first thing that popped into my head, and admittedly I know very, very, very little about draw lowball probabilities, is when an opponent chooses to break a hand or not can have a profound impact on your optimum play. I'd give specific examples, but I don't know the numbers well enough. Essentially, there could be times that an opponent breaking a better hand than what you have might make your best play to stand pat yourself, being a favorite over his rough draw. And clearly, if he raps pat with something like 3456T in Ace to Five, you probably aren't going to want to draw down with your A234T

I'm an overwhelming favorite to be missing something obvious here, btw.

[AaronBrown]

Sklansky says you make money every time your opponents do something they wouldn't do if they knew your hand, not that you always do what you would if you knew your opponents' hands.

The draw order matters tremendously, even with only two players. Let's simplify the problem to focus on the essential mathematics. Suppose we played one-card lowball using only the Spades. Each of us gets one card, and I know yours, but you don't know mine. Each of us has one chance to throw our card away and get another. After that, the lower card wins the pot.

Say you draw first. If you stand pat, you know I'll draw if you have me beat, but not if I don't. So to win you need (a) my card to be higher than yours and (b) the card I draw to be higher. The probability of that is K*(K-1)/132, where K is the number of cards higher than your card.

If you draw, I'm going to draw with an 8 or higher, not with a 6 or lower, and with a 7 if your original card was higher than 7, not otherwise. So you'll throw away an 8 or higher, keep a 7 or lower.

If I have to draw first, you have to pick a maximum card you will hold for the case when I stand pat, and for the case when I draw. If I draw, you're going to draw with a 7 or higher, stand pat with a 6 or lower. If I don't draw, you throw away the 6 or higher, and stand pat with a 5 or lower.

[MarkGritter]

[ QUOTE ]
Sklansky says you make money every time your opponents do something they wouldn't do if they knew your hand, not that you always do what you would if you knew your opponents' hands.

[/ QUOTE ]

His statement of the FToP starts with "Every time you play a hand differently from the way you would have played it if you could see all your opponent's cards, they gain; and every time you play your hand the same way you would have played it if you could see all their cards, they lose." The situation is dual, so if my opponents lose by playing other than they would with knowledge of my hand, I must lose by playing other than I would with perfect information.

[ QUOTE ]

The draw order matters tremendously, even with only two players. Let's simplify the problem to focus on the essential mathematics. Suppose we played one-card lowball using only the Spades. Each of us gets one card, and I know yours, but you don't know mine. Each of us has one chance to throw our card away and get another. After that, the lower card wins the pot.

[/ QUOTE ]

I was actually thinking of the case where all players had perfect information, not the asymmetrical case. I'm sorry I didn't make that clear.

Your example is a good one to think about. Is there a case where position will affect the draw?

Suppose we just have 5 cards, 2-6.

A 2 will never draw. A 6 must always draw.

A 3 will lose to a 2. Against a 4, 5, or 6 it should not draw, since the opponent is only going to hit the 2 1/3 of the time, and the 3 wins otherwise.

A 4 must draw vs. a 5 or 6. If the 4 stands pat he is likely to be outdrawn 2/3 of the time. Similarly, a 5 cannot stand pat against a 6 or he is sure to lose when the 6 draws.

So, in this toy game at least, there is no case where the decision to draw depends on position or not. But I cannot prove why this is the case.

The reason I think the full lowball game is more interesting is that the distribution of cards in the deck might make it so that pat/draw or draw/pat pairs are more advantageous to one player than to another.

[MarkGritter]

Here is a 2-7 lowball hand that partially answers my question.

Player A has 2c 3c 4c 5s Ts
Player B has 2s 3s 4s 5c Tc
Player C has 2h 3h 4h 5d Td
Qd is a dead card.

Now, if all three players stand pat the pot is split 3 ways. (0.333 equity each.)

If just one of the players breaks he increases his equity to 0.343 (according to pokenum) while the others drop to 0.329.

But, if two players break then their equity is 0.290, while the player pat with a T has equity 0.420.

Thus it is advantageous to be either the only player to break or the only player to stand pat.

If the 1st player discards, the 2nd player cannot discard. If he does, the 3rd player will stand pat and gain nearly 9% equity from going last. If the 2nd player does not discard, the 3rd player cannot discard either, or he gives the 2nd player an advantage. Thus if the 1st player discards, the other two players must stand pat (unless they can act in collusion) and the 1st player gains a 1% edge.

If the 1st player stands pat, the 2nd player can gain the 1% edge by discarding.

In this case, there is an advantage to going first because it puts the other two players in a position which neither can exploit (unless they collude.)

So, position does matter with three players.

[Louie Landale]

Going first you can easily contrive an example where you "bluff" by standing pat with say a rough 9, encouraging the opponent to discard is 9 and draw to his 7; whereas if he went first he may be tempted to stand pat.

But generally you want to deny the opponents the information about your draw and want to draw last.

- Louie

[MarkGritter]

I finally found an example lowball hand for two players in which the order matters, even with both players having perfect information.

A has Qc 8c 6s 4s 2s. B has Qs 9c 7d 5h 3s. 5s and 5c are dead.

If both stand pat, A wins. If both draw (A to an 8, B to a 9), A is a small favorite (0.512/0.488). If A draws and B is pat, or A is pat and B draws, A and B are in a dead heat (0.5/0.5).

Thus if B goes first he should draw, and A draws behind him. If A goes first he may either stand pat or draw; B will always do the opposite and gain a 1% advantage over going first.

I just thought it was neat to work out a somewhat paradoxical situation like this, where there is a positional advantage even though both players know all the cards. It's an embedding of the penny-matching game into poker!