Another [0,1] game

[Aisthesis]

For those who missed the "orgy" of [0,1] games a few months ago (started by David, I might add), this game is often used to simulate certain poker situations. The one I have in mind is an enormously loose table, where everyone will pretty much always limp or call a normal raise with any 2 cards.

The game is this: Everyone is dealt a real number between 0 and 1. At the end of the betting, highest number wins.

In this simulation, there are n players at the table, and you are the only player allowed to fold or initiate a raise (to $3--only raise allowed by rule). All the other players must call (in reality they do this by choice, but I'm just trying to simulate a table where they always do) if you raise. But you can also fold. Limping costs $1, and you are on the button.

What is your optimal strategy here in terms of limping and raising hands? That is, what is the minimum number that you need to limp and to raise?

What is the value of this game? (EV)

[fnord_too]

First observation: There is no point in ever limping. Since you are always getting N-1:1 on your money whether you limp or raise, your only decision is whether to play or not, and if you are going to play you raise since that tripples your expected return, bankroll considerations aside.

So, this boils down to win expectation, i.e., what number has a 1/N chance or better of beating > N-1 other random numbers.

My first instinct is that there is an e in this answer (because the second I start thinking about the limit as N-> infinity e just jumps out at me.)

For two people, clearly it is .5
For three people, .5 will win 1/4 of the time, so that is no good.

So, prob of k wining against N-1 opponents is k^(N-1) (call this P(k)).
We want P(k) >= 1/N, so k^(N-1) >= 1/N.

I think you can just take (1/N)^(1/N-1), (i.e. the ninth root of .1 for N = 10, this yields about .77, so my e suspicion was just wrong.)

Lim (1/N)^(1/N-1) as N -> inf = 1,

[Aisthesis]

Yes, that's exactly what I get (Was this one too easy?)

How about value of game?

Additional note: You're clearly right that in this game, if you play, you always raise. From a practical point of view (I actually have a home game in mind where this is pretty close to the way it plays), I think you need to be more selective with the raising hands.

If the players are playing like that, I think actually limping on the percentage of hands given here (by whatever ranking system one chooses to use) is a pretty decent play. But for raising, I'd like to have a hand that I can bet on the flop with greater confidence.

A lot of the top 30% hands just aren't going to give you much on that--e.g., K5o. I am willing to bet the K on that at a table like this (at least if it's 5-handed or so), but there's only one card that you have to hit even for any chance at top pair. This also applies to suited aces and kings, I think.

I just don't see much point in raising hands where there's not a very high probability of flopping a hand you'll genuinely like. (and the [0,1] game idea doesn't really take into consideration that the additional cards drawn suddenly completely change the value of your own hand)

[jason1990]

This is correct. If p = n^{1/(1-n)}, then you should raise if your number exceeds p and fold otherwise. Moreover, if you follow this strategy, your EV/hand appears to be 3(p - p^n), which can also be written as 3(n-1)p^n. For n=10, this gives an EV of about 2.1 units/hand. The limit of this expression, as n -> infty, is 3.

[Aisthesis]

My results agree there, too. By the way, did you get your EV by integrating from p to 1? That's the way I did it anyway.

[jason1990]

Yes, I did. I can't imagine there'd be any other way of doing it. Can you direct me to some of the other [0,1] games you mentioned in your earlier post?

[pzhon]

[ QUOTE ]

I think you can just take (1/N)^(1/N-1), (i.e. the ninth root of .1 for N = 10, this yields about .77, so my e suspicion was just wrong.)


[/ QUOTE ]
If you look for it, you can find e = 2.718...

If you have m opponents and only play the top c/m of your hands, you win (1-c/m)^m ~ e^-c. You want to choose c to set e^-c = 1/(m+1), so c = log_e(m+1).

In other words, with 100 opponents, play the top log_e(101)/100 hands. Actually, this isn't very accurate until the number of opponents is very large, since (1+1/n)^cn is not that close to e^c. For 9 opponents, the approximation says to play anything over 0.7442, but the correct value is 0.7742. The lower value only wins 7% of the time.

[Aisthesis]

Hmmm... Here my values disagree a little with yours (I'm just letting Excel solve the n-th root of n-1).

Just to get an idea of how this looks:

n . . . min. hand . . . value
2 . . . 0.5 . . . . . . 0.25
3 . . . 0.577 . . . . . 0.38
4 . . . 0.630 . . . . . 0.47
5 . . . 0.669 . . . . . 0.53
6 . . . 0.699 . . . . . 0.58
7 . . . 0.723 . . . . . 0.62
8 . . . 0.743 . . . . . 0.65
9 . . . 0.760 . . . . . 0.68

[Aisthesis]

Sorry, the EV calculations are all off by a factor of 3. As was noted, you always raise here, so I just figured it for the $1 game (actual EV hence 3 times what I give in the chart)

[pzhon]

[ QUOTE ]
Hmmm... Here my values disagree a little with yours (I'm just letting Excel solve the n-th root of n-1).

[/ QUOTE ]
There is no disagreement. My m is your n-1. I was talking about a full table, with n=10 people total, so that you have m=9 opponents.