Poker “Riddle”
Posted by Rob Herman at December 8th, 2006
This one is extremely crunchy and might be considered more of a math problem than a riddle. But it caught my interest. From the XKCD forums:
You are playing no-limit Texas Hold’em with one opponent. In this version, you can see each other’s hands.
You hold AK unsuited; your opponent has a pair of queens. After initial betting, each player has $100 in reserve and there is $100 in the pot. The flop comes down Q-J-10 with no flush possibilities, giving you the straight and your opponent three queens. You have first action.
Note: Currently you are winning, but no turn or river cards can improve your hand. Your opponent can win if one of the two hidden cards is a queen (for a four of a kind) or a 10 or jack (for the full house.)
What is your betting strategy?
Hmm. Well, first, consider the position sitting from your opponent’s seat. He has seven outs (1 queen, three jacks, three 10s). This means that of the 45 cards left in the deck, 38 are unhelpful. The chances of flipping two of these cards in a row is
(38/45) * (37/44) = 0.71, so about a 29% chance of winning. This isn’t great, so in order for the right betting strategy to be to continue to play, it must be sufficiently lucrative to win the hand. Specifically, the ratio between the opponents bet and the possible pay-off must be at least as good as 29/71 or roughly 3:7, so the payoff must be 2.33 times the bet. In the long run, if 70% of hands win no money, and 30% win 2.33 times the amount bet, then the opponent will break even, so to entice him to stay in, we want to offer him a potential pay-off at least this good.
Therefore, taking into account the money already in the pot, your bet must satisfy
x / (100 x) = (1/2.33) = .43,
so x = 75.4. Therefore, we should push 75 chips forward. Anything less, and the opponent is being offered too generous a potential payoff. Anything more, and they should fold.
I had similar thoughts, FuLeng, but I believe that they were lost in the ether when I tried to send them. My maths are weak, though, because I came up with different results even though I was working along the same general thought process. Now I’m just trying to figure out why my approach yielded different results. (I said that your opponent has a (7/45) probability of getting a winning hand on the 4th card and a (7/44) probability of getting a winning hand on the 5th card if he doesn’t get it on the 4th card. 7/44 7/45 ~ 0.31.) I think that I must have failed to take into account that getting a winning 4th card reduces the probability of getting a winning 5th card or something like that.
What you neglected was to completely remove double-counting a win generated by both cards on the same deal, looks like.
The last formula in my post above should be
x / (100 x)
Not sure where the plus sign went…
Plus and minus signs don’t display on this site, never have for me. There’s a conspiracy of silence regarding the oppression of these and other implements of punctuation.
Bah. My operaters have escaped into the merciless depths of the internet.
The strange thing is that they _do_ show up in the Preview mode.
Of course, I realize that my previous answer is not complete. What happens in the second round of betting? Let’s assume that your opponent gets no help with the first card. If we repeat the analysis, then we’re offering the opponent a pay-off with existing pot of 250 chips against a 7/44 chance of victory, or about 16%. Against these poor odds, your opponent is hunting for a payoff ratio of .159/.841 = .189. If you push forward 29 chips, that would provide the correct odds, and your maximum possible bet, 25, would be slightly favorable to your opponent. Therefore, you may as well push all the chips in, since they will on their action anyway.
Unfortunately, this entire analysis is probably completely wrong for this situtation.
The total expected value of earnings of this hand is 134 in your favor. The reason you can bet in this way and expect your opponent to play along is that you are balancing your large short-term payoff against the fact that in the long run, your opponent’s behavior is correct. Well, this is the exact correct mindset for _limit_ poker. If this is one hand among many, and the chip counts are relatively irrelevent, since in a limit game, you can buy in for any amount between hands, and assuming you can afford to, then this is exactly the sort of analysis that you must make.
In a no-limit game, especially heads-up, it’s quite different. The short term pay-off is much more important than a correct long-term betting strategy, since there is no long term. If your opponent loses this hand, the best case scenario is that he’ll be down 2:1 in the chip count, which sucks, but is not insurmountable. But the worse case scenario is that he’ll be out of chips, and hence out of the game. If this started as a heads-up (two player) game, that’s just fine, he can console himself in the thought that in the long term, he’ll make money this way, but if this is a tournament-style game, with a large initial wager to the winner, he’s basically betting the game on a 29% chance. If he thinks that his chance of winning against a 2:1 chip lead is better than 29%, then he should fold.
Of course, in an actual game, this is rather difficult to calculate. In our magic perfect-knowledge game, his notional odds for the next hand are 50% if he just goes all-in right away, after which the chip totals will be equal. He’ll then have a 50% chance of victory. So his chance of winning against the 2:1 chip lead is at least 25%. The fact that he can fold bad hands is balanced by the fact that you can do it equally well, so the size of the blinds shouldn’t matter. Let’s assume the skill of the players is equivalent, so 25% is a decent estimate. This is worse than 29%, so your opponent should just push all-in!
After winning a hand at a 2:1 chip deficit, the opponent would then gain a 1:2 chip lead, reversing the win% of the players. Allowing for the stated skill and play assumptions, we are basically dealing with a game of first to 2 consecutive hands won, where the initial 2:1 lead is counted as a hand already won.
That this scenario devolves to a 1/3 chance of victory was my intuitive thought, and is in fact correct. By way of illustration, though:
Call p the chance of the player beginning with the chip lead to win the series.
If he wins the first hand (1/2 probabilty), he wins the series.
If he loses the first hand (1/2 probability), he is now in the position in which his opponent was at the start, ie down 1:2 chip count.
Thus,
p equals 1/2 plus 1/2 (1 minus p)
reducing to p equals 2/3, odds of winning from a 2:1 chip lead (equal to the chip proportion held)
The 33% win chance from folding immediately is of course greater than the 29%
The error which led to a 25% analysis was in the statement that after winning all-in on a hand, the chip totals would be equal. This would require a 3:1 initial ratio (the 3 would lose 1 to the 1, leaving 2:2, on an all-in hand), where again we see the win proportion equivalent to the chip proportion held.