At Jana, we have an occasional poker night after work. Last time we played, I got to thinking about the game theory behind bluffing, and discovered a simplified version of poker called Kuhn Poker that illustrates how an occasional bluff is a win in the long term.
The game goes like this: there are two players (P1 and P2), and three cards in the deck: a jack, a queen, and a king. Each player puts $1 in the pot and is dealt a card from the deck.
- P1 can either bet $1 or check. If she checks, the round is over and whoever has the higher card wins the ($2) pot.
- If P1 bets, P2 can either call or fold. If she folds, player 1 wins the ($2) pot. If she calls, she adds $1 to the pot, and whoever has the higher card wins the (now $4) pot.
Strategy for P1:
- with king: always bet. You’re guaranteed to win the pot either way, so checking only loses money you might have won on a bet/call.
- with queen: always check. In this case P2 knows the outcome, so betting would simply lose an extra dollar if P2 has the king.
- with jack: this is the interesting case. You can either check and always lose $1, or try bluffing, hoping P2 folds a queen. For analysis, let’s say the probability of player 1 bluffing in this case is x.
Strategy for P2 if P1 bets (note again that if P1 checks, the round is over and there is no decision to be made for P2):
- with king: always call, since you’re guaranteed to win.
- with jack: always fold, since you’re guaranteed to lose.
- with queen: this is the interesting case for P2. Calling makes sense is if you think P1 is often bluffing with a jack. Let’s do the analysis.
There are 6 possible hands:
- if P2 calls, loses $2
- if P2 folds, loses $1
- K/J: loses $1
- Q/K: wins $1
- Q/J: loses $1
- J/K: average loss is (x * -$2) + ((1-x) * -$1) = $-1-x
- if p2 calls, average loss is (x * -$2) + ((1-x) * -$1) = $-1-x
- if p2 folds, average loss is (x * $1) + ((1-x) * -$1) = $2x-1
Averaging across these hands and using a little algebra, we get:
- P2 fold strategy: average loss is $x/6
- P2 call strategy: average loss is $(1-2x)/6
Here’s what they look like (negated so that up is more of a win, or less of a loss, for P2):
Naturally if P1 never bluffs (x=0), P2 should never call with a queen, and the players will break even on average. But there is a critical point at exactly x=1/3 where P2 should start calling to minimize her losses, and another at x=1/2 where a call even starts winning on average.
Interestingly, the takeaway for P1 is bluff exactly 1/3 of the time to maximize her profits; she will win $1/18 on average no matter what P2 does.
The same analysis can be done in the other direction as well if you let, say, y be the probability that P2 will call with a queen. This is left as an exercise to the reader (hint: y=1/3 is best if you don’t know anything about P1’s strategy). You might also try analysis with different bet amounts—how do the critical points change if the bet is more than the ante?
Anyway, while this model is much simpler than actual poker, it’s interesting to me to have a model in my mind for why bluffing works in the long term, and why calling to expose bluffs is important as well. And while this isn’t directly related to work, if you enjoyed this dive into my mind then you may be interested in working here!