Every so often I have the pleasure of playing low stakes poker with some friends. Usually we play Texas hold ’em or some variant of it. It’s a tough game to play (at least for unskilled people such as myself), so often near the end of the night when we’re tired (or drunk), we switch to a much simpler game called Screw Your Neighbor.
This game is a form of poker in a very loose sense in that it involves a standard deck of bridge cards and involves betting money, but the similarity stops there. The game is a process of elimination. Each person has a single card and the person or people with the lowest value card (ace high) lose. Three losses and you are eliminated. The play begins with the dealer giving one card to each player. The player to the dealer’s left starts by deciding whether to keep or exchange her card. If she exchanges her card, she trades with the player on her left. As an exception to the rule, if the player to the left happens to have a king, the exchange is blocked and the first player has to keep her card. In any case, play continues to that player on the left in the same way around the table until it is the dealer’s turn. The dealer also has a choice to keep or exchange his card. But because there are no further players, he gets a new card from the deck if he chooses to exchange. At this point everyone shows their cards; the 1 or more people with the lowest value card get eliminated, and the deal shifts to the left.
The strategy for this game is simplistic. Keep high cards; exchange low cards (unless you know there is a lower card). The only real decision comes when you have a between (’tween) card around an 8, in which case you are as likely to get a worse card as a better one in an exchange. When several people are playing, I usually hold on to ’tween cards in the hopes someone will get stuck with a lower card in a bad exchange.
Strategy starts to change, however, as players get eliminated. Eventually (usually) play is reduced to two players: the dealer and the dealer’s opponent. Here things get intense because there is a fair amount of money in the pot and it all comes down to one quick game. Thus, it is helpful to know the odds of winning and the strategy for exchanging.
Here is what my intuition says about the game. First, the dealer has a much better chance of winning this last game. After all, if the opponent exchanges cards, the dealer knows exactly whether he is winning or losing and still gets another chance to win. Second, the opponent should sandbag the ’tween cards because once the exchange happens, the dealer gets the advantage of knowing both cards.
But, hey, why rely on intuition? The game is simple enough to iterate all the possibilities so even I can figure out the probabilities of each outcome. So let’s test this hypothesis and see just how this game is played.
Since each player makes only one choice, we can characterize the strategy of each with one variable. The first is the threshold at which the opponent exchanges cards. The second is the threshold at which the dealer exchanges cards if the opponent does not. (If the opponent does exchange cards, the dealer behavior is obvious and deterministic.)
There are multiple ways to approach the problem. I decided to write a small program that tries all possible combinations. That is, given a strategy for dealer and opponent, it looks at all possible deals to determine who is the winner. You can then count the number of wins and losses to determine the odds. I then repeat that for lots of different strategies. The following table summarizes the dealer odds/opponent odds for a range of viable strategies. (I computed a wider range of thresholds as well, but am only showing the interesting range here.)
Opponent Exchange Threshold  

Dealer Exchange Threshold 

Looking at these odds, we see that the optimal strategies for the players (assuming both use optimal strategies) are for the opponent to exchange cards when she has an 8 or lower and for the dealer to exchange cards when he has a 10 or lower.
These strategies are bit surprising. Exchanging a 10 card is more likely to yield a smaller card than a larger card. However, in retrospect this still makes sense because if the opponent does not switch cards, then she has a card greater than 8. Thus, the dealer is more likely to get a better card than beat his opponent with a 10. Likewise, I expected the opponent to keep lower cards, but again because the dealer can swap cards, his median card is a 10 or better.
But the really surprising thing is that the odds of winning are almost even money between dealer and opponent. The apparent advantage of (sometimes) knowing the opponent’s card does not exist. What gives?
To understand this, let us take a closer look at the odds of winning given a particular dealt hand.
Dealer Starting Card  Dealer Winning Odds  Opponent Starting Card  Opponent Winning Odds  

Ace  0.38  Ace  0.87  
King  0.85  King  0.74  
Queen  0.31  Queen  0.61  
Jack  0.28  Jack  0.48  
10  0.25  10  0.43  
9  0.30  9  0.37  
8  0.34  8  0.31  
7  0.41  7  0.34  
6  0.48  6  0.36  
5  0.53  5  0.38  
4  0.57  4  0.39  
3  0.60  3  0.40  
2  0.62  2  0.40 
The odds for the opponent are about what we would expect. Getting a higher card yields a better chance of winning. But the odds for the dealer are strange. With the exception of the king, which blocks the opponent, getting dealt a higher card actually means a worse chance at winning. There is a better than 50% chance of having to give up a high card to the opponent. It is actually better for the dealer to deal himself a low card in hopes that his opponent exchanges it.
All that said, any rational strategy is within 2% probability of the optimal strategy, and the difference in odds between the two players is similarly close (which is probably insignificant compared to the assumptions I made in computing them). We might as well determine the winner by flipping a coin. It’s a stupid game, but of course that’s why we play it.