A strategy in such a game consists of a set of probabilities governing each player's choice. E.g., the column player might simply always choose the same column, or he might choose various columns with some probability distribution. An optimal strategy for the column player has the property that, no matter what row is chosen, the column player always wins at least some value V (on average), and no other column strategy can guarantee a greater average payoff. Likewise, an optimal row strategy guarantees that the row player loses no more than V. The optimal column strategy and optimal row strategy always have the same expected payoff V.
How one computes the probability distributions for the optimal strategies is beyond the scope of this discussion.
| First Card: 4 | Bid | ||||
|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | ||
| Bid | 1 | 0 | +1 | +1 | +118/1159 |
| 2 | -1 | 0 | +1 | +1/2 | |
| 3 | -1 | -1 | 0 | +1 | |
| 4 | -118/1159 | -1/2 | -1 | 0 | |
| First Card: 3 | Bid | ||||
|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | ||
| Bid | 1 | 0 | +1 | +1/2 | -1 |
| 2 | -1 | 0 | +1 | -1/3 | |
| 3 | -1/2 | -1 | 0 | +4/9 | |
| 4 | +1 | +1/3 | -4/9 | 0 | |
| First Card: 2 | Bid | ||||
|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | ||
| Bid | 1 | 0 | +1 | -19/105 | -1 |
| 2 | -1 | 0 | +17/63 | -1 | |
| 3 | +19/105 | -17/63 | 0 | 0 | |
| 4 | +1 | +1 | 0 | 0 | |
| First Card: 1 | Bid | ||||
|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | ||
| Bid | 1 | 0 | -16/117 | ... | ... |
| 2 | +16/117 | 0 | ... | ... | |
| 3 | ... | ... | 0 | ... | |
| 4 | ... | ... | ... | 0 | |
Some of the individual entries in the above matrices are derived by inspection. For instance, the zeroes down the diagonals can be deduced because they represent games that are still symmetric: if both players bid the same for the first card, then their positions remain identical and they must have equal chances of winning. (Note that a 0 represents not only forced draws, but any situation in which each player has the same chance of winning.) Similarly, the payoff for a row bid of N and a column bid of M must be -1 times the payoff for a row bid of M and a column bid of N.
The "+1"s in the first matrix arise when the column player wins the initial 4 and has an easy winning strategy: he will bid his highest remaining card on the 3, and his 1 on the 1. (The row player either cannot win the 3 and thus cannot possibly beat the column player's total, or in one case he must use his 4 to win the 3 and then cannot win enough other cards to catch up.)
Most of the values, however, were derived by looking at the three subcases that arise depending on what card comes up second. For example, after an initial 3 is won by the column player bidding 4 vs 3, there are three possible second cards, and for each of those we can look at the 3x3 payoff matrix, as shown below. The entries in the 3x3 matrices are in turn derived by looking at the 2x2 payoff matrices, though for those it no longer matters what order the last two diamonds are in. (The bids on either of the last two diamonds determine the bids on the last diamond.)
| 3(4,3) 4 | Bid | |||
|---|---|---|---|---|
| 1 | 2 | 3 | ||
| Bid | 1 | +1 | +1 | +1 |
| 2 | -1 | +1 | +1 | |
| 4 | +1 | 0 | -1 | |
| 3(4,3) 1 | Bid | |||
|---|---|---|---|---|
| 1 | 2 | 3 | ||
| Bid | 1 | +1 | 0 | -1/3 |
| 2 | 0 | +1 | +1 | |
| 4 | +1 | +1 | +1 | |
| 3(4,3) 2 | Bid | |||
|---|---|---|---|---|
| 1 | 2 | 3 | ||
| Bid | 1 | 0 | +1/2 | +1/2 |
| 2 | -1 | 0 | +1 | |
| 4 | +1 | +1 | +1 | |
The middle case (second card is 2) has a "saddlepoint", meaning that each player's optimal strategy contains a single choice. The column player can bid 3 and ensure he never does worse than +1/2, and the row player bids 1 and ensures the column player cannot do better than +1/2. The other two cases do not have saddlepoints, though in each case some of the choices are "dominated", meaning that they should never be selected because another choice or combination of choices always does at least as well. For instance, if the second card is 4, the column player should never bid 3 because, no matter what the row player bids, the column player does at least as well by bidding 2. Likewise, the row player should never bid 1.
Evaluating the 3x3 matrices, we find that the payoff to the column player is +1/3 if the second card is 4, and +1/2 if it is 2 or 1. It is interesting to note that the payoff varies depending on what order the remaining diamonds are in. Within the game, this can be viewed as forcing the row player to commit early. E.g., if the second card is 2, he has to bid 1 on it because otherwise the column player will bid 3 and win for sure (payoff of +1 instead of only +1/2). After that, the row player is behind 5 to 0 and must win both of the remaining cards even to draw; each player will thus bid randomly (50/50) on the last two diamonds. In contrast, if the second card had been the 4, and the two players immediately began choosing bids at random from the same two bids (column player choosing 1 or 2, row player choosing 2 or 4), and the row player happened to win with 2 vs 1, then the row player gets to see that he's won the 4 before having to decide what to bid on the 2. He can then bid 4 to win the 2 and ensure victory with a total of 6 points. Turning this around again, if the second card is 2, the row player cannot afford to bid 4 for it because he doesn't yet know what will happen on the diamond 4. And indeed, once the column player sees the row player bid 4 on the diamond 2, the column player can bid 2 on the diamond 4 and win.
Similarly, from even the partial matrix shown for an initial 1, it is clear that each player should bid 1. Winning the 1 by bidding 2 vs 1 is a net loss, and it is obvious that bidding even higher while the other player still bids 1 cannot be better.
If the first diamond is the 3, the optimal strategy turns out to be to bid 1, 3, or 4 with probabilities of 8/35, 18/35, and 9/35, respectively. Never bid 2. Though bidding 2 leads to a sure victory if the other player bids 1, and an occasional victory if he bids 4, it is guaranteed to lose if the other player bids 3, which (if he follows the optimal strategy) he will do more than half the time. Another way to look at it is that the optimal strategy has an average payoff of 0 if the other player plays any of the bids in the optimal strategy, and pays off better than 0 if the other player is careless enough to bid 2.
Finally, if the first diamond is the 2, you should never bid 4, and should bid 1, 2, or 3 with probabilities of 85/457, 57/457, and 315/457. (Aren't those lovely numbers? Bleah. If you like, you can approximate them as 3/16, 1/8, and 11/16.)
Note that the only time it's ever right to bid 2 on the first card is if the first card is the 2, and even then you should usually bid 3, and should bid 1 more often than 2.
For the high-card question, I looked at whether you come out ahead if you bid 5 on the 5 while the other player bids 1. (If you do, then you certainly come out ahead if he bids 2-4, and it would be clear that both players should bid 5.) Similarly, if the first diamond is the 1 and the other player bids 1, I looked at whether you do worse by bidding 2 than by bidding 1.
Like the earlier analysis, this involved looking at each possibility for which diamond comes up next, and constructing the payoff matrices. If you really want to see them, click here. The upshot was that winning the 5 by a bid of 5 vs 1 is not a positive move; you end up with an expected payoff of -0.073. Likewise, winning the 1 by bidding 2 vs 1 yields a slight positive payoff, +0.174. So much for easily extended results!