In an old post, I half-jokingly suggested that the rules of scrabble should be changed to allow the values of tiles to be determined endogenously by competitive bidding. PhD students, thankfully, are not known for their sense of humor and two of Northwestern’s best, Mallesh Pai and Ben Handel, took me seriously and drafted a set of rules. Today we played the game for the first time. (Mallesh couldn’t play because he is traveling and Kane Sweeney joined Ben and me.)
Scrabble normally bores me to tears but I must say this was really fun. The game works roughly as follows. At the beginning of the game tiles are turned over in sequence and the players bid on them in a fixed order. The high bidder gets the tile and subtracts his bid from his total score. (We started with a score of 100 and ruled out going negative, but this was never binding. An alternative is to start at zero and allow negative scores.) After all players have 7 tiles the game begins. In each round, each player takes a turn but does not draw any tiles at the end of his turn. At the end of the round, tiles are again turned over in sequence and bidding works just as at the beginning until all players have 7 tiles again, and the next round begins. Apart from this, the rules are essentially the standard scrabble rules.
Since each players’ tiles are public information, we decided to take memory out of the game and have the players keep their tiles face up. It also makes for fun kibbitzing. The complete rules are here. Share and enjoy! Here are some notes from our first experiment:
- The relative (nominal) values of tiles are way out of line of their true value. The way to measure this is to compare the “market” price to the nominal value. If the market price is higher that means that players are willing to give up more points to get the tile than that tile will give them back when played (ignoring tile-multipliers on the board.) That means that the nominal score is too high. For example, blanks have a nominal score of zero. But the market price of a blank in our play was about 20 points. This is because blanks are “team players:” very valuable in terms of helping you build words. So, playing by standard scrabble rules with no bidding, if the value of a blank was to be on equal terms with the value of other tiles, blanks should score negative: you should have to pay to use them. Other tiles whose value is out of line: s (too high, should be negative), u(too low), v(too low.) On the other hand, the rare letters, like X, J, Z, seem to be reasonably scored.
- Defense is much more a part of the game. This is partly because there is more scope for defense by buying tiles to keep them from the opponent, but also in terms of the play because you see the tiles of the opponents.
- It is much easier to build 7/8 letter words and use all your tiles. This should be factored into the bidding.
- There are a few elements of bidding strategy that you learn pretty quickly. They all have to do with comparing the nominal value of the tile up for auction with the option value of losing the current auction and bidding on the next, randomly determined, tile. This strategy becomes especially interesting when your opponent will win his 7th tile, forcing you the next tile(s) but at a price of zero.
- Because the game has much less luck than standard scrabble, differences in ability are amplified. This explains why Ben kicked our asses. But with three players, there is an effect which keeps it close: the bidding tends to favor the player who is behind because the leaders are more willing to allow the trailer to win a key tile than the other leader.
Finally, we have some theoretical questions. First, suppose there is no lower bound to your score, so that you are never constrained from bidding as much as you value for a tile, the initial score is zero, and there are two players playing optimal strategies. Is the expected value of the final score equal to zero? In other words, will all scoring be bid away on average? Second, to what extent do the nominal values of the tiles matter for the play of the game. For example, if all values are multiplied by a constant does this leave the optimal strategy unchanged?