## Identifying the Better Player

We have earlier looked at one aspect of game design – fault tolerance. Let us look at another – identifying the better player. This is the extent to which a game allows the better player to win. Each game requires certain specific skills from a player. Given this, to what extent will the better player win the game? Although this sounds pretty obvious, we will sketch it out explicitly, and see that this aspect of the game will influence the game design in an important way.

Consider an archery-based game. Here, either a player hits the target or he doesn’t; there are no partial scores. What will be a good measure of skill of a player? Intuitively, we can say, the probability that he hits the target. A player who hits the target 6 times out of 10 is better than one who hits it 4 times out of 10. Version 1 of the game might be (this looks more or less like a ‘tiebreaker’): a series of rounds; in a round, each player shoots once; if one player hits the target while the other doesn’t, the game ends, with the player who hit the target winning; else (both hit or both miss) the game proceeds to the next round. Version 2: The overall structure of rounds is the same as Version 1, but now, each round consists of three hits instead of one.

Which version of the game is better, with respect to the specific property we are interested in: the better player winning? Lets do some calculations to find out. Assume player 1’s (P1) skill level (probability of hitting the target) is 0.6, and that of player 2 (P2) is 0.4. Consider Version 1. P1 wins in the first round, if he hits (probability 0.6) while player 2 misses (probability, 1-0.4=0.6). The probability of this is (0.6 X 0.6) 0.36. The probability of P2 winning is, similarly, 0.16. Hence the chances are distributed in the ratio 9:4. With the remaining probability (0.48) there is a tie, and the game proceeds to round 2, where again the chances will be split in the same ratio. Hence the final winning probabilities are: 9/13 (~0.69) for P1, and 4/13 (~0.31) for P2. In other words, there is a roughly 30% chance that the poorer player wins the game.

Now for Version 2. Let’s look at the probability that P1 wins in the first round. This can happen in a variety of ways. Looking at the number of hits: {1,0}, {2,0}, {2,1}, {3,0}, {3,1}, {3,2}, i.e., all possibilities in which the number of hits of P1 is greater than that of P2. This probability comes out to (after some painful calculations) 0.54. That of P2 is 0.18. (Probability of a tie = 0.28) The probabilities are split in the ratio 9:3. As above, the final winning probabilities are: P1 = 9/12 (0.75), P2 = 3/12 (0.25). Now, the chances of the poorer player winning have reduced to 25%, making version 2 better than version 1. Also notice that the chances of a tie in each round has come down.

We observe that increasing the number of hits in each round improves the chances that the better player wins. Taking this to the extreme, a round of say, 1000 hits should be really effective in identifying the better player. However, this has to be balanced with the time available, and we can live with some error. We might choose to have, lets say, an initial round of 10 or 20 hits. The P1 winning probabilities for different first round lengths are (computed using some not so painful simulations): 10 rounds = 0.85; 20 rounds = 0.92; 50 rounds = 0.98. After this, the marginal gains by having another full round will be lesser, and we can settle for a tiebreaker in the form of a series of single-hit rounds.