I've seen a roughly similar version of my question asked for describing the "guess the average" games in general, but my question is about a specific step in deletion of weakly dominated strategies.
The questions reads "Consider the following game (often called “guess the average”):1 There are N ≥ 3 players
who try to outguess each other. Each player simultaneously chooses an integer between 1 and
100. The player closest to one-third the average of the guesses wins $100, whereas the others
get nothing. The prize is split evenly if there are ties.
(a) Show that no pure strategy strictly dominates any other.
(b) Show that when there are N = 3 players, and one applies the procedure of iterative weak
dominance, then W1
i = {1, 2, . . . , 14}, W2
i = {1, 2}, and W3
i = {1} for every player i.
For part (b), I am at a loss as to how W1 should lead to a maximum of 14. I can see how you could have a process W1 = {1, 2, …, 33}, W2 = {1, 2, …, 11}, W3 = {1, 2, 3, 4}, W4 = {1}. I don't quite understand the formula used to create W1 as is given though.
Best Answer
This question needs to be formulated more clearly. What are the sets W1,...,W4?
I guess with Wk you mean all strategies that survive k rounds of iterations of weakly dominated strategies. Suppose you are player 3 and the guesses are denoted by (guess player 1, guess player 2, your own guess).
You are right that all numbers > 33 are weakly dominated, because even if the other two are submitting the highest possible number, 100, then you still win by submitting 33. Hence it does not "make sense" to report above 33. However, if you submit, say, 14, you still win against (100,100). Even when the submitted vector is (100,100,1), number 1 wins. So, submitting 100 yourself is weakly dominated by many other guesses.
A strategy (or a guess) X is weakly dominated by guessing Y when Y wins for the same combination of others' guesses and at least one more.
So should 33 be in set W1? No, because there is no combination of the other two players' guesses such that you would win by guessing 33, but not with 32. Moreover, for the combination (33,33,33) it would have been better to guess 32 instead, as with (33,33,32) you win for sure. So guessing 32 weakly dominates guessing 33. You can continue in this fashion and eliminate other candidates.
In contrast, guessing number 1 must be in set W1, because if the other two players guess (1,1), then you have a chance of 1/3 to win by guessing 1 as well, but a chance of 0 by guessing any other number. Number 2 must also be in this set: Suppose the other submit (17,2), then submitting any number other than 2 makes you lose. [...]
To reach the set W2, you accept that no rational player would guess any number not in W1. But if everyone only submits guesses from set W1 (= numbers below 14), then you can further eliminate candidates for the optimal guess. Then you repeat until you arrive at the only rational guess: 1.
Does this help?