Can you find the equilibrium of this 3-player game?
Join Date: Mar 2009
Posts: 6,731
Close, but it is the pure-strategy EV's that you need to equate. So you should not multiply by the probs of playing each choice.
Think of yourself as player 1. Suppose you know the joint probs of player 2 & 3's choices. You are trying to determine whether you should play H, M, or L (purely, 100%). Clearly if any one of those pure-strategy EV's is the highest, you simply play that choice 100%.
It is only when those three pure EV's are equal that you have a mixed strategy Nash Equilibrium, and of course, at a symmetrical equilibrium, all three players play the same strategy.
Join Date: Feb 2017
Posts: 198
Quote:
Originally Posted by whosnext
Close, but it is the pure-strategy EV's that you need to equate. So you should not multiply by the probs of playing each choice.
Think of yourself as player 1. Suppose you know the joint probs of player 2 & 3's choices. You are trying to determine whether you should play H, M, or L (purely, 100%). Clearly if any one of those pure-strategy EV's is the highest, you simply play that choice 100%.
It is only when those three pure EV's are equal that you have a mixed strategy Nash Equilibrium, and of course, at a symmetrical equilibrium, all three players play the same strategy.
ahhhh, easy then.
Join Date: Feb 2017
Posts: 198
I did the same again;
Each player has three choices [h,m,l] and chooses two of those;
Equilibrium will be when our EV for each choice is equal.
we have three choices
hm, which wins when they choose, hl/hl or ml/ml
hl, which wins when they choose, lm/lm or hm/hm
ml, which wins when they choose, hl/hl or hm/hm
it gave me 1/3 for the choice of either [hm,hl,ml]
not to surprising as you win in the same number of ways for any choice you take.
Last edited by akkopower1; 02-22-2017 at 05:26 AM.
Join Date: Mar 2009
Posts: 6,731
Yes, there have to be enough choices so that there a decent possibility of unique values being chosen.
Maybe try it with 6 possible values where each of 3 people have to choose 2. The permutations grow and the math gets messy, but I imagine that it could be done.
Edit: maybe first try it with 4 possible values (this actually seems doable).
Last edited by whosnext; 02-22-2017 at 04:53 PM.
Reason: added PS
