I am new to game theory and recently came across this question:
2 players pick a number from 1 to 100. From the player with the higher number, we subtract 10 and whoever has the higher number then, wins. What is the optimal strategy?
Clearly there is no pure strategy equilibrium, but there should be a mixed one, however I am unsure what it is and how to arrive at it. My thoughts so far are that it makes most sense to mix among the 10 highest numbers where we assigned a higher probability to the larger numbers but I am unsure how to arrive at this formally and what the exact distribution is. Any suggestions would be highly appreciated.
Hm, I’m not into game theory, but I’d say any good strategy should at least 1) be strong against choosing randomly and 2) hard to counter.
If we consider a number $n$ you’ll win as long as your opponent chooses any number smaller than $n-10$, between $n$ and $n+10$. If your opponent has $n-10,n,n+10$ you get a draw, else you loose. This means that as long as your opponent chooses randomly and you pick a number between $91$ and $100$ your chance of winning will be highest (since you’ll reduce the loosing range to the $10$ numbers under your number).
But then this strategy can be easily countered by always choosing $90$ against your number.
So to amend this we can alter the strategy like this: Pick from $90$ and $100$ randomly. This means that as long a your opponent places a number under $80$ he will definitely loose. If your opponent places between $80$ and $90$ you have 50% chance to pick a $100$ and win (or in case of $90$ 100% chance to draw) and 50% chance to lose (or in case of $80$ to draw). If your opponent places between $91$ and $100$ you again have a 50% chance to win and a 50% chance to loose (or in case of $100$ to draw).
So no matter what strategy your opponent plays you’d expect to at least trade equally, so in this sense this is an optimal strategy.