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Originally posted by wolfgang59Good, I'll carry on then 🙂.
mtthw's interpretation of the game is correct.
Here's another step towards an answer.
(I'm ignoring the possibility of mixed strategies at the moment - will try and look at how much that makes a difference later).
Let's say there N players (including me). Let's also assume we all have the same strategy: stop at n. Which value of n maximises my (or any other individual's) chance of winning?
P(I win) = (1/2)^n [1 - (1/2)^n]^(N - 1)
This is maximised by n = ln(N)/ln(2)
So, 4 player, n = 2. 8 player, n = 3 etc.
For other values of N, I cheated and drew up a table in Excel:
N=2 => n=1
N=3 => n=2
N=4 => n=2
N=5 => n=2
N=6 => n=3
N=7 => n=3
N=8 => n=3
N=9 => n=3
N=10 => n=3
N=11 => n=3
N=11 => n=4
etc
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Originally posted by wolfgang59Isn't the goal to beat your opponent rather than maximize winnings? That wasn't clear at all from the first post.
I dont think this is a "mixed strategy" solution since I allow draws.
If you allow for draws (equal to winning) then N players stopping at 1 is a Nash equilibrium. You draw heads, you win and no point in going further.
Originally posted by wolfgang59Do you mean you allow that a draw is a win for the drawing players?
mtthw's interpretation of the game is correct. You cannot win if you toss a tail .. its all about how far you go.
My thinking was stick after a single head (but I'm not 100% sure)
I dont think this is a"mixed strategy" solution since I allow draws.
I do not have a clue what the solution is for 3 players. Thats the real challenge.
Originally posted by mtthw2 questions:
Good, I'll carry on then 🙂.
Here's another step towards an answer.
(I'm ignoring the possibility of mixed strategies at the moment - will try and look at how much that makes a difference later).
Let's say there N players (including me). Let's also assume we all have the same strategy: stop at n. Which value of n maximises my (or any other individua ...[text shortened]... N=6 => n=3
N=7 => n=3
N=8 => n=3
N=9 => n=3
N=10 => n=3
N=11 => n=3
N=11 => n=4
etc
So do you count it as a win if another player or other players and you survive and then all of you stop at n? e.g, 2 players; you both stop ASAP; or 8 players, say 3 including you, survive, and all 3 stop at turn 3. IOW is a draw to be counted as a win? I wouldn't think so.
Is your chance of surviving until turn n, better than anyone else's who plays until turn n? I wouldn't think so.
I'll not argue with the math but I don't see the strategy affecting the odds of winning, if all the players follow this as their guide for maximizing their odds, i.e., if all are rational optimizers.
Originally posted by JS357Q1: No. A win is only if I have more money than anyone else at the end of the game.
2 questions:
So do you count it as a win if another player or other players and you survive and then all of you stop at n?
Is your chance of surviving until turn n, better than anyone else's who plays until turn n? I wouldn't think so.
I'll not argue with the math but I don't see the strategy affecting the odds of winning, if all the players follow this as their guide for maximizing their odds, i.e., if all are rational optimizers.
Q2: No.
The idea is this. If I stop at n, I end up with either 0 or n. If we all have the same strategy, then the only way I win is if I get n and everybody else goes bust.
As n increases - we all have the same expected winnings, but the variance increases, which is why there's a different chance of there being a single winner.
If we start considering mixed strategies, I win if I don't go bust and everybody either sticks before I do or they go bust. It's not that hard to formulate the problem, but solving it is another matter!
Originally posted by mtthw
Q1: No. A win is only if I have more money than anyone else at the end of the game.
Q2: No.
The idea is this. If I stop at n, I end up with either 0 or n. If we all have the same strategy, then the only way I win is if I get n and everybody else goes bust.
As n increases - we all have the same expected winnings, but the variance increases, which is ...[text shortened]... they go bust. It's not that hard to formulate the problem, but solving it is another matter!
Originally posted by mtthwI believe I see my mistake. I have neglected maximizing the frequency of games that SOMEONE wins.
Q1: No. A win is only if I have more money than anyone else at the end of the game.
Q2: No.
The idea is this. If I stop at n, I end up with either 0 or n. If we all have the same strategy, then the only way I win is if I get n and everybody else goes bust.
As n increases - we all have the same expected winnings, but the variance increases, which is ...[text shortened]... they go bust. It's not that hard to formulate the problem, but solving it is another matter!
Sorry to have taxed anyone's patience.