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Here's how the game works:

I get a pay bonus. I'm so happy I decide to give away $100$ dollars to one of my $n>1$ friends. My friends try to bribe me to get the money for themselves; I'm a sucker for bribes so I give the money to the highest briber (if there's a tie, I split evenly among those who tie; also, you can choose to bribe nothing). If you don't win, you still lose your bribe (i.e. everyone pays). Assume my friends are rational and only care about their own self interest.

I'm interested to see how this game plays out. First of all it's obvious that no one will give me more than $100$ dollars since that guarantees their expected value is negative. There's also no reason to give me $100$ dollars: at best, they get paid off in full and make nothing (at worst, you tie and only get paid partially), so bribing nothing is unconditionally better.

I then thought maybe everyone bet nothing, but that's not true because any player in that situation can bribe me a dollar and get the entire sum.

Is there a definite strategy that is employed by the players, or is it possible that an array of strategies are possible? I'm not too familiar with game theory, so if there are assumptions about the players I'm missing then feel free to add them in if they make sense.

MT_
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    I think this is an all-pay auction. – f'' Jul 18 '16 at 04:07
  • @f'' Looks like it. Can you elaborate what wikipedia means when it says that players play a mixed strategy with zero expected payoff? Which mixed strategy in particular? – MT_ Jul 18 '16 at 04:12
  • A mixed strategy means each player plays one of a number of choices depending on a random number. One such strategy would be to flip a coin and offer $10$ if you get heads and $1$ if you get tails. The Wikipedia article does not say which mixed strategy they should follow. It undoubtedly depends on $n$, or (if your friends do not know $n$) your friends' guesses on what $n$ is. – Ross Millikan Jul 18 '16 at 04:17
  • You didn't specify whether the bribe is only accepted if they are one of the highest bidders, and that other bidders do not lose anything. – user21820 Jul 18 '16 at 04:49
  • @user21820 Yes, I meant to clarify that -- you lose your money if you bribe no matter what – MT_ Jul 18 '16 at 04:52
  • @RossMillikan Can you elaborate on that? Say $n=10$, what is such a mixed strategy? If the expected pay-out is zero, I don't see why a pure strategy of bribing nothing isn't a viable (and non-mixed) strategy – MT_ Jul 18 '16 at 04:54
  • As I said, a mixed strategy is bribing an amount that depends on a random number. It is in contrast to a pure strategy where you always bribe the same amount. Knowing $n$ just allows you to choose the correct mixed strategy. I listed one in my previous comment, another would be to bribe a number from $1$ to $10$, each with probability $1/10$. Finding the optimal mixed strategy is often not easy. – Ross Millikan Jul 18 '16 at 05:07
  • @RossMillikan: No mixed strategy that never offers both 0 and 100 is optimal, since it is outdone by a strategy that always offers 100. – user21820 Jul 18 '16 at 05:17
  • @MichaelTong: A strategy that always offers less than 99 is not optimal either, since it is outdone by a strategy that always offers 99. – user21820 Jul 18 '16 at 05:20
  • @RossMillikan I know what a mixed strategy is. I was asking what an optimal mixed strategy is. Sorry if that was unclear... – MT_ Jul 18 '16 at 05:41
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    My guess is that you should throw a random number $x$ in $[0,1]$, then offer $100x^{n-1}$ where there are $n$ friends, but it is just a guess. It gets the correct average of $\frac {100}n$ – Ross Millikan Jul 18 '16 at 13:48

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