2 players start with a total of 1 each. Each simultaneously places one wager at fair odds of their choosing risking up to their total. The goal is to maximize your chance of having the highest total. What is the Nash Equilibrium? It's fine to ignore betting strategy and simply choose a final total distribution which is achievable by some betting strategy.

An example which is clearly not at equilibrium would be standing/betting 0. The other player can win more than half the time by making any wager with 1:x odds, with x > 1. His winrate approaches 100% as x approaches infinity.

Let p be Hero's fair probability of winning a bet at Hero's chosen odds. Choosing those odds is synonymous with choosing this p. Let q be Villain's counterpart to p.

If you always choose p, Villain can exploit you at least one of two ways:
1. By choosing q such that p/(1+p) < q < p
2. By choosing q such that q > p/(1-p)

Thus, any pure strategy is exploitable, so if a solution even exists, it must be a mixed strategy. Since this isn't a finite game, there's no guarantee that an equilibrium exists.

Is this a genuine question or just a challenge that you know the answer to? If the latter, does a solution exist? I'll be more willing to spend time on this if I know it's not a dead end lol. So far I haven't had luck finding an unexploitable mix.

I don't know the answer. It's just something I thought of which felt interesting enough to try to find the answer to. I will attempt to answer it myself but thought others might find it interesting as well (or not, who knows!)

From the wikipedia article on NE: "Nash equilibria need not exist if the set of choices is infinite and non-compact." Compactness is a new term to me but I believe the criterion is met for a NE to exist? The set of choices is 0 <= p <= 1, which is bounded and closed.

I think it can be solved with integral calculus which I need to brush up on.

I don't know how to articulate what I mean by this, but I feel the answer will be "interesting." Perhaps having some broader mathematical significance? That's just my intuition.

Then I stand corrected, didn't realize compactness guaranteed a solution. Definitely interesting, just wanted to make sure it wasn't a trick question. I too suspect the answer will be worth the effort.

I've ruled out mixing between every possible choice. That's easily exploitable, for instance it loses to pure 1/2. Pure 1/2 would get it in good 5/6 of the time.

The solution must involve both halves of the interval. Only picking from [0, 1/2) is defeated by always picking 1. Mixing within [1/2, 1] is defeated by always picking 1/2. Pure 1/2 is crushed by pure .49; pure 1 is crushed by pure .99

Cool problem!

I have found an equilibrium for this problem for n players.

I am worried you may have misunderstood before. At the time I was a little confused by your replies and put off clarifying until I forgot to.

Unless you meant to look at a specialization of the game, simply choosing 'p' is not enough. You'd also need to specify the bet amount to build a discrete distribution with 2 outcomes. For example, p=.5 and b=1 could describe an even money wager where you bet one unit. The resulting total will be 0 50% of the time and 2 50% of the time. However, you could also do p=.5 and b=.5, and your resulting total will be .5 50% and 1.5 50%. Your wager could also be a half-normal distribution with mean 1. It can follow any distribution, continuous or discrete, as long as the mean is 1 and the domain contains only real values >= 0.

Maybe I haven't understood the problem, but this is what I get:

a = player A's chosen bet proportion.
b = player B's chosen bet proportion.

p = player A's chosen bet win probability.
q = player B's chosen bet win probability.

(1) If a>b then the fraction of games won by A is simply p.
(2) If a<b then the fraction of games won by A is simply q.
(3) If a=b then the fraction of games won by A is: p*(1-q) + (1/2)*p*q + (1/2)*(1-p)*(1-q) = (1/2)*(1+p-q)

So lets say player A chooses these parameters:

a = 1
p = 1-epsilon

How can player B do any better than setting his parameters to:

b = 1
q = 1-epsilon

?

---------------------------

This solution kind of makes sense when you think about the "virtual stock market" games of the late 90s / early 00's where the objective was to end up with the biggest gains at the end...

There was no skill involved at all: you simply had to filter to find the highest variance stock(s) and then go all-in! You either crushed everybody or went broke... The only way to compete with this strategy was to try and find even higher variance stocks.

Juk

Ah, never mind - I see (1) and (2) are wrong because they don't account for the opponent decreasing their bet win probability...

Juk

I should have read some of the replies as heehaww pointed out it couldn't be a pure strategy anyway...

So this limits the search to CDFs which satisfy:



I can't prove it but I'm pretty sure it's this for the 2-player case at least:



It's easy to disprove by finding a PDF g(x) such that:



and for the n-player case:



Too late here now but I'll look again at the n-player case and see if I can prove what I think is correct for the 2-player case later in the week.

Juk

You are correct for the 2 player case.

Shouldn't the comparison be >, not >=?

Some bonus questions after you find the full solution:

1. Is the Nash Equilibrium unique?
2. Which properties of a strategy, if any, cause it to lose EV against the NE strategy?

Yeah, I was thinking along the lines of there possibly being other optimal strategies but I agree finding one of these wouldn't disprove the optimality of the existing strategy.

I can see the distribution must be different for the n-player case:



I'll think some more on this over the next few days.

It's certainly an interesting problem!

Juk

Again, I can't prove it but I'm pretty sure it's this for the n-player case:



EDIT: It' can't be that as it doesn't have a mean of 1...

Juk

Rather than type it all out in Latex I'll just paste a screenshot from Mathematica:



Is this along the right lines?



Juk

Nope can't be right as p=(1/2), b=1 clearly wins more than 1/n



Juk

Pretty sure this is it now (really! ):



Juk

Yup, that's correct. I think this is the most straightforward way:



How about the bonus questions?

That's certainly a lot more elegant than my brute-force approach

1. Is the Nash Equilibrium unique?



2. Which properties of a strategy, if any, cause it to lose EV against the NE strategy?



Juk

The problem wasn't initially well worded on my part. I intended for the solution to be represented by the distribution, so I wouldn't consider those distinct if they form the same final distribution.

It's likely trivial that there is only one distribution that represents the solution. I just wanted to make sure I understood the problem and the math involved since I'm not very familiar with PDF/CDF functions or continuous random variables.

That's what I concluded also.

Thanks for posting this - it was a really interesting problem!

The continuous distribution needed to solve it was something I've never come across before either.

Juk