A method for fair bargaining over odds in 2 player bets!
post by Agrippa Kellum (asktree) · 2020-01-11T01:18:23.185Z · LW · GW · 9 commentsContents
9 comments
Alice and Bob are talking about the odds of some event E. Alice's odd's of E are 55% and Bob's are 90%. It becomes clear to them that they have different odds, and being good (and competitive) rationalists they decide to make a bet.
Essentially, bet construction can be seen as a bargaining problem, with the gap in odds as surplus value. Alice has positive EV on the "No" position for bets at >55% odds. Bob has neutral or better EV on the "Yes" position for bets at <90% odds.
Naive bet construction strategy: bet with 50/50 odds. Negative EV for Alice, so this bet doesn't work.
Less naive bet construction strategy: Alice and Bob negotiate over odds. The problem here, in my eyes, is that Alice and Bob have an incentive to strategically misrepresent their private odds of E in order to negotiate a better bet. If Alice is honest that her odds are 50%, and Bob lies that his odds are 70%, so they split the difference at 60%, Bob takes most of the surplus value.
If both were honest and bargaining equitably, they'd have split the difference at 72.5% instead. So I'll call 72.5% the "fair" odds for this bet.
A nicer and more rationalist aligned bet construction strategy wouldn't reward dishonesty! So, here it is.
1. Alice and Bob submit their maximum bets and their odds.
2. Take the minimum of the two maximum bets. Let's say its $198.
3. Construct 99 mini bets*; one at 1% odds of E, 2% odds of E... 99% odds of E. Each player automatically places 2$ on each mini bet that is favorable according to their odds ($198/99 = $2).
*99 chosen for simplicity. You could choose a much higher number for the sake of granularity.
So, in this case, Alice accepts the No position on all bets at =>55% odds, and Bob accepts the Yes position on all bets at =<90% odds, so they make 35 $2 bets, the average odds of which are 72.5%, which is the fair odds.
Observe that there is no incentive for either player to have misrepresented their odds. If Alice overrepresented her odds as 60%, she would just deny herself the ability to bet on bets at 56% through 59%, which have positive EV for her.
Note that Alice and Bob only bet $70 -- less than half of the maximum bet. If Bob wanted Alice to bet more money than she was really willing to risk, he might try to convince her that his odds were close to hers, such that a high maximum bet would still lead to a low actual bet. Does this seem like a problem to you? I think this method is still an improvement.
*The mini bets can be abbreviated analytically as one bet at average odds, I just like the mini bets concept for making the intuition clear.
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I am not sure how exciting this method is to anyone. I like it because misrepresentation of value is a core problem in 2 player bargaining, and I realized betting is a bit of a special case. Other special cases exist too-- anything where players would be willing to accept uncertainty over the size of the trade (and are trading a continuous good such that that's even possible).
9 comments
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comment by Scott Garrabrant · 2020-04-24T20:18:33.933Z · LW(p) · GW(p)
https://www.lesswrong.com/posts/aiz4FCKTgFBtKiWsE/even-odds [LW · GW] is another proposal that gives incentive compatible betting by having the bet be smaller than the maximum. (maybe its the same, haven't checked.)
comment by Bucky · 2020-01-12T00:45:53.719Z · LW(p) · GW(p)
This is an interesting idea. Essentially you penalise dishonesty by making the pot smaller.
This works provided one player doesn’t predictably have a lower maximum bet and can then increase their maximum bet (and therefore the overall pot) while simultaneously misrepresenting their believed odds.
Did you consider using the Kelly Criterion for the mini-bets instead of using a flat rate? I’m not sure how this would affect the result but I suspect it might have some nice properties.
Replies from: asktree↑ comment by Agrippa Kellum (asktree) · 2020-01-12T17:28:29.013Z · LW(p) · GW(p)
I like your summary of the method. Good point as well. Perhaps you would want a norm that players don't discuss their maximum bets before entering them.
I'm not familiar with the Kelly Criterion so I'll check that out.
Replies from: Bucky↑ comment by Bucky · 2020-01-13T00:14:59.717Z · LW(p) · GW(p)
Zvi made a reference post [LW · GW] to the Kelly Criterion a while back which might be a good starting point.
comment by Dagon · 2020-01-13T20:18:38.190Z · LW(p) · GW(p)
How does this actually work mechanically? You don't actually make an even-money bet about a probability, you make a weighted bet about an outcome, right? There's no such thing as a $2 bet that 66% is correct. There's a $3-against-$2 bet that the thing happens. (that is, the person who says "more likely than 66%" wins $2 if it happens, and loses $3 if it doesn't happen. The person saying "less likely than 66%" is the exact opposite.)
I'm guessing the "max wager" is what each wants their maximum loss to be, and if you distribute that evenly across the range, it ends up exactly equivalent of a single bet at the midway-point between the stated beliefs.
And that means, assuming you know the direction of disagreement (if you're A and give something a 60% probability, and you know B thinks it's more probable), you're incented to OVERSTATE your difference from your opponent (A is going to win if the event doesn't happen, so wants to get better odds on more bets, so claims a 0% probability, and gets a way better distribution of actual wagers).
Replies from: asktree↑ comment by Agrippa Kellum (asktree) · 2020-01-16T18:31:37.970Z · LW(p) · GW(p)
By $2 bet at 66% odds, I mean that the Yes position costs $2*66% and the No position costs $2*34%.
You're right that "max wager" is meant to be maximum loss. I think you're picking up on the fact that I made a mistake in calculating loss for each player. I was calculating the potential loss for "$2 bet at 66%" as 2 dollars for both players, but that's obviously wrong, and no reason afaik that the players should have the same maximum loss. Thanks for the observation.
I don't understand your observation about the incentive to overstate.
Let's say A gives event E 60% odds and B gives E 90% odds. For a bet at even odds:
EV_A(YES) = .4 * -.5 + .6 * .5 = .1
EV_A(NO) = .6 * -.5 + .4 * .5 = -.1
From A's perspective the no position on the 50/50 bet (or any bet where the no position costs more than 40 cents on the dollar) is negative EV. So if A submitted 0% odds, they'd be forcing themselves to take a lot of negative EV bets.
Replies from: Dagon
comment by migueltorrescosta · 2020-01-14T14:53:47.348Z · LW(p) · GW(p)
Lovely idea.
Minor point: it feels to me the average bet isn’t the usual average but instead the harmonic mean of all bets taken. The difference might be small and more importantly there’s no reason why the arithmetic average is fairer than the harmonic average, but it was just a small thing I noticed 😜
Replies from: asktree↑ comment by Agrippa Kellum (asktree) · 2020-01-16T18:32:53.639Z · LW(p) · GW(p)
Good point.