Calculating gambling odds/payout

Factual Questions
1

Suppose that there are three persons, A, B, and C, who assign different probabilities to the likelihood of a particular event occurring, and wish to bet each other. Person A believes that the event will occur with 90% probability, person B assigns a 60% probability, and person C only a 20% probability. So persons A and B will bet against person C.

Now, they can sit and wrangle back and forth about it, but it seems that there must be a procedure to set payouts so that all three will bet, feeling that the bet is, if not optimal, at least fair. So person A would bet if he would receive ~$1.12 or more for every dollar bet (expects to lose only 10% of the time) and person C will bet only if she will get more than $1.25 for every dollar bet (expects to lose 20% of the time). Meanwhile, person B, who expects to lose nearly half the time, will only bet if the payout is more than $1.66 per dollar bet.

How do I get from there to determining how much each person has to put up to make it all work out? Intuitively it seems like persons A and C have to put up more than person B, but I don’t know how to calculate it, or how to generalize it to more than three persons.

2

Probabilities and odds (payouts) are equivalent so any probability between 20% and 60% (equiv., odds between 2-3 and 4-1) will work. Or the probability could be set between 60% and 90% to get B to bet the other way.

A or C will always have more incentive to bet than B does. I think you’ll need more constraints to get an answer more definite than the above. How much money do the players want to bet? You can phrase it as “What odds maximize total betting?” but will then need player’s utility functions.

3

C needs to get $4 a win, no?

4

Neither A nor B will pay that. C will need to take the opposite side of any wager.

5

I think this is equivalent to trying to find one value of odds that advantages or disadvantages all of them to the same degree, and I don’t think it is possible, if you consider what makes a bet atrractive or not.

Odds have an implied probability, which is basically 1/odds (exactly so, if you use the mathematically sensible decimal odds system - i.e. the one in which evens is equal to 2.0)

A bet in which your estimate of actual probability is higher than that implied by the odds has positive value. It is, in your estimation, a bet worth taking. A bet with implied probability higher than your estimated probability has negative value (as is sadly the case with most real-life bets).

Since A, B and C have different estimates of the probability, no one value of odds can have equal value to all of them. The implied probability will either be higher than all of their estimations, in which case it will hurt A the least, or it will be lower than all their estimations in which case it will advantage C the most, or it will be in between, in which case some will gain and some will lose.

6

Yeah, what the OP apparently confuses is that every bet has two sides–one that the event will happen and one that it won’t.

Having three players complicates the logic without clarifying the situation. If A thinks the event will happen at 90% probability A will bet any amount at even money that the event will happen; if C thinks the event will NOT happen with 80% probability C will bet any amount at even money that the event will NOT happen. A and C will happily bet each other.

B would bet against C at even money but would not bet against A unless he was given odds–which A would give, once C had committed all available funds at even money.

The OP might have had the idea that there was another party, D, that was taking the “NOT event” side of the bet. In that case, D would also have an estimate of the probability and would offer odds based on that probability; A, B, and C would take or not take the bet based on those odds.

7

Thanks for the replies so far. I admit that the OP was not well worded. Let me try again and see if that helps:

I’m trying to determine if a particular gambling scenario can be ‘optimized’ mathematically; by that I mean that all the players would agree that a change would result in a worse outcome for them personally. I don’t know if the problem is solvable, even with an iterative process.

The players have two motivations: to gain as much as possible if they win, but also to bet as little as possible while still inducing others to bet. The players publicly and honestly state the probabilities that they attach to the event; therefore everyone knows what the others’ minimum acceptable payouts are. So in the situation in the OP, player C knows that A will bet for a return of more than $1.11 and B will bet for a return of more than $1.66, etc. A and B will be betting against C in this case. I put in 3 players because, well, that’s the heart of the problem. A and B can independently make bets with C much more easily by accepting different odds; I’m trying to come up with a three-way solution in which the amounts that A and B make depend on the proportion that they contribute to the bet with C, not on different payout odds as would be the case in two independent bets with C.

Here’s a more concrete example of what I’m thinking: A, B, and C could each throw a dollar in the pot. If C loses, A and C would split $1 in winnings: A getting $0.50 meets his own win criteria, but B getting $0.50 doesn’t, so B wouldn’t agree to this. Similarly, if C won he would get $2.00, in which case A and B would complain that C had not bet enough, since he ‘would have’ accepted a bet from A and B combined of only $0.25 against C’s $1 bet; to be ‘fair’ C should have put up more like $8 against (A+B)'s $2. So the 1//1/$1 proposal is rejected.

So next they propose that A and B each put forward $1, and C puts forward $2. If A and B win, they both get $1 which meets their win criteria, and if C wins he gets $2 which meets his win criteria as well. But again, A and B argue here that they have had to bet too much, since C would have taken a combined bet from them of $0.50 for his $2. So both A and B’s victory conditions have been met, but their loss conditions haven’t. C can make the opposite claim: if he wins he meets his victory condition, but, he argues, a $2 bet is too much if he loses - A would have accepted $0.11 and B would have accepted $0.66, so why did he have to put up more than $0.77?

I can’t think of a way to balance these two motivations. I can imagine that there are some values that A, B, and C would agree to, where any change would increase winnings at the expense of increased possible loss or vice versa. I think Ximenean grasped the gist of what I’m trying to figure out. No one solution is optimal for all, but is there a least-bad compromise?

8

OK, maybe it is possible if we’re talking about equalising their respective expected returns. The actual returns will be unequal, since they can’t all be right about the probability.

Having three people complicates things, so let’s consider just two people, A and B. A wants to bet that the event will happen. B is going to bet that it doesn’t happen. That is, A estimates a probability p that is greater than 0.5, while B estimates a probability q that is lower than p.
They put some money into the pot. If the event occurs A takes the pot, otherwise B takes it.

Let’s say that for every dollar that A puts into the pot, B puts in some fraction b dollars. Therefore, if the event occurs A gains b dollars while B loses b dollars. If the event doesn’t occur, A loses $1 and B gains $1.

According to my algebra, their expected returns are equal when b = (2 - (p + q))/(p + q) .

Example:
A estimates the probability at 0.9 (90%), B estimates 0.4 . Therefore p + q = 1.3. For every $1 that A puts in the pot, B must put in (2 - 1.3) / 1.3 = ~0.54 dollars .

A’s expected profit: 0.9(~0.54) - (1-0.9)(1) = ~0.386
B’s expected profit: (1-0.4)(1) - (0.4)(~0.54) = ~0.384
If my reasoning and algebra is correct (always quite an assumption) I guess you could extend this to more than two players.

9

I’d ask the OP to first define the problem or solution clearly for the two-person case, which is, I think, already ambiguous. Extending to three persons will be difficult in any case. To see this, suppose A and C make the chances 80% and 20% respectively, and consider B estimates of 49%, 50%, 51%. Note that, with the right payoffs, B can bet either way.

Three-person games are much harder to solve than two-person games which in turn are harder than one-person games. I’ll describe a simple “one-person gambling game” which has a particularly elegant solution.

Suppose there are N possibilities to win a horse race (or election) and a bookie (or Intrade) offers odds on each. For simplicity, assume there is no vigorish; i.e. that the implied odds (p[sub]1[/sub], p[sub]2[/sub], p[sub]3[/sub], …) sum to 1. Suppose that you estimate the odds differently (i.e., as some q[sub]1[/sub], q[sub]2[/sub], q[sub]3[/sub], …, again summing to 1). Given a $1000 bankroll, how should you bet to maximize the logarithm of your post-election bankroll?

(Maximizing not the bankroll but its logarithm leads to the solution’s elegance, but isn’t arbitrary: this criterion has been correctly applied in gambling problems for centuries.)

10

I’ve been thinking about this some more: I reckon that the following formula is fair:

A, B, and C estimate the probabilities at a, b, and c respectively. C’s is the lowest estimate. A and B are betting that the event will happen, C is betting that it won’t. Every round they should put the following amounts into the pot:

A: a
B: b
C: (6 - a - b - 4c) / 3

If the event occurs, A and B split the pot equally
If the event doesn’t occur, C receives 2.0 and the remainder is split equally between A and B.
Concrete example with a = 0.9, b = 0.6, c = 0.2

A puts in 0.90, B puts in 0.60, C puts in 1.23. Total pot 2.73.

If the event occurs, A and B get back 1.37 each, C gets nothing.
If the event doesn’t occur, A and B get back 0.37 each, C gets back 2.0 .

A’s expected return per round: 0.9 * 1.37 + 0.1 * 0.37 = 1.27
B’s expected return per round: 0.6 * 1.37 + 0.4 * 0.37 = 0.97
C’s expected return per round: 0.2 * 0.00 + 0.8 * 2.00 = 1.60

That is, everybody expects to make 0.37 profit per round.
More generally, with n players who estimate the probability at p[sub]1[/sub], p[sub]2[/sub], …, p[sub]n[/sub], of which p[sub]n[/sub] is the lowest estimate,

player 1 puts in p[sub]1[/sub]
player 2 puts in p[sub]2[/sub]
…
player n puts in ( n(n-1) - sum(p[sub]1[/sub] to p[sub]n-1[/sub]) - p[sub]n/sub[sup]2[/sup] ) / n

If the event occurs, players 1 to (n-1) split the pot equally. If the event doesn’t occur, player n takes (n-1) and the remainder is split equally between the other players.
Since player n’s estimate of probability is lower than everybody else’s, he is happy to take their bets at what he believes to be too-short odds. He would expect to make a profit. This formula effectively shares out his expected profit equally among all of them.

It works even if the lowest estimate of probability is greater than 0.5. There is a positive expected profit as long as they don’t all have the same estimate. If they do, the expected profit is zero.