Suppose 3 friends want to bet \$100 on whether candidate John Doe will win the next election. They state their perceived likelihood that the event will occur:

  • Alice believes John Doe will win with probability $35\%$
  • Bob believes John Doe will win with probability $50\%$
  • Charlie believes John Doe will win with probability $40\%$

How would you go about setting up this bet?

In other words, what amount should each friend put down (all three amounts totalling \$100) and what should the payoffs be if John Doe is elected, and what should the payoffs be if John Doe loses?

(Asking for a friend)


2 Answers 2


This is an interesting problem. I thought about it for some time and I believe I have a complete solution.

First, let's see how we would answer the question if we had only 2 people. Say, Alice believes that an event will happen with probability $90\%$, while Bob believes that it will happen with probability $80\%$. How do you make a bet out of this event? It is not obvious.

Intuitively we can think that the two people must take opposite positions, otherwise they will both be winners or losers at the same time. Moreover, the one with more confidence in the event (Alice) should take the for position and the other person should take the against position. We will prove this later, but there is also another major question: on what odds should they bet? Let's explore the problem:

Assume that Alice bets $\$1$ on the for position (i.e. the event actually happening) How much should she make back to consider this a favourable bet for her? Let's name the return (i.e. the amount won) as $x$. Just to be clear, $x$ is the amount returned above the $\$1$ she's already put in. In other words, if Alice is betting $\$1$ and wins, she is getting back $\$1+\$x$. To find $x$ of a favourable bet, we just make the expected value of the bet positive: $$0.9\cdot x - 0.1\cdot 1 > 0 \iff x > \frac{1}{9}$$

So if Alice bets $\$1$, she would like her opponent to bet at least $\$1/9$. If we write this in the typical form of odds as opponent_bet:your_bet then Alice is looking for a $1/9:1 \iff 1:9$ bet or better.

What if Alice wants to bet on the against position (i.e., the event not happening)? A favourable bet of $\$1$ would be one with return $x$ where $$0.1\cdot x - 0.9\cdot 1 > 0 \iff x > 9$$

The odds now are $9:1$. It is the reciprocal as we expected.

Doing the analysis for Bob we find that

  • Bob betting on the for position wants odds $1:4$ or better
  • Bob betting on the against position wants odds $4:1$ or better

How would we create a fair bet between the two? Let's make a table with all the possible position combinations and conditions for each combination

$$ \begin{array}{l|c|c} & \text{Alice for} & \text{Alice against} \\ \hline \text{Bob for} & \text{not a bet} & \begin{array}{c} & \text{Alice wants}\space x:1, \space \color{red}{x>9}\\ & \text{Bob wants}\space 1:x, \space \color{red}{x<4} \\ \end{array}\\ \hline \text{Bob against} & \begin{array}{c} & \text{Alice wants}\space 1:x, \space \color{green}{x<9}\\ & \text{Bob wants}\space x:1, \space \color{green}{x>4} \\ \end{array} & \text{not a bet} \\ \end{array} $$

As we see from the table, the only possible bet is when Bob bets against and Alice bets for. This is inline with our intuition that the person most confident should take the for position. We also know that an $x$ such that $4<x<9$ is acceptable for both Alice and Bob (in the sense that they both expect a positive gain). What is the most fair value however? Should we just take the middle value ($x=6.5$)? This might seem reasonable, but remember that odds are "multiplicative" entities, and taking "additive" averages might not work. Indeed the most fair selection of $x$ is the one that yields the same expected values for both parties: $$0.2\cdot x -0.8 \cdot 1 = 0.9 \cdot 1 - 0.1 \cdot x\iff 0.3 \cdot x = 1.7 \iff x = \frac{17}{3} \approx 5.666$$ So Alice should make the for bet with odds $3:17$ and Bob make (the opposite) bet with odds $17:3$. And if we wanted for both of them to bet $\$100$ in total, the bets would be $100\cdot 17/(17+3) = \$85$ for Alice and $100\cdot 3/(17+3) = \$15$ for Bob. Notice how Alice's bet is the average of their perceived likelihoods.

So if we know the perceived likelihoods for each of the two players, we can construct a fair bet. It's interesting to note that we can always construct a fair bet for two people, and that the closer the likelihoods are to each other, the smaller the gain both players expect. So if both Alice and Bob believe that the event happens with probability $80\%$, then the only fair bet is $1:4 \space\space(4:1)$ and the expected gain for both of them is $0$. If, on the other hand, Alice believes the event will happen $90\%$ and Bob believes it will happen $10\%$ of the time, then the fair bet is $1:1$ and the expected gain of both players is $\$0.8$ for each $\$1$ played. Just to be clear, this is the expected gain based on a player's perceived likelihood, it does not have to agree with the actual expected gain (in other words: people can have mistaken beliefs)

The case for 3 players

Let's examine what happens with 3 players. Let's take the particular example given in the question.

  • Alice believes the event happens with probability $35\%$
  • Bob believes the event happens with probability $50\%$
  • Charlie believes the event happens with probability $40\%$

This gives us the following favourable odds for each person:

  • Alice wants to bet for with $13:7$ (or better), or against with $7:13$ (or better)
  • Bob wants to bet either for or against with $1:1$ (or better)
  • Charlie wants to bet for with $3:2$ (or better), or against with $2:3$ (or better)

Let's build the betting table again. With $3$ people we have $2^3=8$ combinations. Similarly to the 2-people case, combinations where everyone bets for or everyone bets against are not valid bets. So we are left with 6 betting combinations. What should the conditions be for each combination? Things are more complicated now that we have 3 people. First let's revisit the conditions for the 2-person case. In the analysis above, I stated without much explanation that conditions for the 2-person case look like this: $x:1$ for Alice and $1:x$ for Bob. It seems intuitive, but it would be helpful to derive it so we can understand the restrictions we are taking and the assumptions we are making. The initial conditions for the 2-person example should have been: $$ \begin{array}{rc} \text{Alice wants odds } \space x:a, & \frac{x}{a}>\frac{1}{9} \\ \text{Bob wants odds } \space y:b, & \frac{y}{b}>\frac{1}{4} \\ \end{array} $$ But now we have 4 variables instead of one! This is because these two conditions describe only what the favourable bets are for each player. So if Alice and Bob were betting any amount against a bookmaker, these are the conditions they would use. But if we want to make them bet against each other then we need to impose more restrictions. We need to restrict these quantities so that whatever one player bets is the (potential) winning of the other player. This means that $a=y$ and $b=x$. With these restrictions we are down to two variables. How do we eliminate another variable? Since odds are essentially ratios, we can simply normalise one of them by setting $a=1$. This is what we have implicitly done when analysing the 2-person case above. Another (perhaps better) option is to treat the $x,y,a,b$ variables as the actual quantities being betted and won. In this case, if we know the total amount we want our players to bet (say $\$100$) we simply set $a+b=100 \iff a+x = 100$ which together with the equation of making the expected gains equal, yields a $2\times 2$ system. Its solution for the 2-person example we gave earlier results in $a=85$ and $x=15$ as we have found before.

How does all these translate to the 3-person case? Let's take one particular combination where Alice bets against, while Bob and Charlie bet for. Also assume, as the question states, that all together are betting $\$100$. Here are the initial three conditions for favourable bets, along with the three extra restrictions that make this a valid bet among Alice, Bob, and Charlie : $$ \begin{array}{rl} \text{Alice wants odds } \space x:a, & \frac{x}{a}>\frac{7}{13} \\ \text{Bob wants odds } \space y:b, & \frac{y}{b}>\frac{1}{1} \\ \text{Charlie wants odds } \space z:c, & \frac{z}{c}>\frac{3}{2} \\ a &= y+z\\ x &= b +c\\ a+b+c &= 100 \end{array} $$

If we make the expected gains to be equal to each other in order to ensure the fairness criterion, we will get $2$ equations which along with the $3$ restriction make an underdetermined system with $5$ equations and $6$ variables. We could bring another restriction to force the system to have a unique solution (unique for a given betting combination out of the possible $6$), but there is no strong restriction I can think of. There are desirable properties such as maximising the product $a\cdot b\cdot c$ which means that we are trying to make the 3 bets as equal as possible (in other words, make it unlikely to have one person betting $\$98$ and the other two betting $\$1$). Or we can maximise the (equalised) expected gain, but note that this does not mean that everyone will get their best odds. In any case, I do not see these as necessary restrictions. We can have one or not. If not, we will have a infinite number of solutions (under the remaining restrictions) for each betting combination. The solutions ensure that everyone has a perceived positive gain, and that all the perceived gains are equal, so we have a fair solution.

Here are the 5 equations and the parametrised solutions for the case where Alice bets against, while Bob and Charlie bet for, and all together are betting $\$100$:

$$ \left. \begin{array}{rl} 0.65\cdot x - 0.35\cdot a &= 0.5\cdot y - 0.5\cdot b \\ 0.5\cdot y - 0.5\cdot b &= 0.4\cdot z - 0.6\cdot c \\ a &= y+z\\ x &= b +c\\ a+b+c &= 100 \end{array} \right\} \iff \begin{array}{rl} x &= 39.62 - 0.077 \cdot t\\ y &= 48.85 - 1.231 \cdot t\\ z &= 11.54 + 1.310 \cdot t\\ a &= 60.38 + 0.077 \cdot t\\ b &= 39.62 - 1.077 \cdot t\\ c &= t \end{array} $$ (Solution provided by this online tool)

Our parameter (free variable) is $t$, and we can notice that in this particular case it mostly affects $b,c,y,z$ and minimally affects $a,x$. By changing the parameter we are essentially changing the relative "weight" between Bob and Charlie as they both take on Alice. For example, try $t=19$ for a fairly balanced "weight" between Bob and Charlie (where Bob and Charlie bet almost equal amounts). This results in the following bet:

  • Alice bets $\$61.85$ against. If she wins she'll earn $\$38.15$ (so she'll receive back $\$38.15 + \$61.85 = \$100$)
  • Bob bets $\$19.15$ for. If he wins he'll earn $\$25.46$ (so he'll receive back $\$25.46 + \$19.15 = \$44.61$)
  • Charlie bets $\$19.00$ for. If he wins he'll earn $\$36.39$ (so he'll receive back $\$36.39 + \$19.00 = \$55.39$)

Notice how the sum of players' bets is $\$100$, and what winners get back also totals $\$100$ (as it should).

We can find similar solutions (families of solutions) for the five remaining betting combinations. It turns out that with this example (these likelihood beliefs) all six combinations can yield fair solutions, albeit some combinations have a smaller range of solutions than others. You can explore the other families of solutions yourself, doing the same analysis and solving equivalent $5\times6$ systems.

I believe this is a quite interesting result and it fully solves this betting problem.


I include here my earlier work on the 3-people case in order to show some of the progress of my ideas. I created a table with all the betting combinations and tried to provide the conditions for each combination. I arbitrarily restricted 3 of the variables to take specific values (but still abiding by all restrictions), and this resulted in a $3\times3$ equation system. For example, for the case we have been analysing above I took $x = 7$, $b = 3$, $c=4$, satisfying the condition $x=b+c$. I do not assume that $a+b+c=100$. This simplified some expressions and I could check the conditions of all the combinations manually. For three of the combinations there are no solutions with this arbitrary choice that I made (impossible conditions are written in red at the table below). Here's the table (note: I changed the original variable names used, so that they agreed with the names of the rest of the analysis above.)
$$ \begin{array}{l|c|c} & \text{Alice for} & \text{Alice against} \\ \hline \text{Bob for, Charlie for} & \text{not a bet} & \begin{array}{c} & \text{Alice wants}\space 7:a, \space \color{green}{a<13}\\ & \text{Bob wants}\space y:3, \space \color{green}{y>3} \\ & \text{Charlie wants}\space z:4, \space \color{green}{z>6} \\ & \color{green}{a = y + z} \end{array}\\ \hline \text{Bob for, Charlie against} & \begin{array}{c} & \text{Alice wants}\space x:7, \space \color{red}{x>13}\\ & \text{Bob wants}\space y:1, \space \color{red}{y>1} \\ & \text{Charlie wants}\space 8:c, \space \color{red}{c<12} \\ & \color{red}{x+y = c} \end{array} & \begin{array}{c} & \text{Alice wants}\space 7:a, \space \color{green}{a<13}\\ & \text{Bob wants}\space y:9, \space \color{green}{y>9} \\ & \text{Charlie wants}\space 2:c, \space \color{green}{c<3} \\ & \color{green}{a+c=y} \end{array}\\ \hline \text{Bob against, Charlie for} & \begin{array}{c} & \text{Alice wants}\space x:7, \space \color{red}{x>13}\\ & \text{Bob wants}\space 9:b, \space \color{red}{b<9} \\ & \text{Charlie wants}\space z:2, \space \color{red}{z>3} \\ & \color{red}{x+z=b} \end{array} & \begin{array}{c} & \text{Alice wants}\space 7:a, \space \color{green}{a<13}\\ & \text{Bob wants}\space 1:b, \space \color{green}{b<1} \\ & \text{Charlie wants}\space z:8, \space \color{green}{z>12} \\ & \color{green}{a+b=z} \end{array}\\ \hline \text{Bob against, Charlie against} & \begin{array}{c} & \text{Alice wants}\space x:7, \space \color{red}{x>13}\\ & \text{Bob wants}\space 3:b, \space \color{red}{b<3} \\ & \text{Charlie wants}\space 4:c, \space \color{red}{c<6} \\ & \color{red}{x=b+c} \end{array} & \text{not a bet} \\ \end{array} $$

Applying these arbitrary restrictions we can see that only three combinations (the ones with conditions in green) offer viable solutions. A minor side note: these restrictions were not so arbitrary. They were chosen so that we get integer values in the conditions, so we can check easily and quickly if the conditions hold.

If we take the first of these valid betting combinations: Alice against, Bob for, and Charlie for, and apply the rule of equal expected gains we get:

$$ \left. \begin{array}{rl} 0.65\cdot 7 - 0.35\cdot a &= 0.5\cdot y - 0.5\cdot 3\\ 0.5\cdot y - 0.5\cdot 4 &= 0.4 \cdot z - 0.6 \cdot 4\\ a = y + z \end{array} \right\} \iff \begin{array}{rl} a &= \frac{1179}{103} \approx 11.45\\ y &= \frac{421}{103} \approx 4.09\\ z &= \frac{758}{103} \approx 7.36 \end{array} $$

(Solution to the above $3\times3$ system provided by this online tool)

Which means Alice is betting against with odds $7:11.45$, Bob betting for with odds $4.09:3$,and Charlie betting for with odds $7.36:4$. How much should each person put in, if the total initial bet is $\$100$?

For Alice it should be $\frac{100 \times 11.45}{11.45+3+4} = \$62.06 $, for Bob $\frac{100\times 3}{11.45+3+4} = \$16.26 $, and for Charlie $\frac{100\times 4}{11.45+3+4} = \$21.68$.

This solution corresponds to our general/parametric solution where $t = 21.68$

  • $\begingroup$ Thank you for the awesome write-up, I will have to let it sink in for a moment. Our initial approach was to set it up as a series of 2-way bets, but we were suspicious of the results. If you wish to see the spreadsheet, we'd love to hear your comments $\endgroup$ Nov 9, 2016 at 16:54
  • $\begingroup$ @RafaelMagalhães thank you, I was happy that I finally got to the bottom of this problem! I might keep improving parts of my answer in the next couple of days. I looked at your spreadsheet. I see what you are doing, but I can't say I understand the reasoning behind it. Moreover, I am not sure how this information is supposed to setup a bet. How much money each person is supposed to put down? You have payout values for the two outcomes (event happening or not), but these have different ratios, so you cannot use them as a guide to find out what should each person pay upfront. $\endgroup$
    – Thanassis
    Nov 10, 2016 at 11:17
  • $\begingroup$ Wow, thank you (anonymous donor) for the bonus bounty award! $\endgroup$
    – Thanassis
    Sep 24, 2017 at 14:04
  • 1
    $\begingroup$ Stuck at the 2-player case, specifically that the expected value should be the same for both being equivalent to $0.2\cdot x -0.8 \cdot 1 = 0.9 \cdot 1 - 0.1 \cdot x$. As Alice is betting "for", isn't her expected value $0.9x - 0.1$? Edit: Ah nevermind, I just realized you changed the meaning of $x$ from the amount Alice would win, to its reciprocal (I think). Let me think more, maybe using a different name for the two senses, to keep things clear in my head :-) $\endgroup$ Apr 30, 2018 at 0:07
  • $\begingroup$ @ShreevatsaR Yes, you are right to be confused, I got confused myself looking at it now :) Indeed I changed the meaning of $x$ without me realising, and you are right that it is the reciprocal (just for Alice). What I did was looking at the bet odds 1:x and x:1 and interpreting them (correctly) as amount_you_earn:amount_you bet . If you see it like this, the formula equating the expected gains makes sense. A better way to write these formulas (without making assumptions) is what I present a little later, after I begin examining the 3-person case. $\endgroup$
    – Thanassis
    Apr 30, 2018 at 1:08

I don't know what exactly you mean by "setting up this bet", but assuming that each individual would bet only if she perceived the bet as fair, and that the bet amount was returned along with the prize, as is customary in UK, the odds should be

  • $ 65:35 = 13:7\; against\; for\; A\; to\; bet,$

  • $50:50 = 1:1 \; against\; for\; B\; to\; bet,$

  • $60:40 = 3:2\; against\; for\; C\; to\; bet.$

The above odds would ensure no loss no gain for them, so represent the worst odds respectively at which they should play, e.g. at the odds of $3:2$ against, $C$ would get $3+2$ dollars for every dollar bet, and just break even, and have a net gain only with better odds, e.g $\;3.5:2 = 7:4\;$ against.

  • $\begingroup$ Thank you, that seems to answer it. To make sure i get this right, would you care to include the payoffs for a $100 bet? (e.g. how much each friend wins/loses in that scenario) $\endgroup$ Nov 9, 2016 at 12:10
  • $\begingroup$ As I have remarked at the outset, idk what exactly you mean by "setting up this bet," so I have just given whatever could be gleaned from the figures. $\endgroup$ Nov 9, 2016 at 14:54
  • $\begingroup$ Yes, these are the odds that each one of them would agree to play, but this does not tell us how to construct a bet with all 3 of them. For example we cannot have all 3 of them betting against (someone has to take the opposite position, otherwise no money will exchange hands). Even if they take opposite positions, there could be a range of acceptable bets. Which is the most fair one? I believe I have found a solution and describe it in my answer. @RafaelMagalhães take a look and let me know if this is what you were after. $\endgroup$
    – Thanassis
    Nov 9, 2016 at 15:46
  • $\begingroup$ @trueblueanil Maybe I wasn't specific enough. My question is, if John Doe is elected, how much money does each friend wins/loses? I will update the question to reflect this, but do let me know if it isn't clear yet $\endgroup$ Nov 9, 2016 at 16:38

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