# Intuitive Odds; or: The Expectation of Fairness

When you make a bet at 9:1 odds, you stand to make a profit of $9 for every $1 that you risk. But odds are a measure of *probability* - so why is the payout so closely related?

This post will hopefully give you a solid intuition of why things work the way they do. We’ll see that a simple requirement of “fairness” let us arrive at this payout scheme with just a couple lines of algebra.

This post is targeted at those not familiar with probability. If you know how odds relate to probability, and what it means to take the expectation of a random variable, you can just skip straight to the summary. If not, read on!

First, why does the payout vary at all? In the next section, we’ll examine a simple scenario showing why this is important.

## The Even Wager

Alice and Bob are betting on a coin flip. It’s a fair coin: the probability it lands “Heads” is \(\frac{1}{2}\). Alice is betting the coin will hand Heads up, Bob is betting on Tails.

They’ve agreed to both wager $1, and the winner will take all of the money. That is, the winner receives a total of $2, for a profit of $1. The loser will receive nothing - a loss of $1 (or a “profit” of \(-1\)).

This seems fair - but what happens if we have a biased coin? For example, we’ll say the coin has a \(\frac{9}{10}\) chance of landing Heads.

If the payout remains the same, then Alice has a huge advantage! Consider that Bob will win only \(\frac{1}{10}\) of the time, but he’ll win the same amount as Alice - who’ll win \(\frac{9}{10}\) of the time!

The problem is that Bob made a much more risky bet, but his reward didn’t change despite taking on a bigger risk. If they repeat the bet many times, it’s clear that Alice will have a far larger profit: On average, Alice will win $0.9 for each game they play, and Bob will only win $0.1.

We want the payouts to be *fair*: if the bet is repeated a number of times, then Alice and Bob should profit about the same.

But how exactly should this work? The next section is a quick review of what odds are, so we can relate them to probabilities.

## Never tell me the odds

Let’s say the odds of getting a telephone call in the next hour are 2:5. This means that 2 times out of 7, we’ll get a call, and 5 times out of 7, we won’t.

We can think of these odds as areas - on the left is the area of size 5 representing when we don’t get a call, and on the right is the area of size 2 representing when we *do* get a call.

```
┌───────────────────┬───────┐
│ No Call = 5 │Call=2 │
└───────────────────┴───────┘
```

The total area is 7 - so the probability of getting a call is the fraction of the total area taken by the “Call” region - that is \(\frac{2}{7}\).

That’s all there is to it. Odds are just a simple representation of the probability of an event with two possible outcomes.

## The Expectation of Fairness

Previously, we decided that in the interests of fairness, if Alice and Bob repeat the bet a large number of times, they should end up with roughly equal profits.

To make that a precise mathematical statement, we say that the *expectations* of profits should be equal. If \(R_a\) is Alice’s profit, and \(R_b\) is Bob’s profit, we simply say:

\[ E[R_a] = E[R_b] \]

Intuitively, you can think of \(E[R_a]\) as the average profit made by Alice if the bet were to be repeated a very large number of times.

Note that \(R_a\) and \(R_b\) are *random* quantities. They depend on the outcome of the biased coin flip. Let’s call the amount wagered by Alice \(a\), and the amount wagered by Bob \(b\). Then, Alice will either gain Bob’s wager (for a profit of \(b\)) or lose her own wager (for a “profit” of \(-a\)).

We summarise the outcomes in the table below:

Outcome | Winner | \(R_a\) | \(R_b\) | Probability |
---|---|---|---|---|

Heads | Alice | \(b\) | \(-b\) | \(p = \frac{9}{10}\) |

Tails | Bob | \(-a\) | \(a\) | \(1 - p = \frac{1}{10}\) |

We can calculate \(E[R_a]\) by multiplying the profit of each outcome by the probability of it happening:

\[E[R_a] = p b + (1 - p) (-a)\]

\[E[R_b] = (1 - p) a + p (-b)\]

Note that Bob wins when alice loses - and the probability of Bob winning is \(1 - p\). Bob wins when the coin is Tails, and \(p\) represents the probability of Heads, so \(1 - p\) is the probability of Bob winning.

So now we can rewrite the equation \(E[R_a] = E[R_b]\) to give us the following:

\[ pb - (1 - p) a = (1 - p) a - pb \]

You can read the left side as

\[ \text{Bob's wager} \times \text{P(Alice Wins)} - \text{Alice's wager} \times \text{P(Alice Loses)} \]

We want to figure out how the ratio of wagers - \(\frac{a}{b}\) - relates to the probability of the event. This lets us answer the question “How much more should Alice bet compared to Bob?”. To do this, we simply rearrange the above equation:

Moving \(p\) terms to the left and \(1 - p\) terms to the right,

\[ 2pb = 2(1 - p)a \]

Grouping \(a\) and \(b\) terms on the right, and \(p\) terms on the left,

\[ \frac{p}{1 - p} = \frac{a}{b} \]

Now we’ve written \(a\) and \(b\) in terms of the probability - or in other words, we’ve expressed how the wagered amounts should relate to the probability of the outcome.

Let’s answer the “biased coin” question: For every $1 that Bob bets, how much should Alice have to bet?

Given the agreed odds of 9:1, we know that \(p = P(\text{Alice wins}) = \frac{9}{10}\). We can fill in our equation above to give us:

\[ \frac{0.9}{0.1} = \frac{a}{1} \]

And therefore,

\[ 9 = a \]

Alice should have to bet $9 for each $1 bet by Bob - because she is nine times more likely to win.

## Summary

In summary, we make the simple assumption that each bettor should profit the same on average *assuming the odds are correct*.

We state this mathematically as

\[E[R_a] = E[R_b]\]

Which expands to

\[ pb - (1 - p) a = (1 - p) a - pb \]

Simplifying to

\[ \frac{p}{1 - p} = \frac{a}{b} \]

Which can be interpreted to mean

If Alice is \(n\) times more likely to win, then she must bet \(n\) times more money than Bob.

## Final Note: Agreed-upon odds in the real world

Of course, we have no way of knowing the true odds of a real world event like a horse race. In this case, the agreed-upon odds of a bet may not be the true odds of the outcome. In fact, the odds may not even reflect the true beliefs of each bettor!

For example, I may have inside information that my horse is *definitely* going to win, but I’m happy to agree on even odds, because I’m guaranteed to make a profit.

The betting odds merely state a probability at which both parties are happy to make the bet, and the payout scheme says “as long as these odds are correct, the payouts will be fair.”

In fact, there is no point to take a bet at odds you believe are exactly accurate, because your expected profit is zero!

Paul —