# Are you 100% sure? Betting for the same team.

Betting is often a good way of settling arguments: “I bet you so much that my sports team will beat yours”, can quickly lead to a resolution. But some of the most heated arguments are between people who completely agree with each other. So what happens when you want to gamble with someone else, but both of you want to bet on the same outcome?

You could bet on a particular score, or the timing of an event. But how about if the options are limited: is the correct direction to turn left or right? Or if you have identical views: the game will end in a nil-nil draw, or this party will win the election?

Here, the bet can go ahead as long as one of you is more sure than the other that they’re correct. This isn’t that strange in the world of commerce. Here an analogy would be that two memorabilia traders each have a copy of a certain piece of memorabilia, perhaps a Ram Man figurine from He-Man. One trader thinks it’s worth £10, but the other thinks it’s only worth £5. As long as they have no emotional attachments to their figurines, it makes economic sense for the second trader to sell his Ram Man figure to the first at some price between £5 and £10. If they agree on £7.50, then each will believe they are making a profit of £2.50. At least one of them will be ‘wrong’, in some way, however: the first trader might know of a naive collector or have some extra information or skill that the second doesn’t, or one might simply be deluded.

Let’s consider two gamblers, Blaise and Pierre, who both think that the Red team will beat the Blue team, but the first is 95% confident, and the other is only 85% confident. Then the bet can proceed. Here’s one possible way to go about it: on reflection I was probably inspired by the bidding system in the dice game Perudo.

Blaise starts by offering even odds: if Red wins Pierre gives him £1, otherwise he’ll pay £1 to Pierre. If there’s more than two possible outcomes, say $n$, Blaire might start at £ $n-1$ to £1, or any other agreed value.

Pierre now has two choices:

1. Pierre could make a counter-offer: if Red wins Blaise has to give him £1, otherwise he’ll pay £2 to Blaise. Or £1 for a win against £3 for a loss, or any higher amount for the outcome neither thinks will happen.
2. Pierre could keep the bet the same, but raise the stakes: he can double, triple, quadruple, etc. the bet. If the Red team wins Blaise must give him £2, and he’ll pay £2 otherwise.
3. Or Pierre may accept the bet.

Unless Pierre chose to accept, then it’s Blaise’s turn again and he can make the same two choices, and so it continues until someone gives in.

We need one further rule: a player may not use rule 2 twice on the same bet. That is, if Pierre multiplies the stakes by some amount, and Blaise then does the same, then Pierre must now change the bet using rule 1.

Let’s have an example:

• Blaise offers £1 to £1 against Red winning.
• Pierre offers £1 to £2 against Red winning. This is how the terminology works: Pierre thinks Red is more likely to win. Alternatively you could say “2 to 1 for Red winning”, but I find it easier to remember that “a million to one against” is the odds betting shops would offer for extremly unlikely events (which perversely frequently happen in fiction). For instance, Paddy Power offers 500 to 1 odds against the first word that the Pope says in his 2011 Christmas Speech being ‘Britney’.
• Blaise is very confident and offers £1 to £5 against.
• Pierre doubles to offer £2 to £10 against (betting sites would regard this as two units of 1 to 5).
• Blaise also raises to offer £3 to £15 against (three units of 1 to 5: even though he’s technically multiplying by 1.5, it does stay within the integers, so we’ll allow it).
• Pierre cannot raise the stakes again, and now has to increase the odds and offers £3 to £16.
• Blaise offers £3 to £18.
• Pierre accepts these odds.

We can actually try to extract some information from this gambling auction:

• Blaise is more certain than 50% ($\frac{1}{1+1}=1/2=50 \%$).
• Pierre is more confident than 66% ($\frac{2}{1+2}=2/3 \approx 66.67 \%$).
• Blaise thinks Red is more than 83.33% likely to win.
• Pierre may not be 85.71% confident of Red winning as he did not raise to 1 to 6, instead doubling the stakes.
• Blaise does not raise either.
• Pierre is at least 84.21% confident.
• Blaise is at least 85.71% confident ($\frac{18}{3+18}=6/7 \approx 85.71 \%$).
• Pierre accepts, so, unless he was bluffing somewhere, is between 84.21% and 85.71% certain that the Red team will win, as I perhaps rather unrealistically stated above.

So both Blaise and Pierre will expect to win some money, if they could keep making this bet in the long run.

Blaise, being 95% confident, if he made this same bet many times, would expect, on average to win $0.95 \times \pounds 3 - 0.05 \times \pounds 18 = \pounds 1.95$ each time, if his judgement is correct.

Pierre, being only 85% confident gets a much worse deal. He expects to win $- 0.85 \times \pounds 3 + 0.15 \times \pounds 18 = \pounds 0.15$ each time.

To put this more generally, Blaise should only offer a bet of $\pounds x$ to $\pounds y$ against some outcome, if $px - (1-p)y \geq 0$ and Pierre should only accept if $-qx + (1-q)y \geq 0$, where $p$ is how sure Blaise is (ie. his percentage surety over 100, 0.95 in the example above), and $q$ is how sure Pascal is, where $p>q$. For both to be true, rearranging, we see that:

$\frac{p}{1-p} \geq \frac{y}{x} \geq \frac{q}{1-q}$.

So if someone ever claims to be 100% sure about whether they locked the door, or whether Claudia Wells was in Back to the Future Part II, make them put their money where their mouth is and offer them a £1 to £100 bet against it, to see whether they’ll take you up on it. If they refuse, perhaps they were only 99% certain, after all.