31 July 2025
When two people have had three beers, and one comes up with something the other does not want to do, they play What are the odds. The game goes as follows:
This is a fairly simple and fun game. It serves two functions. First and most notably, you can use it to make someone do something they do not want to. On the other hand, you can also use it to do something you want to do, without anyone knowing that you actually do (e.g. "Oh no, do I really need to have spoonful of Lao Gan Ma crispy chili sauce? Seriously? Oh no...").
Despite its simplicity, I have always had one doubt regarding the rules: during step 3 (and 5), are the players allowed to say 0?
Before going any further, lets clarify what odds are. The odds of an event $A$ are
$$odds~of~A = \frac{P(A)}{1-P(A)}.$$
Or in other words, how many times more likely is $A$ happening than $A$ not happening?
The offerer asks the other what are the odds that he does $X$. Thus, the rules of the game should ensure that the odds of him doing $X$ are indeed the number he gives (i.e. $N$). Right?
Well, not really. There is a small issue with What are the odds. Namely, we ask for the odds that the other does $X$, but we actually want to know the odds that the other does not do $X$. This is apparent from the observation that the bigger $N$ is, the less likely the other is of doing $X$, but the higher the alleged odds of the other doing $X$. This is absurd, so we must be talking about some other odds. The only other candidate are the odds of not doing $X$.
Thus, the rules should ensure that the odds of the other not doing $X$ are equal to $N$. To achieve this, should we include or exclude 0 in step 3?
Lets call $A$ the event that the other does $X$. If we exclude 0 there are $N$ outcomes that satisfy $A$ (i.e. both say 1, both say 2..., both say $N$), and there are $N^2$ possible outcomes. Thus there are $N^2 - N$ that satisfy $\neg A$, and so:
$$P(\neg A) = \frac{N^2-N}{N^2} = \frac{N-1}{N}.$$
Knowing $P(\neg A)$ we can plug it in the odds formula to get that the odds of $\neg A = N -1$. Not what we wanted.
Including 0 there are $N+1$ outcomes that satisfy $A$ (i.e. same as before plus the possibility of a matching 0) and $(N+1)^2$ possible outcomes. Thus there are $(N+1)^2-(N+1)$ that satisfy $\neg A$, and so:
$$P(\neg A) = \frac{(N+1)^2-(N+1)}{(N+1)^2} = \frac{1}{N+1}.$$
Plugging it in the odds formula we get that the odds of $\neg A = N$. Which is what we wanted 🎉.
Although the discussion is less straightforward than one may anticipate, we do have an answer. Things are not made easier by What are the odds being actually about the odds of not doing $X$. Nevertheless, we can happily conclude that the players are allowed to say 0. And if someone tells you otherwise, link them to this post.
It has come to my attention that What are the odds is often played differently. Namely, in step 2 the other says "1 in $N$", where $N$ is a number larger or equal to 1. Here, odds do not mean what I described above. In the context of betting and chance games, this word may have different definitions, one of them having this "$X$ in $Y$" form. Simply enough, it means that event $A$ will tend to happen $X$ times out of $Y$, in other words $P(A) = \frac{X}{Y}$.
This version of the game has a clear advantage: the answer given in step 2 does indeed correspond to the odds of him doing $X$. In this sense, it seems clear that this is the original and correct version of the game. The one I described above likely arises from efficiency in communication: since all answers in step 2 contain "1 in...", we can assume it and simply drop it. Unadvertedly though, this changed the rules of the game! So much so that, under this definition the players are not allowed to say 0. And, in true math textbook fashion, I leave the proof to the reader.