# Who Was On The Winning Team?

My wife had a small competition at her work. Fifteen of her 19 colleagues participated, divided into three teams of five. When she came home, she smiled and asked me, “Guess who was on the winning team?”

Given that my wife asked me the question, it’s highly likely she was on the
winning team. But I like to practise thinking in terms of the base rate and not
just the conditional probability^{1} See Meehl and the research inspired by his
for why this is a good thing to do. so I answered “Peter”. Why Peter? I happen
to know there are four (!) people named Peter at her job.

But what was the probability of Peter being on the winning team? At first, this sounded like a tricky question to answer, for at least two reasons:

- Not all of her colleagues participated. Maybe all Peters did, but maybe none of the Peters did, or, most likely, something in between.
- It matters how the Peters are distributed among the teams. If all the people named Peter are in one team, then Peter has a much lower chance of being on the winning team than if each team has at least one Peter in it (in which case there’s a 100 % chance that a Peter was on the winning team.)

If we know how to compute with conditional probability, these complications
don’t actually make the problem harder, only more annoying. This is because we
can compute the probability of the Peters being distributed various
ways^{2} What is the probability that no Peter competed? What is the probability
that exactly one Peter competed and was on the winning team, etc. and then
combine the results.

However, there’s an easier way to do it by thinking combinatorically^{3} And
this is the reason I wrote this up in the first place: I didn’t expect to be
able to use the enumerative combinatorics I learned a few months ago, but thanks
to spaced repetition it was still fairly accessible in my brain!: There are
\({19 \choose 5}\) ways to select the members of the winning team from all 19
colleagues. There are \({15 \choose 5}\) ways to select the winning team without
picking any Peter^{4} If we restrict the valid candidate pool to only those
whose names are not Peter, there are 15 colleagues to choose from.. Thus, the
probability that Peter is in the winning team is

\[1 - {15 \choose 5} \bigg/ {19 \choose 5} = 74 \%.\]

It was a fairly good guess!

# Motivation: Picking the Winning Team First

We seemingly ignored a bunch of information in that calculation. In fact, the only things we did pay attention to were

- Number of colleagues
- Number of colleagues not named Peter
- Team size

What reason do we have to ignore everything else? It rests on a few assumptions.
The most important of which is that the winning team is – conditional on the
knowledge we have about the situation, which is very little – independent on who
is actually on the team. That independence means we can pretend that a winning
team was picked first, and only then was participants randomly assigned to the
team.^{5} This sort of feels like we’re reasoning with time flowing
backwards, which is not a bad intuition. Stochastic independence between events
means their relative location in time becomes meaningless, and in fact, time
itself ceases to be a meaningful concept. At some point I want to write about
how time and causality are a phenomena of logical/stochastic dependence but this
is not that day.

Since we don’t have any reason to suppose that the Peters (or non-Peters) as a
group were particularly likely to not be present at the competition at all, we
can also reason as if every colleague has equal probability to end up in any
team – including the winning team.^{6} This sort of thinking applies to many
competitive situations, which unfortunately has the side effect of making those
competitions rather uninteresting.

# Conditional Probability of Peter winning

When I had guessed, my wife rolled her eyes and said “I was.”^{7} She may
also have made a comment about me being boring. Now we’re in a position with
more information: how confident should we be now that Peter is on the winning
team? The reasoning is very similar, except now there are only four places left
to fill on the winning team, since we know the identity of one of the people in
it, and we have one fewer colleague to put there.

\[1 - {14 \choose 4} \bigg/ {18 \choose 4} = 67 \%.\]

This is still a good chance, but it turns out Peter was not on the winning team. I should probably have guessed “You!” after all.