There was 20 seconds left, Vanderbilt had just scored a layup to go ahead by 1 and Northwestern’s Bryant Mcintosh was racing to midcourt to set-up a final chance to regain the lead and win the game.  Vanderbilt’s Matthew Fisher-Davis intentionally fouled him, sending McIntosh to the line and the commentators and all of social media into a state of bewilderment.  Yes, we understand intentionally fouling when you are down 1 with 20 seconds to go, but when you are ahead by 1?

But it was a brilliant move and it failed only because the worst-case scenario (for Vanderbilt) realized:  McIntosh made two clutch free throws and Vanderbilt did not score on the ensuing possession.

(Before we get into the analysis, a simple way to understand the logic of the play is to notice that intentionally fouling late in the game very often is the right strategic move when you are down by a few points and there is no reason that should change precipitously when the point differential goes from slightly negative to slightly positive.The tactic is based on a  tradeoff between giving away (random) points and getting (for sure) possession.  The factors in that tradeoff are continuous as a function of the current scoring margin.)

Let $p$ be the probability that a team scores (at least two points) on a possession.  Let $q$ be the probability that Bryant McIntosh makes a free throw.  Roughly, the probability that Vanderbilt wins if they do not foul is $1-p$ because Northwestern is going to play for the final shot and win if they make a field goal.

What is the probability that Vanderbilt wins when Fisher-Davis fouls? There are multiple, mutually-exclusive ways they could win.  First, McIntosh might miss both free-throws.  This happens with probability $(1-q)^2$.  The other simple case is McIntosh makes both free-throws, a probability $q^2$ event, in which case Vanderbilt wins by scoring on the following possession, which they do with probability $p$. Thus, the total probability Vanderbilt wins in this second case is $q^2p$.

The third possibility is McIntosh makes one free-throw.  This has probability $2q(1-q)$. (I am pretty sure McIntosh was shooting two, i.e. Northwestern was in the double bonus, but if it was a one-and-one this would make Fisher-Davis’ case even stronger.)  Now there are two sub-cases.  First, Vanderbilt could score on the ensuing and win.  Second, even if they don’t score, it will be tied and the game will be sent into overtime. Let’s say Vanderbilt wins with probability $1/2$ in overtime, a conservative number since Vanderbilt had all the momentum at that stage of the game.

Then the total probability of a Vanderbilt win in this third case is $2q(1-q)\left[ p + \frac{1-p}{2}\right]$.  Adding up all of these probabilities, Vanderbilt wins using the Fisher-Davis foul with probability $(1-q)^2 + 2q(1-q)\left[ p + \frac{1-p}{2}\right] + q^2p$

Fisher-Davis made the right move provided the above expression exceeds $1-p$.  Let’s start by noticing some basic properties.  First, if $p = 1$ then fouling is always the right move, no matter what $q$ is.  (If Northwestern is going to score for sure, you want to foul and get possession so that you can score for sure and win.)  If $q = 0$ then again fouling is the right strategy, regardless of $p$.  (If he’s going to miss his free-throws then send him to the line.)

Next, notice that the probability Vanderbilt wins when Fisher-Davis fouls is monotonically increasing in $p$. Since the probability $(1-p)$ Vanderbilt wins without fouling is decreasing in $p$, the larger it is the better the Fisher-Davis gambit looks.

Finally, even if $q = 1$, so that McIntosh is surely going to sink two free-throws, Fisher-Davis made the right move as long as $p > 1/2$.

Ok so what are the actual values of $p$ and $q$.  McIntosh is an 85% free-throw shooter so $q = .85$.  Its harder to estimate $p$ but here are some guidelines.  First, both teams were scoring (at least two points) on just about every possession down the stretch of that game.  An estimate based on the last 3 minutes of data would put $p$ at at least $.7$, in any case certainly larger than $1/2$.

More generally, I googled a bit and found something basketball stat guys call offensive efficiency.  It’s an estimate of the number of points scored per possession.  Northwestern and Vanderbilt have very similar numbers here, about 1.03.  A crude way to translate that into the number we are interested, namely probability of at least 2 points in a possession, is to simply divide that number in half, again giving $p > 1/2$.  (This would be exactly right if you could only ever score 2 points.  But of course there are three-point possessions and one-point possessions.)  A third way is to notice that Northwestern was shooting a 49% field goal percentage for the game.  This doesn’t equal field goals per possession of course because some possessions lead to turnovers hence no field goal attempt, and on the other side some possessions lead to multiple field goal attempts due to offensive rebounds.

So as far as I know there isn’t one convincing measure of $p$ but its pretty reasonable to put it above $p = 0.5$ at that phase of the game.  This would be enough to justify Fisher-Davis even if McIntosh was certain to make both free throws.  (I used Wolfram Alpha to figure out what $p$ would be required given the precise value $q = .85$ and it is about .45).

Finally, even if $p$ is below $.45$ say around $.4$ it means that the foul lowered Vanderbilt’s win probability but not bvery much at all. Probably less than every single time in the game that someone missed a shot.  Certainly less than a few seconds later when LaChance missed the winning shot on the final possession.  Its interesting how in close games the specific things we focus our attention on when in fact pretty much every single play in the game turned out to be pivotal.