The Game Theory of HORSE

HORSE may seem like a simple backyard game, but, surprisingly, there is a strategy which improves your chance of winning — and it probably isn’t what you’d expect.

HORSE is a popular basketball game where players take turns attempting the same shot. With two players, the game works like this:

1. Player A attempts any basketball shot.
2. If they make it, then player B has to make the exact same shot. Otherwise, it’s player B’s possession.
3. If player B misses, they acquire a letter. Either way, Player A gets to shoot again.
4. Letters are acquired in order: H-O-R-S-E. Once a player acquires all five letters, they are a horse and lose the game because horses can’t play basketball.

It’s a simple game, but strategy isn’t well understood by most players. I wanted to dig into the metagame of HORSE and work out if there are optimal ways to play using game theory, strategy, and a little math. Of course, being better at shooting will always be advantageous, but let’s take a look at the decisions you can make.

The Setup

There’s only one element of strategy in the game — shot selection. You could take a free throw, a layup, or a left-handed, eyes-closed, 3-pointer skyhook. Conventional wisdom is to take a shot that you’re good at but that your opponent will probably miss. For instance, Steph Curry can just attempt three-pointers and watch his opponent miss.

For a given shot, there’s a probability that you make the shot and a probability that your opponent makes the shot. Given these values, we can calculate the expected value of the number of letters your opponent will get on your possession (or the average number of points on each possession if you were to play it many times). The best strategy maximizes this expected value. Remember, in HORSE, a possession lasts until you miss. So, you could have multiple shots per possession.

E = expected number of letters your opponent gets on your possession
p = probability that you make the shot you select
o = probability that your opponent makes the same shot

In order to give your opponent a letter, you need to make a shot and your opponent needs to miss their shot. The probability of this is:

p(1 - o)

This is also the expected value of letters your opponent gets on one shot. But, if you make it, then you get to shoot again. The expected value of future shots is the probability you make the first shot multiplied by the expected number of letters your opponent gets on a possession (since if you make your shot, you’re in the same situation as if you just started).

p * E

To get the expected value of letters your opponent gets in an entire possession you just add the expected value of the first shot and the expected value of future shots.

E = p(1 — o) + p * E

Then, we can solve for E:

E — p * E = p(1 — o)
(1 — p)E = p(1 — o)
E = p(1 — o) / (1 — p)

Now, we have a formula for the expected value of a possession based on the chance you can make a shot and the chance your opponent makes the same shot.

A Basic Scenario

Let’s assume both players have the same skill and ability for all shots. So, if Player A has a 40% chance of making a shot, then Player B also has a 40% chance of making that same shot. This is a big assumption and is never true in practice, but it’ll reveal whether shot selection makes a difference. We’ll look at some more complex scenarios later.

Player B’s strategy is to repeatedly take shots with a 50% make rate. Let’s look at the expected value of that strategy.

E = p(1 — o) / (1 — p)
Solving for p = 0.5 and o = 0.5
E = (0.5)(1 – 0.5) / (1 – 0.5)
E = 0.5

Since the expected value is 0.5, on every turn for Player B, Player A will get on average half a letter. This means Player A is likely to get H-O-R-S-E in 10 turns.

Optimizing for Player A

Let’s generalize this for any shot and try to find the optimal strategy. In this scenario, p = o (the probability you each make a shot is equal), so we can just substitute it in the formula:

E = p(1 — o) / (1 — p)
E = p(1 — p) / (1 — p)
E = p

The expected value of a certain shot is equal to the probability of making that shot. To maximize E, you just need to maximize p. This means you should always choose the easiest shot you can! Just keep taking layups and hope your opponent misses before you do.

If you keep taking a shot that has a 99% chance of making your shot, the expected number of letters your opponent will get per possession is 0.99 which means your opponent will lose in about 5 possessions. So this strategy is twice as good as selecting a 50% shot. Just keep attempting layups, and you will have an optimal strategy.

What if you are much better at shooting three-pointers?

The previous scenario assumed equal abilities, which, of course, is never true. Let’s assume that you’re playing against Steph Curry who dominates three-pointers. Steph has a 40% success rate, and you have 10%. You both have a 99% chance of making a layup (probably low for Steph, but let’s go with it). Since Steph is 4x more likely to make a three-pointer, surely that’s a better shot to take, right? We know that the expected value for the layup is 0.99. The expected value for the three-pointer:

E = p(1 — o) / (1 — p)
E = (0.4)(1 – 0.1)/(1 – 0.4)
E = 0.6

If Steph Curry were to shoot three-pointers, you’d get on average 0.6 letters per turn. But, he’d be getting 0.99 letters per turn! So, even if you have a much better chance (4x) of making that three, the layup strategy is still better! It turns out that keeping possession (making your shot and shooting again) is much more important than ensuring a better shot than your opponent.

So wait…did I just ruin HORSE?

Personally, I was disappointed by my results. It looks like I’ve reduced a great game to a layup competition which stretches on for hours. So, I feel obligated to try to fix the game. I have three suggestions:

1. Don’t allow the player to keep the ball if they make their shot

The main problem is that the player with the ball keeps the ball if they make it. Let’s take a look at what would happen if we remove this rule. This means the expected value is simply:

E = p(1 — o)

If you have two identical players, then this simplifies to p(1 — p) and the graph for p vs E looks like this.

E is maximized at 0.25 when p is 0.5, so the optimal strategy is to take shots with a 50% make-rate. That’s already sounding like a better game. If you’re four times more likely to make a three-pointer (like Steph Curry in our previous example), then E becomes 0.36. This better outcome would reward being a better three-point shooter. With this rule change, we’re much closer to making the strategy result in a fun game. But sadly, low percentage shots are still not a good idea — meaning your patented 360º, underhand, off-the-backboard, ten-foot jumper is still far from optimal.

2. Set different point limits for different shots

If you assign more letters to harder shots, there would be a higher incentive to try shooting them. For instance, if you make three-pointers worth double (two letters instead of one), then the expected value from our earlier example becomes 2 * 0.6 = 1.2, making it slightly better than the layup strategy where the expected value is 0.99. You could also ban easy shots (requiring a minimum distance).

However, these approaches are not practical. Assigning a letter value to every shot is difficult and subjective. Also, some shots close to the basket are difficult even if they’re close (e.g. shots from behind the backboard).

3. Just play the game

Maybe the best solution is to ignore everything in this article and just play the game. Is it really that important to win HORSE? Maybe you should play for the beauty of the game. Maybe you shouldn’t subject your friends to 200 layups in a row. You could optimize by taking the easiest shot every time, but maybe the real goal is to become the legend who makes that behind-the-back, eyes-closed, half-court shot.

If you have any better ideas for how to fix HORSE, please let me know in the comments.

Thanks to Alice Avery, Brad Svenson, Hari Arul, Ranee Bhutani, and Alastair Livesey for your feedback and comments.