By Ben Blatt

(You can also a version of this post in the Wall Street Journal here)

For someone with so little basketball skill, I’ve been playing a lot of H-O-R-S-E in the basketball/squash court of Lowell House basement. I’ve come to the conclusion that in most H-O-R-S-E games the shots called are excessively difficult. While hookshots and crazy bounce shots can be fun, there must be an ideal strategy to maximize the amount of letters your opponent(s) receive and therefore your own chances of winning.

We’ll start with the assumption that the odds of making a shot are defined by the location and type of shot, and not based at all on the player. This means any shot you call that you have a *p* chance of making, your opponents will also have a *p *chance of making. This is obviously never true, but if you are playing people of roughly the same skill level the numbers should hold roughly the same.

In order to maximize your chances of winning, you want to maximize the total amount of letters your opponents receive. Under standard HORSE rules if the shot caller makes the shot and the opponent misses, the opponent will receive a letter. The person calling the shots only loses control of calling the shots if (1) he/she misses or (2) all other players make the called shot.

Let’s start by imagining a game of H-O-R-S-E with *n* players. In round one, the expected number of letters just one of your opponents will receive is the probability you make the called shot times the probability that your opponent misses the shot, *p(1-p).* By linearity, the number of expected letters received by all of your opponents in round one will be *p(1-p)(n-1).* In round two, the probability expression is the same, except in order for a round two to even occur you must make your second called shot and at least one of your opponents must miss their shots. Therefore, the number of letters your opponents will receive in round two is *p(1-p)(n-1) * p(1-p ^{n-1}*). We want to find the expected number of letters your opponent will receive in each individual round and sum this total. When we do this we get

Luckily, this can be evaluated since it is a geometric series. The expected value for our game becomes *p(n-1)(1-p)/(1-p(1-p ^{n-1}*)). Once we pick the number of players, we can calculate the value of

*p*that maximizes the number of letters your opponents receive. When n=3 the value of p that maximizes the expected value is .531. This means the best strategy is to call a shot you will make 53.1% of the time. If you do this, you will give your opponents a combined total of .81 letters per possession. If you called a shot that was more difficult, such as

*p*=.25, you would give your opponents about .49 letters per possession. Leaving.32 letters per turn might not seem like a lot, but its effects will be clear later in this post.

Below is a chart depicting the ideal percentage shot and the expected number of letters all of your opponents will receive based on the number of players.

Ideal Shot Percentage | Expected Number of Letters | |

Two Player | 50.0% | .33 |

Three Player | 53.1% | .81 |

Four Player | 56.4% | 1.37 |

This is the number of expected letters per turn, but what we really want to know is likelihood of winning overall. This can be done by determining the round at which your opponent reaches five letters and the round you reach five letters. The probability of winning in round ‘k’ was calculated by determining the probability of winning with each different possible combination of letters. For instance, to win in three rounds you could give your opponents 0 letters in the first two rounds and 5 in round three, 1 in the first two rounds and 3 in round three, etc. This was done for all possible combinations to reach five letters and for up to 345 rounds. The probability of player one winning in a particular round is the probability of the opponent reaching five letters in that round multiplied by the probability that player one reaches five letters in a round afterwards. Once the probability for each round is calculated, they can be summed to find the total probability of player one winning. Below is another table for two players showing the probability of winning based on different ‘p’ strategies. Using the value 42.5% in the first row and second column as an example, the chart can be read as follows: “If player one always calls shots with .3 chance of success and player two always call shots with .4 chance of success, player one will win 42.5% of the time.”

P2=.3 | P2=.4 | P2=.5 | P2=.6 | P2=.7 | |

P1=.3 | 51.6% | 42.5% | 39.7% | 42.5% | 51.6% |

P1=.4 | 60.9% | 51.9% | 49.0% | 51.9% | 60.9% |

P1=.5 | 63.7% | 54.8% | 52.0% | 54.8% | 63.7% |

P1=.6 | 60.9% | 51.9% | 49.0% | 51.9% | 60.9% |

P1=.7 | 51.6% | 42.5% | 39.7% | 42.5% | 51.6% |

There are a couple of neat things that can be seen by looking at this chart. Firstly, the first player to shoot has a noticeable advantage. Secondly, it is usually equally unwise to take a shot of too hard difficulty as it is to take one that is too easy. Lastly, .5 is clearly the dominant strategy as it always outperforms other values of p.

I stated earlier that as long as you were playing opponents of roughly the same skill level the numbers should hold about the same. The chart below, similar to the first chart, shows the ideal p-value when your opponents’ *p*-value is different than yours. For example, -10% indicates that your opponents’ *p-*value is .1 less than yours and thus makes the shot 10% less than you. (This model doesn’t work well when p is close to zero or one but serves as a good model for all other values.)

-25% | -10% | Even | 10% | 25% | |

Two Player | 62.5% | 55% | 50.0% | 45% | 37.5% |

Three Player | 69.9% | 59.5% | 53.1% | 47.0% | 38.5% |

The values show that if you are better than your opponents, it is wise to take shots that are easier for you and let their mistakes hurt them. If you are worse, it is actually to your advantage to take harder shots that will be more difficult for both of you. Below is yet another chart showing the results of a game if you’re player one and your opponent is 10% better than you.

P2=.25 | P2=.35 | P2=.45 | P2=.55 | P2=.65 | |

P1=.25 | 33.6% | 21.6% | 16.5% | 15.1% | 16.5% |

P1=.35 | 44.8% | 30.1% | 24.5% | 22.7% | 24.5% |

P1=.45 | 48.3% | 34.0% | 27.4% | 25.4% | 27.4% |

P1=.55 | 44.8% | 30.1% | 24.5% | 22.7% | 24.5% |

P1=.65 | 33.6% | 21.6% | 16.5% | 15.1% | 16.5% |

Once again the calculated ideal p-values, in this case .45 for player one and .55 for player two, have turned out to be the dominant strategies. The bad news is if you’re a weaker shot than your opponent it can be very difficult to win even when you use superior strategy.

While it will help if you call your shots based on these calculations, at the end of the day the best way to improve your H-O-R-S-E odds is to become more familiar with a basketball and not a just with a calculator.

*Ben Blatt can be contacted at bbblatt@gmail.com*.

By linearity, the number of expected letters received by all of your opponents in round one will be p(1-p)(n-1).This makes several assumptions. First of all, it assumes you hit your shot. Second, it assumes everyone else misses.

However, I’m not sure you’ve carried those assumptions through to the rest of the example.

Thanks for commenting Gabe.

Those assumptions did not actually need to be made , I’m not sure if the confusion comes from the fact I explained that part quickly or by the fact that many people have different rules for playing HORSE (in which case this post only applies to the rules I established).

The model doesn’t assume that you hit your shot as the ‘p’ in front accounts for that probability.

It also doesn’t assume everyone else misses. If I had assumed this then the expression would be p(1-p)^(n-1). The expression p(1-p)(n-1) comes from the fact that the expected value of one person receiving a shot in one particular round is p(1-p). We can multiply this by (n-1) by linearity. The reason linearity did not come up later in my post is because linearity applies to expectation only, not probability. The rest of my post dealt with the probability of winning and therefore it wouldn’t have been appropriate.

The probability

prepresents the chances of you making your shot.What do you mean by “the expected value of one person receiving a shot”? I’m not sure what it means to “receiv[e]” a shot.

Also, I thought

nrepresented the number of participants, not the number of rounds?Sorry I made a typo in the previous comment (the part that you quoted). I will try to break it down again so you and anyone else reading this can see my methodology once more.

Assume you are player one. p is the probability that you make your called shot. The probability that player two (or any other player) misses is (1-p). Therefore the probability he receives a letter from you in round one is p(1-p). Therefore the expected amount of letters (this is where my typo came in) that player two will receive in round one is p(1-p). This only gives us the expected value of player two but we also want to factor in all other players. The expected number of letters that player three receives is also p(1-p). That will be the expected value for any player. Although the events of player two receiving a letter and player three receiving a letter are not independent, linearity still holds. We can simply sum up all the expected values which is how we get (n-1)p(1-p). It is n-1 instead of n because player one obviously cannot get any letters. n does represent the number of players and k represents the number of shots. We cannot use linearity to determine the expected number of letters over multiple rounds since each round has a decreasing probability of occurring.

That makes sense. Thanks.