Target practice and conditional probability

Bruce Ediger of Information Camouflage solved a nice little probability problem on his blog the other day. He got the problem from Greg Ross at Futility Closet, a blog I hadn’t run across before. Here it is:

Marksman A hits a certain small target 75 percent of the time. Marksman B hits it 25 percent of the time. The two of them aim at that target and fire simultaneously. One bullet hits it. What’s the probability that it came from A?

Implicit in this setup is that the two shooters are statistically independent. Why? First, because that makes the most sense; and second, because you can’t solve the problem otherwise.

Ross solves the problem in the traditional way, working out the intersectional probabilities

P ( A hits B misses ) = ( 3 4 ) ( 3 4 ) = 9 16

and

P ( B hits A misses ) = ( 1 4 ) ( 1 4 ) = 1 16

Since these two intersection events, which we’ll call AB and BA to make things more compact,1 are mutually exclusive, the probability of their union, which is when exactly one of the shooters hits the target, is just the sum of these two probabilities. Therefore, the conditional probability we want is

P ( AB   |   AB BA ) = 9/16 9/16+1/16 = 9 10

Ediger does a Monte Carlo simulation and gets roughly the same answer2 but is unsatisfied:

This is a completely unintuitive result to me.

I can’t disagree. Luckily, this problem is simple enough that it can be visualized in a way that might make it more intuitive.

We’ll sketch out a Venn diagram of the problem, but we’ll use boxes instead of the usual ovals, and we’ll size the boxes so their areas represent the probabilities of the corresponding events.

The universe of all possibilities will be a 1×1 box, and we’ll consider each axis to represent one of the shooters. Event A is a blue box that takes up ¾ of the horizontal range. Event B is a red box that takes up ¼ of the vertical range. Where they intersect (both shooters hit the target) is purple—it’s both blue and red. The uncolored portion represents neither shooter hitting the target.

Graphical probability

By subdividing the space into grids, we see that there are 10 squares representing exactly one shooter hitting the target and 9 of them are blue. There’s our answer, and you can see why it’s so.


  1. Overbars are typically used to mean “not,” and putting events adjacent means taking their intersection. 

  2. “Roughly” because there’s almost always some fuzz around a numerical solution.