This riddle hit the internet as I was leaving on my 6th Annual Michigan Beer Tour. Being the minimalist that I am, I wasn’t about to bring my laptop and type up anything on the road. However, that didn’t stop me from thinking about it.

Here is the riddle for this week quoted for convenience:

A basketball player is in the gym practicing free throws. He makes his first shot, then misses his second. This player tends to get inside his own head a little bit, so this isn’t good news. Specifically, the probability he hits any subsequent shot is equal to the overall percentage of shots that he’s made thus far. (His neuroses are very exacting.) His coach, who knows his psychological tendency and saw the first two shots, leaves the gym and doesn’t see the next 96 shots. The coach returns, and sees the player make shot No. 99. What is the probability, from the coach’s point of view, that he makes shot No. 100?

I was proud to say I was able to come up with the correct answer right away. Because of this, however, I naturally thought I was missing something and would have to rigorously analyze it.

My initial thinking was as follows. The coach has seen three shots, two of which he has made, so from the coach’s perspective, the player has a conditional probability of 2/3 of making his next shot. This is correct.

But doesn’t knowing that it is the 100th shot coming up give you extra information? It turns out that it doesn’t.

Let X3 be a random variable that will be 0 if shot 3 is missed and 1 if shot 3 is made. Define X4-X100 analogously.

Let p3 be the probability of making the 3rd shot, and q3 = 1-p3 the probability of missing the 3rd shot. Define p4-p100 analogously. We know from the problem statement that p3=q3=1/2.

If we are given that X3=1 (that he makes the 3rd shot), then the expected value of p4 is 2/3. In fact, it is a sure thing.

Suppose the coach didn’t see the 3rd shot, but saw that the player made the 4th shot (X4=1). Given that, what is the expected value of p5?

What could have happened? There is a 1/2 probability that he made the third shot, leading to a 2/3 probability that he made the fourth shot. Multiplying these together gives us that the probability that he made the 3rd AND 4th shots is 1/3.

There is also a 1/2 probability that he missed the 3rd shot leading to a 1/3 probability that he made the 4th shot. Multiplying these together gives us that the probability that he missed the 3rd and made the 4th is 1/6.

So, the *conditional *probabilities of each of these events if (1/3)/(1/3+1/6)=2/3 and (1/6)/(1/3+1/6)=1/3 respectively.

All the coach knows right now is that one of those two events happened since he only knows that the 4th shot is made. If the first of these had occurred, then there is a probability of 3/4 that he will make shot 5 (since 3 of 4 shots have been made thus far). If the second of these had occurred, then there is a probability of 1/2 that he will make the shot 5 (since 2 of 4 shots have been made thus far).

So, the conditional expectation of the probability he will make his 5th shot given he made his 4th is found by taking a weighted average of these probabilities: (2/3)(3/4)+(1/3)(1/2) = 2/3.

Missing what happened on the third shot and witnessing the 4th shot made had no effect of the probability of making the 5th shot *from the coach’s perspective. *This is the bizarre nature of conditional probabilities.

Let’s explore one more just for kicks. Suppose the coach missed shots 3 and 4 and came back in to witness that he made shot 5. There are four possible scenarios: (X3=1,X4=1,X5=1), (X3=1,X4=0,X5=1), (X3=0,X4=1,X5=1), or (X3=0,X4=0,X5=1). These four scenarios have probabilities (1/2)(2/3)(3/4)=1/4, (1/2)(1/3)(1/2)=1/12, (1/2)(1/3)(1/2)=1/12, and (1/2)(2/3)(1/4)=1/12 respectively. Their sum is 1/2.

Since we are under the condition that one of them has occurred we can find the conditional probabilities by dividing them all by 1/2. This gives 1/2, 1/6, 1/6, and 1/6 respectively.

In the four scenarios, the probability that he will make the 6th shot is 4/5, 3/5, 3/5, and 2/5 respectively. Taking the weighted average gives the conditional expectation of the probability he will make the 6th shot given that he made the 5th: (1/2)(4/5)+(1/6)(3/5)+(1/6)(3/5)+(1/6)(2/5)=2/3.

Missing what happened on the third and fourth shot and witnessing the 5th shot made had no effect on the probability of making the 6th shot *from the coach’s perspective.*Still not convinced? Then you’re stubborn.