# Helen and Joe play guitar together every day at lunchtime.

Helen and Joe play guitar together every day at lunchtime. The number of songs that they play on a given day has a Poisson distribution, with an average of 5 songs per day. Regardless of how many songs they will play that day, Helen and Joe always flip a coin at the start of each song, to decide who will play the solo on that song. If we know that Joe plays exactly 4 solos on a given day, then how many solos do we expect that Helen will play on that same day?

My attempt: If the average is $$5$$ songs a day and Joe performs $$4$$ solos on one day. I thought we should expect Helen to perform $$1$$ solo on the same day $$(5-4=1)$$

But The answer given to me is: $$2.5$$ solos we expect Helen to play

My question is why? What is the way of thinking that gives me $$2.5$$? Is it cause of the coin flip? so $$5 \cdot .5 = 2.5$$? What does Joe's $$4$$ solos have to do with anything then?

Thank You for any help.

• Helen and Joe must have played $4$ or more solos on that day. In other words, you should be thinking about a conditional probability. – Toby Mak May 15 at 2:42

Let $$X$$ denote the total number of games played, and let $$J$$ denote the number of games Joe played. The conditional distribution of $$X$$ given $$J=4$$, namely $$X|J=4$$, is supported on $$\{4,5,\ldots\}$$ and has pmf $$P(X=k|J=4)=\frac{P(J=4|X=k)P(X=k)}{\sum_{k=4}^{\infty}P(J=4|X=k)P(X=k)}$$ which is non zero whenever $$k\geq 4$$. Using the facts that $$X\sim \text{Poisson}(5)$$ and $$J|X\sim \text{Binomial}(X,1/2)$$ we get $$P(X=k|J=4)=e^{-5/2}\frac{1}{(k-4)!}\bigg(\frac{5}{2}\bigg)^{k-4}$$ This means $$X-4|J=4\sim \text{Poisson}(5/2)$$ and so $$\mathbb{E}(X-4|J=4)=\mathbb{E}(X|J=4)-4=5/2$$ Hence $$13/2=E(X|J=4)$$ so expected number of times Helen plays is $$5/2$$

While Matthew Pilling and tommik provide answers that show the mathematics of getting the answer, I will provide the intuition involved.

1. We know that, on this specific day, Joe played 4 solos. This provides a minimum number of songs - specifically, there must have been at least 4 songs. Note that it is very possible for the Poisson distribution to produce 0 songs or 1 song. However, we have been given information that constrains the set of possible numbers of songs - it must be at least 4.

2. We know that Joe played exactly 4 solos - this is a lot more likely if there are 8 songs (~27%) than if there are 4 songs (~6%). This changes the likelihoods of each of the possibilities, compared with the basic Poisson distribution, given this information.

3. How many solos we expect Helen to have played can then be worked out from the new probabilities, which have incorporated the additional information (that Joe played 4 solos).

To see why the 50% information can't be directly used to conclude that Helen is expected to have played 4 solos as well, consider a slightly modified version of the problem. Rather than the number of songs following a Poisson distribution, we will assume that they follow a uniform distribution of between 1 and 7 songs.

Now, we know that Joe played 4 solos. How many solos do we expect Helen to have played? Well, it can't be 4, because that would mean they may have played 8 songs, which can't have happened - the maximum is 7 songs.

To work out the correct answer, we turn to Bayes' Theorem, which is explicitly used in Matthew's answer, and is hidden by proportionality in tommik's answer. Think of the fact that Joe played 4 solos as a "new piece of information". Bayes' Theorem (at least by Bayesian thinking) lets you update your probabilities given the new information.

The answer of @Matthew Pilling is perfect (+1). A Bayesian approach will lead to the same solution avoiding a lot of calculations: (constants are not considered until the end of the process)

$$\mathbb{P}[X|J=4]\propto \mathbb{P}[X]\cdot \mathbb{P}[J=4|X]\propto\frac{5^x}{x!}\cdot \binom{x}{4}\left(\frac{1}{2}\right)^x\propto\frac{\left(\frac{5}{2}\right)^{x}}{(x-4)!}\propto\frac{\left(\frac{5}{2}\right)^{(x-4)}}{(x-4)!}$$

Setting $$Y=X-4$$ we immediately recognize the kernel of a Poisson distribution with mean 2.5

Here $$Y$$ is the distribution of the solos played by Helen

$$\mathbb{P}[Y=y|\text{Joe }=4]=\frac{e^{-2.5}\cdot 2.5^y}{y!}$$

Here's some intuition:

The Poisson distribution is the limiting case for a binomial distribution where the number of trials goes to infinity while the success probability shrinks proportionally to keep the total expectation constant. So we can imagine Helen and Joe spending their lunch break making a very large number of Bernoulli trials, with each giving a small probability of "play a song now".

When the number of trials grows large we can fold the coin flip into that process, deciding that when "play a song now" comes out at trial number $$n$$, Helen plays the solo if $$n$$ is odd, and Joe plays the solo if $$n$$ is even.

But this effectively means that each of Helen and Joe might as well be making their own separate series of trials, with half as many trials but the same probability, and therefore half the expected value. The number of solos played by each is independent of the other, and still Poisson distributed.

The number of solos Helen plays is independent of the number of solos Joe plays, so his $$4$$ solos is not actually relevant information.

An alternative (and perhaps slightly more rigorous) phrasing of the same reinterpretation:

In each step, instead of first asking "should we play a song now?" and then "who should play the solo if we do play?", in each step we can ask "should Joe play now?" and "should Helen play now?" each independently with half the probability. That's almost the same, except that there's a finite chance the answer would be that both should play. But when we go to the limit of infinity many steps, the risk of that happening in any given step goes to zero as $$1/N^2$$, and so the probability of it ever happening during the entire lunch break goes to zero as $$1/N$$. Therefore the risk becomes irrelevant in the limit where the distributions become Poisson.