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.

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.
