What to do when you can’t find a coin

Imagine — you’ve just finished a nice long walk and pass by your favourite bakery on the way home. The pistachio croissant and the pain au chocolat both look delicious, but you should only have one and you just can’t decide! You didn’t bring a coin with you, so how will you choose?

The solution I have been using in situations like this is to look at the step-count on my watch. If the number of steps is even, I will have the pistachio, and if it is odd, I will have the chocolate. I took thousands of steps during my walk, so surely the parity of my step-count should be 50/50! Let’s test that assumption.

Let’s say the number of steps X I take up to a given time of day is distributed Poisson with rate parameter \lambda. This is a terrible assumption as it assumes my steps arrive independently of each other, but it’s a common first guess for an unbounded count statistic, so let’s go with it.

The probability that X is even is then:

\mathbb{P}(X \equiv 0 \pmod{2}) = \sum_{i=0}^{\infty} \frac{\lambda^{2i}e^{-\lambda}}{(2i)!} .

And of course the complement gives us the probability of an odd step-count. Below I’ve desmos’ed up an animation of the probability mass function of X for various values of \lambda, with red dots representing the evens and black dots representing the odds.

If the above embedded animation does not appear for you, you can see it in desmos here.

Meditate on whether you think there will be a bias, and if so, which way. Then put your nickel down below:

My initial intuition was that as we vary \lambda, we would sometimes get more evens and sometimes get more odds, and this would give us a bunch of crossover values for \lambda where the two outcomes are equilikely. But it turns out that this is not true.

I will leave it as an exercise to show that:

\mathbb{P}(X \equiv 0 \pmod{2}) = \frac{1}{2} + \frac{e^{-2\lambda}}{2} .


This of course is strictly greater than \frac{1}{2}, so no matter how many steps I average on my walk, I am always ever-so-slightly more likely to choose the pistachio croissant. This bias does vanish rather quickly as \lambda increases, so I can live with this bias for my purposes. But the general takeaway is that all Poisson distributions are biassed towards even outcomes, and I find that rather surprising! I leave to those who are interested the following questions:

Question 1: Is the Poisson distribution reduced \pmod{n} biassed towards a certain remainder for other values of n?

Question 2: Do other distributions with the same support have this bias? If X \sim \mathbf{Geometric}(p), then \mathbb{P}(X \equiv 0 \pmod{2}) = \frac{1}{2-p}, but what about more complex distributions?

Question 3: Is there an intuition behind why this bias exists?

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