In yesterday’s post, I argued that, when flipping two unfair discs (or coins), there is a greater chance that both discs will land with the same side up than different sides up. As pointed out in the comments, I was assuming that the probability of heads is equal for both discs:

Aren’t you assuming that p (and q=1-p) are the same for both discs? But isn’t it more reasonable to assume that, while no disc has a perfect p=0.5 probability of landing ‘heads’, the p’s of no two discs are likely to be the same? (Assume, perhaps, that each disc’s p is drawn independently from some kind of larger distribution, maybe dependent on something like manufacturing or the way the person throwing it holds it, before throwing).

What happens if we allow for different probabilities of heads on the two discs?

This is a graph of the probability of same for various probabilities of heads for each disc. The probabilities of heads on disc 1 is shown on the X-axis, and the probability of heads on disc 2 are 0.1, 0.3, 0.5, 0.7, and 0.9 (the five different lines). As you can see, the probability of same is greater than 0.5 when the two discs are biased in the same way (i.e., both have a probability of heads greater than 0.5, or both have a probability of heads less than 0.5). On the other hand, the probability of different is greater than 0.5 when the discs are biased in opposite directions (one with a probability of heads greater than 0.5, and one with a probability less than 0.5).

So, I say to all ultimate players out there, CHOOSE SAME if you think the discs have the same biases, and CHOOSE DIFFERENT if you think they are both biased, but in different directions.