## Microequiprobability

If you spin a wheel of fortune, the outcome -- red or black -- depends on the speed with which you spin. As you increase the speed, the outcome quickly cycles through the two possibilities red and black. As a consequence, any reasonably smooth probability distribution (or frequency distribution) over initial speed determines an approximately equal probability (frequency) for red and black. Here is an example of such a distribution, taken from Strevens.

The stripes under the curve indicate whether a given velocity leads to red or black. As you can see, the shaded area under the curve is approximately equal to the non-shaded area, which means that the probability of red and black are both about 50 percent.

Of course, not all possible curves determine an equal probability for
red and black. Poincare avoided having to specify the right kind of
input distribution by considering the limit case as the stripes become
infinitely narrow, but if we want to learn something about actual
gambling devices, this is hardly an illuminating move. The usual story among
those who keep the stripes as they are (von Kries, Reichenbach, von Plato, Rosenthal, Strevens, etc.) is that realistic input
distributions *assign roughly equal probability to neighbouring
stripes*; that is, they are roughly uniform over small
intervals. Strevens calls this property *microequiprobability*.

The microequiprobable distributions are not the only distributions that determine an equal probability for red and black, but the idea is that actual distributions have a strong tendency to be microequiprobable, which explains why in fact red and black come up equally often.

But is this true? In the example distribution pictured above, the probability on neighbouring stripes is for the most part not all that uniform, but maybe it is "roughly uniform" enough for the distribution to count as microequiprobable. If so, then the following distribution also counts as microequiprobable. Here I have simply repeated a small section from the original distribution.

The chance of red in the permuted distribution (as I know from its histogram) is over 55 percent. I take it that that's not "approximately 50 percent". So if microequiprobability is supposed to be sufficient to determine an approximately 50 percent chance of red, the second curve shouldn't count as microequiprobable. But then neither does the first.

The problem is that the first distribution looks fairly plausible: it's the kind of distribution I would expect to get if someone were to tabulate initial speeds for a large number of spins. (Perhaps it should be shifted to the right a little, but that doesn't affect the present point.)

So it looks like microequiprobability isn't quite the relevant feature of actual distributions for explaining the probability of outcomes. That is, when we assume that there's a 50 percent chance of red, we assume not only that the distribution is microequiprobable to whatever extent the above two distributions are, but also that the distribution looks more like the first than the second -- which is not a difference in microequiprobability.

If the stripes are all of equal width--unlike in your first diagram--and the density is differentiable, and there are an even number of them, then an upper bound on the difference in probabilities between red and black is:

(1/n) integral |f'(x)| dx,

where f' is the derivative of the density function, and n is the number of stripes, and the integral is over the whole line. Note that this goes to zero with n as long as the absolute value of the derivative is integrable.

Now, in your first example, the derivative is relatively small except in an area to the left of the bump and an area on the right side (the area that, not coincidentally, you cloned to make your second distribution).

But in the second example, the first derivative is pretty big everywhere. (And it's also like a Dirac delta around the bumps.)

So in the first example, the integral of the absolute value of the derivative is much smaller than the integral of the absolute value of the derivative.

If the density function isn't differentiable, presumably you want to replace the integral of the absolute value of the density with the total variation of the density.

If the stripes aren't equal width, you can reparametrize to make them equal width.