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.