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.

An allegedly microequiprobable distribution

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.


# on 11 November 2014, 22:18

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.

# on 12 November 2014, 18:41


Interesting. One could even sell that as another way to cash out the idea
that the relevant functions are approximately flat over microsized
regions: their average absolute slope over each stripe is
approximately zero. I don't follow why (1/n)\int |f'(x)| dx is an
upper bound on P(red)-P(black), though. What if the density has one
large peak that is mostly located within a red band, and then trails
off at alomst zero across the next million bands -- isn't
(1/n)\int |f'(x)| dx then approximately zero, but P(red) is much
larger than P(black)?

(Also, in reality the number of stripes is often unbounded, and they
are almost never of equal size. Reparameterization here looks a bit
like a hack.)

# on 17 November 2014, 20:00

Note that (1/n)int |f'(x)| dx is invariant under linear rescaling of x, as it should be.

So now suppose our million bands are all in [0,1]. If most of the mass of f is contained in [0,1/n], there will be an enormous slope as we go from something of the order of n to something of the order of 1 (say) over the course of a stripe.

# on 18 November 2014, 10:29

Ah, right, thanks!

# on 19 November 2014, 17:59

Note that my upper bound can be rewritten as (\Delta x) (1/L)\int |f'(x)| dx, where L is the width of the interval the integral is over and \Delta x = L/n is the width of a stripe. (Assuming equal width. Without equal width, of course you can get weirdness. After all, it could be that all the red stripes are twice as wide as black stripes, if things go badly.)

Under appropriate assumptions about how nice f is, this formula will carry over to the case where the number of stripes is unbounded, and we will have the upper bound:
(Delta x) limsup_{L\to\infty} (1/2y)\int_{-y}^y |f'(x)| dx.
I don't know what exactly the appropriate assumptions will be.

I conjecture that for any (Lebesgue integrable) f we will have the bound
(Delta x) limsup_{L\to\infty} (1/2y) TV(f|[-y,y]),
where TV(f|[-y,y]) is the total variation of f restricted to the interval [-y,y]. Intuition says this shouldn't be hard to prove, but I need to prepare for class. :-)

I suspect these bounds will be fairly sharp.

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