# Differing conclusion on a coin's side given two logically equivalent pieces of information

The following scenario was posed by my philosophy professor:

Consider the scenario where one day God decided to flip infinitely many fair coins (not stated if countably or uncountably infinite), with each flip being independent. Exactly one of this coin has your name on it. After He is done, He sent an angel to provide you the following information: miraculously, out of all the coins, only a finite number of them turned out to be heads. If you have to bet whether the coin with your name has head or tail facing up, what would you bet on?

By classical or naive probability, one would be tempted to say "tail", as the probability of the coin being a head is $$\frac{k}{\infty} = 0"$$, where $$k$$ is the (finite) number of heads in total. Those who have studied probability theory may argue that the scenario forms an immeasurable space, so the probability of the coin being a head cannot be established. On the other hand:

Suppose the angel told you the following information instead: out of all the coins except yours, only a finite number of them turned out to be heads. Now, what would you bet on?

In that case, one would easily conclude that the probability of the coin being a tail is $$\frac{1}{2}$$, by fairness of coin and independence of tosses.

At first glance, these two conclusions seem reasonable. However, even though the two pieces of information seem different, they are logically equivalent - there are finitely many heads in all coins except yours iff there are finitely many heads in all coins including yours.

What is the logical fallacy here? Could it perhaps be that the problem was phrased in a (subtly) paradoxical way? Any insights provided would be appreciated.

• If I were told that infinitely many "fair" coins were flipped and only finitely many of them landed on heads, I'd conclude that whoever said the coins were fair was a liar. Dec 26, 2019 at 23:22
• @JMoravitz please explain why. Dec 26, 2019 at 23:23

The word "tempted" is doing a lot of work here. We are being "tempted" to forget that the coin is fair (i.e. $$p=\frac{1}{2}$$ where $$p$$ is probability of heads, and, in this case, also tails).

First off, there is no problem with measurability if the number of flips $$X_i$$ is countable. In that case, we identify the sequence of heads and tails with the digits after the decimal point in the binary number $$X=0.X_1X_2X_3\dots$$. Then $$X$$ is a normal number (there is an interesting discussion on this random variable for general $$p$$ here: Derivative and calculus over sets such as the rational numbers). The normal numbers are a set of full measure in $$[0,1]$$, so equivalently we consider $$X$$ to be uniformly distributed on $$[0,1]$$.

The probability that only finitely many $$X_i$$ are 1 is the same as the probability that $$X$$ is rational (since the binary expansion terminates). This event has probability 0, since the rationals have Lebesgue measure $$0$$. This is why JMoravitz states in the comments that one would be tempted to consider it unlikely that $$p=\frac{1}{2}$$: the likelihood that that is the case, given our one observation, is $$0$$. Apparently, we are meant to make the leap that $$\mathbb{P}(p=\tfrac{1}{2}\mid X\text{ is rational})=0.''$$

But wait, that actually makes no sense. We never said that $$p$$ was a random variable and that it has any chance of being anything other than $$\frac{1}{2}$$. We are told $$p=\frac{1}{2}$$ deterministically. There is no Bayesian update of priors. So in this model, actually $$\mathbb{P}(p=\tfrac{1}{2}\mid X\text{ is rational})=\mathbb{P}(p=\tfrac{1}{2})=1.$$ In other words, $$p$$ does not depend on $$X$$ at all. It's just the constant $$\frac{1}{2}$$.

So, there is no problem. If $$p=\frac{1}{2}$$, then in both of the scenarios above, the probability of your coin coming up heads is $$\frac{1}{2}$$.

In general, there is usually something like this going on in these kinds of "philosophical" "probability" questions. The professor is either confused or trying to confuse you. If $$p=\frac{1}{2}$$, why are we suddenly being asked to consider changing that? Are we supposed to construct a new model of the coin from empirical facts? If so, that would be a different model of the coin than the one we started with, and the paradox here is just the conflation of these two different model coins.

The narrative aspect adds an element of smoke and mirrors to a simple question. There would be no confusion if instead the question was written:

Let $$X_i$$ be independent Bernoulli($$\tfrac{1}{2}$$) random variables for $$i\in\mathbb{N}$$. What are:

• $$\mathbb{P}(X_1=1)$$,
• $$\mathbb{P}(X_1=1\mid \sum_{i>1}X_i<\infty)$$,
• $$\mathbb{P}(X_1=1\mid \sum_{i} X_i<\infty)$$?
• +1 great explanation esp. with the last example, identifying "your coin" with $X_1$. I think that actually is the key. If "your coin" is always the $1$st coin (or the $1073$rd one, etc), then the ${k \over \infty} = 0$ argument doesn't apply. However, like (haha) Monty Hall, if you're offered a chance to switch to a random coin (uniform $n \in \mathbb{N}$??) before betting H/T, you should! ;) Dec 27, 2019 at 0:50
• Excellent explanation. Thank you. Dec 27, 2019 at 1:06