## Monday, July 27, 2020

Suppose I have infinitely many coins and I toss them. I then load each of them equally in favor of heads and toss them all again. All the tosses are independent.

Trick question: Was it more likely that they all landed tails on the second toss or on the first?

Although it seems more likely on the first, the correct answer has to be neither. Here’s one way to see it. Let’s imagine for simplicity that the second set of coins is physically distinct from the first, and that all the coins—the loaded and the unloaded—are tossed simultaneously. That shouldn’t make any difference to the comparison. But now let’s suppose that the loaded coins are such that the probability of heads is 3/4 for each coin. Now, line up the coins in such a way that two fair coins are put beside each loaded one. The probability that both of the fair coins land tails is 1/4. The probability that the loaded coin lands tails is 1/4. To get all the fair coins landing tails thus shouldn’t be any more likely than to get all the loaded coins landing tailings, or vice versa. And if the loading is different, we just tweak the arrangement.

So, either the event of all the fair coins landing tails is equal in probability with the event of all the loaded coins landing tails, or the two events cannot be compared in probability.

Martin Cooke said...

LOL

I haven't seen this paradox before (I like it :-)

Martin Cooke said...
This comment has been removed by the author.
IanS said...

Great example!

Martin Cooke said...

I think that we can compare, in probability, the event of all the fair coins landing tails with the event of all the loaded coins landing tails. For example, if there were infinitely many copies of this possible world (of infinitely many coins), in some absolutely huge multiverse, then as you have shown, we should not expect more such outcomes from the fair coins (or from the loaded coins). That means that they are equal in probability (I think).