# Probability “Paradox” I've been thinking of

Last night I thought about a probablity "paradox" that I would like if someone could clarify for me.

I couldn't find this version of it anywhere (I thought about it myself) so I hope it isn't something that has been answered before.

suppose we have 2 friends, Alice and Bob, and we let them play a game: the "game master" puts a note with a natural number on Alice's forehead, and the same goes for Bob (Those two numbers can be equal, or different). Each of them is requested to state a natural number that is larger (or equals to) the number on their forehead. They win if at least one of them is correct.

In the first version of the game, they are both in seperate rooms, and don't know anything about their friend in the other room.

In the second version of the game, they are sitting in front of each other, and can see the number on their friend's forehead.

Now, intuitively, there should be no difference between their chances of winning the game in both versions, because the number on their friend's forehead gives them absolutly no information about their number.

But this is not true. In the second vesrion, they can use the simple strategy of just stating the number on their friend's forehead. It assures winning because one of the two numbers is larger (or equal to) the other one.

However, I think it is quite obvious that there is nothing that they can do in the first version. I don't know a lot about probability, but intuitively I believe their winning chance is actually $0$ (is it even something that makes sense?) - stating "$1$" isn't really better than stating "$10^{10^{100}}$" in any way.

So the paradox is: what exactly is the extra information they recieved in the seocnd version? How does this make sense?

I'd love to hear your explanations. I think it is a beautiful paradox but I can't think of anything convincing to say about it.

• The win condition involves both of their numbers, and in the second version they receive some information about both of their numbers. What is unintuitive about this? It is also not true that the probability of winning in the first case is zero, assuming things are set up so that it exists at all. Alice and Bob can both randomly choose a number in such a way that arbitrarily large numbers have positive probability. – Qiaochu Yuan Aug 14 '16 at 9:10
• The probability of winning when unable to see each other is not zero because the strategy of choosing the highest practical number can be looked at as a partition which is at least 50% likely to succeed, resulting in at least 75% of success in the problem. This is because even with zero information about what number selection is "practical" by whoever wrote on your forehead, you can take what is practical for you as your best estimator of what was practical for them, and by choosing the maximum such number you guarantee that your choice is biased in favour of success. – samerivertwice Aug 17 '16 at 17:18

$P(A)+P(B)=P(A \text{ or } B)+P(A \text{ and } B)$
this means that even when $P(A)$ and $P(B)$ remain the same it might be possible to increase $P(A \text{ or } B)$ so long as we decrease $P(A \text{ and } B)$. Note that when Alice and Bob can see each other the probability that they both win is 0%.