This week the #puzzle is: Can the Archers Coordinate? #probabilities

Two logicians are trying to earn a fabulous prize as a team. There are three targets, and, to win the prize, each logician must fire a single arrow and hit the same target as the other. Two of the targets are closer but are otherwise indistinguishable; the logicians know they each have a 98 percent chance of hitting either of these targets. The third target is farther away; the logicians know they each have a 70 percent chance of hitting that target.

The logicians can’t cooperate or consult in advance, and they have no knowledge of which target their counterpart is aiming for or whether they are successful.

What is the probability they will win the prize?

And for extra credit:

As before, there are still three targets, but their respective probabilities of being struck have changed. That said, two of the targets remain indistinguishable from each other and have the same probability of being struck. Moreover, all three probabilities are rational.

After doing some mental arithmetic, the logicians realize that it doesn’t matter which target they aim for—their probability of winning the prize is the same no matter what.

What is their probability of winning the prize?

Can the Archers Coordinate?

Possibly incorrect solution:

If the logicians could’ve consulted beforehand, they could have chosen 1 of 2 strategies. The 1st one involves aiming for one of the easy targets, the 2nd involves aiming for the hard target. Let’s look at the probability a strategy succeeds.

The 1st strategy includes, that the 2nd logician won’t know, which easy target the 1st logician aimed at, so there’s a 50% chance of aiming for the same one.

Both logicians will realize, that the 2nd strategy is superior. They both employ this strategy. The probability they will win the prize is 0.49 = 49%.

And for extra credit:

This time the probabilities depend on unknown numbers for the underlying probabilities. Based on integer choices for all the unknown numbers (so that the resulting probabilities are rational), this must hold:

It doesn’t matter which target is chosen, the probabilities will be the same. This means.

As the square root isn’t an integer, this must mean the rest of the equation is 0.

Therefore the probability of hitting anything is 0.

My only problem is, this doesn’t seem to require any mental arithmetic?

Another solution could be, that actually all 3 targets are indistinguishable. But again, this wouldn’t require any mental arithmetic. Still, in that case the probability would be, for some integers:

The chance they choose the same target, and the chance of hitting it.

Result: m^2/3n^2, m and n integers, at least 0, at most 1/3.