It seems Bob cannot win, and his pace during each mile does not matter. According to the conditions, Alice must reach 25.2 miles before Bob (Alice covers this in 25.2 x 8 minutes, Bob in 25.2 x (8 minutes 1 second)).

Alice will run the last 1 mile in 8 minutes. Bob MUST run the total distance in 8 minutes 1 second, not matter how he varies his speed.

But Bob was behind Alice with 1 mile to go, exactly 25.2 seconds behind in fact. So, in order for Bob to beat Alice, he must cover the last mile in less than 480-25.2 seconds. Since the distance is a mile, Bob must take 481 seconds to cover the distance. Which places him 26.2 seconds behind Alice at the end. So he cannot win.

He could win if you allow his 481 seconds to cover a mile to include him going past the mile point, turning around and coming back. Thus, he runs each mile in 300 seconds, keeps going past the mile point for a distance, turns around and comes back, and the total time, including the extra distance run, is 481 seconds. Then he is still behind Alice at 25.2 miles, but he runs much faster, gets to the finish before Alice. Of course to satisfy the 481 second mile condition, he would have to keep going past the mile, then turn around and come back. For this to work, you have to define his time per mile based on when he gets back for all but the last mile. For the last mile you time only until he reaches the finish line the first time.

Of course, this is similar to saying that Alice runs at a constant speed, but wanders off course, and ends up running more than 26.2 miles, while Bob runs slower but on course, so he gets there first. Either of these solutions ignores the implications of the problem that both contestants are running in an otherwise conventional manner.

The probability I win = p^2 (1/2) +2pq + (1-p)^2 (1/2)

i.e. I win if you choose 1 positive and 1 non-positive number and have a 50-50 shot otherwise.

I whacked myself in the head when someone pointed out the the state bordering deleware mnust also have the same border.

Re: Neil

Not quite – you can use the cumulative distribution function for the normal curve, but what that would mean is guessing higher with probability

98% if quoted 2,
84% if quoted 1,
50% if quoted 0,
16% if quoted -1, etc.

That way if A = 1 and B = 2, you win 98% when shown 2, and 16% when shown 1. Overall, you win 1/2 * (98+16)% = 57% when A = 1 and B = 2.

Just guessing higher for positive numbers fails, for when A = 1 and B = 2, you don’t do better than even.

If Bob runs >11.0086 mph in the first .2 miles of each mile segment including 200-26.2 his time will be less than Alice’s.

If you claim the border lies on the surface of the Earth then you run into the problem that the Earth isn’t perfectly spherical. Since this would disqualify every border, including the given example, we should probably exclude local irregularities (hills and valleys). But there’s still the issue that the Earth isn’t really a sphere, but (to much higher approximation) an oblate spheroid. If we consider the Earth to be oblate then this disqualifies the longitudinal lines because they would be elliptical sections and not circle sections. But the latitudinal lines would still be circle sections, as they are drawn around the Earth’s oblate axis.

Of course, if you treat the Earth as non-sperical then the Delaware / Pennsylvania border is disqualified too, as it would be slightly taco or potato chip shaped. Maybe the latitudinal lines might be the best possible answer.

Finally, if we wanted to be really hard about it then we could include tidal distortions from the moon and the sun, which I’m pretty sure would make it impossible to draw a circle anywhere on the idealized surface of the Earth.

To prove that this works, suppose A is less than B. If shown B, the chance you win is f(B). If shown A, the chance is 1 – f(A). So the chance you win is .5 + .5 * (f(B) – f(A)). But f(x) is increasing, so this is better than even odds.