Solved by Siqi Chen

Question:

Monte Hall Strikes Back

You're a contestant yet again facing the wrath of Monty Hall. This time
Monty has thrown the goats out and other budget constraints have forced
him to only use two doors... Door #1 and Door #2.

Before the show, Monty went behind Door #1 and wrote on a sheet of paper
a real number. He then went behind Door #2 and wrote on a sheet of paper
a different real number. Nobody knows what numbers he wrote, least of
all you -- but Monty is an honest man, and he really did write the
numbers down, and there will not be any last second shenanigans.

Now it's lights, camera, action, and you're back on stage facing the
studio audience and Monty. The last time you were here, you were trying
to shove a goat in your backseat to take home. That stupid goat chewed
up your calculus book!!!

Monty is ready to give you a substantial prize (NOT a goat) if you can
choose the door behind which is the sheet of paper on which the larger
real number is written.

Monty, of course, wants to give you a sporting chance. "If you please,
gallant contestant, choose a door!" You choose a door, and the number
behind that door is revealed.

Now comes the moment of truth. The lights are bright, the studio
audience is on the edge of their seat, and Monty says to you, "Is THIS
the larger number that I wrote down today, or is the larger number
written on the sheet of paper behind the door you did not pick?"

What do you do? What DO you do?

Can you beat Lady Luck? You talked to her last week when you were at the
racetrack, and she told you that your chances of beating Monty were
1/2... but as she said it, there was a little bit of a twinkle in her eye.

Could she know something you don't yet know? Is there a way of doing
better that boring old 1/2?

Prove or disprove: there is a way to assure that you will get the prize
with probability greater than 1/2.

Answer:

Let function F(x) be any cumulative probability function such that for any two reals
a > b, we have F(a) > F(b), and let the real shown from the picked door be t. Then
the optimal strategy would be to choose to stick with the already chosen door with
probability F(t) and switching to the other door with probability 1-F(t). This
strategy yields a greater than 0.5 chance of winning since the probability of a win
is:

P(win) = P(being shown larger number a)*P(we stick with the current door given a) +
P(being shown smaller number b)*P(we switch doors given b)

P(win) = 1/2 * (F(a) + 1 - F(b))
P(win) = 1/2 * (1 + (F(a) - F(b)))

Since F(a) > F(b) by assumption, P(win) > 1/2.


- Siqi Chen