correct strategy in THE PRICE IS RIGHT

[imfatandugly]

What is the correct strategy for players 1,2,3 in the price is right? What is the probability that player 1,2,3 will win given that each player plays correctly. If there is a tie, i.e. player 2 and 3 both spin 100's, then they do a spin off, whoever gets the next highest roll wins.

If it makes it easier you can make it two player or they are given a random number in [0,1] instead of spinning if the sum of the two numbers is greater than 1, then they lose.

Intuitively i'd think that player 2 and player 3's best strategy is to just get a higher score than the previous guys, is this always correct? But where player 1 should stop is not so clear to me.

[Wyman]

It's not quite true that p2 only wants to get a higher score than p1. Say p1 spins .05 twice, and p2 spins .15 on his first spin. I know I'm spinning again...

That said, these things are usually easier if we look at them on [0,1], thus eliminating the chance of a tie. One could do these calculations with sums instead of integrals if one wished to write some code to grind it out (and thereby handle the cases of ties). We'll also assume all spins are independent, which is almost certainly not true.

Let's look at p3's strategy. Again simplifying, let's assume that p3's only goal is to make it to the showcase showdown (when really he may wish to maximize his profits; if he were particularly bad at pricing, and he knew his opponent was particularly good, and if he was going to be OOP in the showdown, he may wish to take another spin at the wheel for the $1k, even if he had, say $0.95). Then p3 only needs to beat the current high score. If that is x, then he will win with probability 1-x^2. [Why? Well, what's the prob that he loses? He has probability x of spinning too low on spin 1, and probability x of spinning too high or too low on spin 2. Spins are independent, so p3 loses with prob x^2.]

Ok, we've established that if p2 has the high score with x, he'll win with probability x^2. So, say he spins an x. Should he spin again? Well, we compute the expected value of his probability to win if he spins again:

[Integral from x to 1] (t^2 dt) = (1-x^3)/3

And we ask, which is bigger, x^2 or (1-x^3)/3? Turns out the latter (spinning again) is better until x > 0.532.

So what's p2's strategy? Well, he's going to spin again unless he's got $0.55 or better AND he's beating p1.

So what about p1? Well, the simpler calculation is as follows. If we assume your way of thinking, that p2 and p3's strategies are going to be "spin 'til I'm in the lead, then stop", then (by similar reasoning as above), p1 will win with probability x^4 if p1 spins x and stops, and will win with probability (1-x^5)/5 if he spins x and spins again. This time, the "it doesn't matter" value of x is 0.649. So p1 should spin unless he gets a $0.65 or better.

But we said above that p2's strategy is not that simple. He should spin unless he BOTH beats p1 and has at least $0.55. This gives p1 a little bit of an extra chance; namely, p2 might outspin p1, but remain under $0.55, spin again, and bustoooo. So, the probability of p1 winning (assuming he only spins once) is instead:
x^4 if x >= 0.55
(x^2)(x^2 + (0.55-x)(0.55-x/2)) if x < 0.55
[the first x^2 is the prob that p3 does not outspin p1, the 2nd x^2 is the prob that p2 does not outspin p1, and the (0.55-x)(0.55-x/2) is the avg probability that p2's first spin is between x and 0.55 and that p2's second spin makes him busto].

Note that the piecewise function above is continuous (the pieces agree at x = 0.55), so we can integrate. This is how we figure out the probability of p1 winning if p1 spins
again (after an initial spin of x). Let p(x) be the piecewise function above. Then the probability that p1 wins after spinning twice is:

[Integral from x to 1] (p(t) dt), which is

(1-x^5)/5, x >= 0.55
.190 + [Integral from x to .55] ((1.5t^4 - 1.65t^3 + 0.55t^2) dt) if x < 0.55, which simplifies to

(1-x^5)/5, x >= 0.55
.198 - 0.3x^5 + 0.4125x^4 - 0.1833x^3, x < 0.55

The downside of typing such a response on the fly (as I am doing) is that you do this computation and realize all too late that adding this tweak to p2's strategy doesn't change p1's strategy at all. So, the (rough) strategies are as follows:

p3: Spin until you get at least the high score so far.
p2: Spin until you get at least p1 and at least $0.55.
p1: Spin until you get at least $0.65.

I hope this not only answers your question, but sheds some light on how to approach such a problem.

[provisions]

That's great!