# Can You Play 'The Price Is Right' Continuously?

## This Week's Fiddler

Let's consider Player B and their chance of winning as a function of $a$, Player A's roll.

There is an $1-a$ chance B's initial roll exceeds A's, allowing them to outright win. Otherwise, there is an $a$ chance that B needs to roll again, and then there is a $1-a$ chance that B exceeds $a$ without exceeding 1. This makes B's chance of winning $a(1-a)+(1-a)=(1+a)(1-a)=1-a^2$.

Now consider Player A. Given their first roll being $a$, they have a choice to roll again, depending on whether it increases the odds of winning. In the first scenario, there is a $1-B(a)=a^2$ chance of winning. In the second, the new roll is uniformly distributed from $[a, 1-a)$, though we can only count rolls below 1. So there is a $\int_a^1 (1-B(x))\ dx=\int_a^1 x^2\ dx=\left. \frac{1}{3}x^3 \right|_{x=a}^1=\frac{1}{3}(1-a^3)$ chance of winning.

So between these options, we want to choose the maximal value for any initial value of $a$. Since this is a uniform distribution, this makes the expected value $\int_0^1 \max(a^2, \frac{1}{3}(1-a^3))\ da$.

Between the two expressions in the max function, there is a value of $a$ at which the first expression will strictly exceed the second. It works out to $a\approx0.532089$ being the root, making $\int_0^1 \max(a^2, \frac{1}{3}(1-a^3))\ da\approx\int_0^{0.532089} a^2\ da+\int_{0.532089}^1 \frac{1}{3}(1-a^3))\ da\approx0.453802$.

Hence, A's chances of winning are about **45.38%**.

**Answer:** 45.38%

## Extra Credit

As expected, things get more complicated with Player C in the fray. Similar to the first part, Player C's chances of winning, as a function of $a$ (Player A's roll, if legal) and $b$ (Player B's roll, if legal) is $1-(\max(a,b))^2$. For the remainder of the problem, we can denote this as $C(x)$, where $x$ is simply the maximum legal roll between Player A and B, or 0 if neither are legal.

Player B has an $a$ chance of *needing* to spin twice, contributing a $a\int_a^1 (1-C(b))^2\ db=a\int_a^1 b^2\ db=\frac{1}{3}a(1-a^3)$ chance to win. There is a $1-a$ chance of having the choice to spin again, contributing a $(1-a)\int_a^1 \max(b^2, \int_b^1 x^2\ dx)\ db$, which we know from the first part to be about $(1-a)\bigl(\int_a^{\max(a, 0.532089)} b^2\ db+\int_{\max(a, 0.532089)}^1 \frac{1}{3}(1-b^3))\ db\bigl)=\frac{1}{3}(1-a)\biggl(\frac{1}{4}\max(0.532089, a)+\max(0.532089, a)^3-\max(0.532089, a)-a^3+\frac{3}{4}\biggl)$ (final integral calculation).

Finally, Player A has to defeat both Player B and Player C; Player A also has a choice of whether to spin again or not. Without spinning again, the odds are $(1-B(a))(1-C(a)$ for a given roll $a$, which works out to $\biggl(1-\frac{1}{3}(1-a)\bigl(\frac{1}{4}\max(0.532089, a)+\max(0.532089, a)^3-\max(0.532089, a)-a^3+\frac{3}{4}\bigl)\biggl)a^2$. When rolling again, we actually integrate this expression from $a$ to $1$ to get $\int_a^{\max(0.532089,a)} (1-\frac{1}{3}(1-x)(\frac{0.532089^4}{4}+0.532089^3-0.532089-x^3+\frac{3}{4}))\ dx+\int_{\max(0.532089,a)}^1 (1-\frac{1}{3}(1-x)(\frac{a^4}{4}+a^3-a-x^3+\frac{3}{4}))\ dx$. Honestly, the integrals were getting too verbose to keep typing, and I was running low on time, so I plugged it all in here. Not sure if it's right, but we got 0.489523. Seems suspicious, since I would expect lower chances with more players.

**Answer:** 48.95%?