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 aa, Player A's roll.

There is an 1−a1-a chance B's initial roll exceeds A's, allowing them to outright win. Otherwise, there is an aa chance that B needs to roll again, and then there is a 1−a1-a chance that B exceeds aa without exceeding 1. This makes B's chance of winning a(1−a)+(1−a)=(1+a)(1−a)=1−a2a(1-a)+(1-a)=(1+a)(1-a)=1-a^2.

Now consider Player A. Given their first roll being aa, 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)=a21-B(a)=a^2 chance of winning. In the second, the new roll is uniformly distributed from [a,1−a)[a, 1-a), though we can only count rolls below 1. So there is a ∫a1(1−B(x)) dx=∫a1x2 dx=13x3∣x=a1=13(1−a3)\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 aa. Since this is a uniform distribution, this makes the expected value ∫01max⁡(a2,13(1−a3)) da\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 aa at which the first expression will strictly exceed the second. It works out to a≈0.532089a\approx0.532089 being the root, making ∫01max⁡(a2,13(1−a3)) da≈∫00.532089a2 da+∫0.532089113(1−a3)) da≈0.453802\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 aa (Player A's roll, if legal) and bb (Player B's roll, if legal) is 1−(max⁡(a,b))21-(\max(a,b))^2. For the remainder of the problem, we can denote this as C(x)C(x), where xx is simply the maximum legal roll between Player A and B, or 0 if neither are legal.

Player B has an aa chance of needing to spin twice, contributing a a∫a1(1−C(b))2 db=a∫a1b2 db=13a(1−a3)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−a1-a chance of having the choice to spin again, contributing a (1−a)∫a1max⁡(b2,∫b1x2 dx) db(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)(∫amax⁡(a,0.532089)b2 db+∫max⁡(a,0.532089)113(1−b3)) db)=13(1−a)(14max⁡(0.532089,a)+max⁡(0.532089,a)3−max⁡(0.532089,a)−a3+34)(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)(1-B(a))(1-C(a) for a given roll aa, which works out to (1−13(1−a)(14max⁡(0.532089,a)+max⁡(0.532089,a)3−max⁡(0.532089,a)−a3+34))a2\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 aa to 11 to get ∫amax⁡(0.532089,a)(1−13(1−x)(0.53208944+0.5320893−0.532089−x3+34)) dx+∫max⁡(0.532089,a)1(1−13(1−x)(a44+a3−a−x3+34)) dx\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%?