The Price is Right Wheel

I was thinking about this as I was watching this show (I know, once you read the rest you will realize how weird I am). Let's say that you are on the Price is Right, and it's you and two other people preparing to spin this wheel. The wheel I presume is 20 sided with numbers from 0-100 by 5. Now when you spin it goes flying around and lands on a number. You can stay or spin again, but the object is to spin a higher number or a higher number through sums (if you elect to spin again) than the other two. Now I was thinking that there has to be some way to calculate the lowest # that you can safely claim a mathematical edge, so any # lower you spin again, and higher you stay. Can this be generalized to n people vs. you? Let's see it boys!

P.S. - Don't kid yourselves....I don't have the solution. You are on your own.

P.S.S. - I don't think the wheel has a zero, so maybe it's 19 sided. Feel free to post the actual number of sides if you know.

It may not have a zero but did it not have a nasty bankrupt thingy on it?
It made a sound lke someone breaking wind down a trombone when you landed on it.

Or am i getting confused with the wheel of fortune?

P.S.S. - I don't think the wheel has a zero, so maybe it's 19 sided. Feel free to post the actual number of sides if you know. [/b]

if it doesn't have a zero, only then is it 20-sided.

count with me:

side01:005
side02:010
side03:015
. . .
side20:100

Very interesting question. I did a little Excel simulation using the following model:

Step 1 - P1 spins a number from 5-100. If the number is less than P1's cutoff (variable), then P1 spins again. If not, P1 sticks with the first spin. All results greater than 100 are treated as 0.

Step 2 - P2 spins a number from 5-100. If the number is less than P2's cutoff (different variable), or if it's less than P1's result, P2 spins again. If not, P2 stick with the first spin. All results less than P1 or greater than 100 are treated as 0.

Step 3 - P3 spins a number from 5-100. If the number is greater than the max of P1/P2, P3 spins again. If not, P3 sticks with the first spin. All results less than the max of P1/P2 or greater than 100 are treated as 0.

Step 4 - Throw out all ties, and count the number of wins by each player.

Step 5 - Repeat simulation a bunch of times (not very exact, I know, but this is preliminary after all!).

Step 6 - Change the cutoffs for P1 and/or P2 and run the simulation a bunch more times.

So far, P1's cutoff seems to make the most difference. I found that P1's winning percentage is highest (about 30% ) when P1's cutoff is 75 cents. Setting P2's cutoff doesn't make much difference, but it seems to be highest (about 31-32% ) at 75 cents also. Even with optimization, P3 still gets the best winning percentage (about 38-39% ).

So there you go. I think Excel will throw up if I try to account for ties (unless I can pare down the algorithm), but it's not so bad. I think the main errors will come from taking small samples of the entire range (5<P1 cutoff<100, 5<P2 cutoff<100), and from throwing out the ties.

If anyone can come up with an analytical solution for this, I will be very impressed (and yet strangely repulsed).

Good hunting!

I think we need to draw some sort of graph or mathematical fomula to figure this one out. That's as far as my brain cells can take us. WHo would likee to elaborate on that.

Lets get rid of 5 to 100 straight away and call it 1 to 20. with discrete steps of 1 in between.
so we cannot go over 20.
p1's first number = a
p2's first number = b

If a = 20, number of 'bust' numbers = 20
If a = 19, number of 'bust numbers' = 19
etc. Aha we see a pattern emerginf here.

So probability of p1 busting on second go is a/20

also odds of p2 beating p1 with first spin
Hmmm. some complicated mathtmatical functions required here. can some take over please.

the probability of p1 busting on the second go is a/20, but I don't know how useful that is considering the fact that at some number of a which we are trying to find he stops....

Originally posted by !~TONY~!
the probability of p1 busting on the second go is a/20, but I don't know how useful that is considering the fact that at some number of a which we are trying to find he stops....

I think we need a graph that shows us the optimum number to stick on at an intersection. Some way of outweighing the chance of p2 catching him against him going bust himself. At some point the graph should start soming back on itself like a parabola. But i can't figure out how to do it.

Originally posted by PBE6
Step 4 - Throw out all ties, and count the number of wins by each player.

I think Excel will throw up if I try to account for ties (unless I can pare down the algorithm), but it's not so bad.

It's been a while since I've watched Bob Barker & Co., but I believe that in the case of a tie they do a spin-off - highest man wins, repeat until they stop tieing. Given that, for all ties you can put in a value of 50/50.

OK, here's a slightly different problem solution, but it's a huge step forward in answering this one. I must admit, I got some help from this website: http://mathproblems.info/group4.html (problem No. 69)

This solution involves balancing risk and reward to find the optimum strategy for a continuous wheel from 0 to 1, then calculating the probablility given the strategy. This will give us a good ballpark for the quantized wheel. I'll break the solution into parts to make it easier to digest.

Part I - Probability of winning with 1 spin

Let's start with a two player game, and say Player 1 (P1) spins A1 as a first spin. If P1 stays with his first spin, the probability of losing to Player 2 (P2) with it is as follows:

P2 can win with his first spin --> P = 1-A1
P2 can win with two spins --> P = [1-(1-A1)]*(1-A1) = A1*(1-A1)

(here, A1 comes from spinning under the first time, and (1-A1) comes from not busting on the second spin)

So the total probability of losing with A1 is (1-A1) + A1*(1-A1) = 1-A1^2. Therefore, the probability of winning with A1 is 1-(1-A1^2) = A1^2.

For a three player game, P1 can lose to P2 or not, and lose to Player 3 (P3) or not, so we get the probability of losing as follows:

P = (1-A1^2) + [1-(1-A1^2)]*(1-A1^2) = 1-A1^4

(here, the first term is losing to P2, and the second term is not losing to P2 times losing to P3)

In general, the probability of losing with the first spin in an "n" player game is 1-A1^(2n-2), so the probability of winning with a first spin is A1^(2n-2).

Part II - Probability of winning with 2 spins

Let's start with the 2-player game again. We already know that the probability of winning with one spin is A1^2. This can be generalized to the probability of winning with a given result being R^2 (0<=R<=1). So the probability of winning with 2 spins is (A1+A2)^2.

Now, after the first spin any numbers between 0 and 1-A1 will be OK, and any numbers above 1-A1 will bust P1. So the probability of winning with 2 spins will be the average of all values of A2 between 0 and 1 times the winning result. Since the winning result of A2 above 1-A1 is 0, we can get rid of it (but we must still average over the range 0 to 1). The average winning result is then the integral of [(A1+A2)^2] dA2 from A2=0 to A2=1-A1. After some algebra we get P = 1/3*(1-A1^3). For the three player game, P = 1/5*(1-A1^4). In general, the probability is 1/(2n-1)*(1-A1^(2n-1)).

Part III - Risk vs. reward to find optimum strategy

Two player game again. OK, so the probability of winning with the first spin is A1^2, and the probability of winning with 2 spins is 1/3*(1-A1^3). Now, at some value of A1, call it C1 for "cutoff", these probabilities will be equal, because the risk of busting will balance the advantage of having a bigger result. So we set A1=C1, and set the probabilities equal:

C1^2 = 1/3*(1-C1^3)

So: C1^3+3*C1^2-1 = 0

In general, the optimum cutoff will be the solution to C^(2n-1)+(2n-1)*C^(2n)-1=0.

Using Excel (my favorite piece of MS crapware), we get C1 = 0.5321 (approx.), so if the first spin is less than 0.5321, P1 would spin again. For a 3 player game, C1 = 0.6487 (approx.).

Part IV - Probability of P1 winning

Two player game again. Now we find the probability of winning using this optimum strategy. To do this, we integrate in 2 pieces. First, if the first spin is under C1, P1 will spin again so we integrate 1/3*(1-A1^3) from 0 to C1, and if it's over C1, P1 will stick so we integrate A1^2 from C1 to 1. The overall expression is:

P = int[1/3*(1-A1^3)] (0,C1) + int[A1^2] (C1,1)

The final answer is P = 45.38% (approx.). For the 3 player game, the answer is 30.43% (approx.).

I checked these results on my spreadsheet from the other day. For C1, I used 65 cents, and for C2 I used 55 cents (closest integers). P1's winning percentage is still around 30%, P2's is about 33% (which is about 33/70 = 47% of what's left), and P3's is about 37%. So, the results are pretty close, which is comforting.

Using this method on integer wheel values will involve lots more counting, but I'll give it a go (after some sleep!!!).

🙄