|Price Is Right Wheel has 20 sectors|
This is the 97th Riddle proposed by The Riddler from FiveThirtyEight.
The three contestants that win all go head to head at this wheel at a chance to advance to the Showcase Showdown. The one with the least amount of money goes first (the worst position in this game), while the one that won the most goes last.
The first player must get to as close to 100 as they can without going over. They are allowed a maximum of two spins. The wheel has 20 slots, with the lowest being 5, and the last being 100, all the rest increase in increments of 5 and are distributed all around the wheel.
If they go over, they are automatically out of the game.
The second player must beat the first player’s number, and have a number that the last player cannot beat.
The last player, if the first two bust, automatically advances. Otherwise, they have to beat whoever is in front of them.
What strategy should the first player adopt to give them the best chance at advancing to the Showcase Showdown? Specifically, at what value (or above) should the first player definitely not spin again, and therefore below what value should the first player definitely spin again?
An Attempt at a Solution
To answer this question, we are interested in two conditional probabilities:
- Under the condition that we stay with this number, what is the probability that we win?
- Under the condition that we spin again with this number, what is the probability that we will not bust?
If the probability calculated in 1 is larger than the probability calculated in 2, than we should stay with that number. Otherwise, we should spin again. Our goal is to find the largest such number for which the probability of 1 is greater than the probability of 2.
Before going any further, let’s make calculations a little easier by assuming that the values 1-20 are on the wheel instead of 5-100 in increments of 5. So now, we must try and add to 20 and not go over.
Player 2’s Perspective
Suppose for the moment that player 1 has busted, and that player 2 only has to worry about player 3. Also, let’s assume their first spin is a 10. Then the conditional probability for #2 above is 50%, since they will not bust only if a 1-10 is spun again.
The probability that player 3 will win is more than 50% since that is the probability that they win on the first spin.
Thus, in this situation, they should spin again. So, we will only consider values above 10. Let the value of player 2’s first spin be X. For probability 2, we will need to also consider S, the sum of their two spins.
Let the value of player 3’s first spin be Y and the sum of their two spins be S (if they end up spinning twice).
The notation for the probabilities in #1 and #2 above are given as:
- P(Y < X) * P(S 20 | Y < X) + 0.5*(P(Y = X)+P(Y < X)*P(S = X | Y < X))
- P(S < 21 | X) = (X-20)/20
The first probability can be explained in this way. If I stay with X, then in order to win, player 3 must spin a value less than X, and then given that… when they spin a 2nd time, their sum must be lower than X or above 20, OR, their first spin must match X or their sum must match X given their first spin was lower and they win the spin-off.
The second probability is easier. If X is 15 for example, the probability of not busting, is the probability of spinning a 1, 2, 3, 4, or 5, which is 5/20. This is the same as (20-X)/20 for when X=15. This works for all other values of X.
For those mathematically inclined, I hope you can confirm that
- P(Y < X) = (X-1)/20, and that
- P(S 20 | Y < X) = (X-1)^2/400, so that the product is
This is then summed with 0.5 multiplied by
- P(Y=X) = 1/20, plus
It is OK to skip that part. Let’s compare these probabilities for some different values.
When X=13, the probability in #1 is 0.276 while the probability in #2 is 0.35. Therefore, player 2 should spin again.
When X=14, the probability in #1 is 0.316 while the probability in #2 is 0.3. Therefore, player 2 should stay.
Now, player 2 can only do this if player 1 has spun something less than or equal to 14, or has busted. They must beat our value otherwise. Assuming player 2 will play in this optimal way, let’s continue with the calculations for player 1.
Player 1’s Perspective
From watching several YouTube videos involving spin-offs, it seems that spin-offs occur among two contestants at a time. So, if the first two tie, they spin-off to go against the third contestant, on which another spin-off can occur.
So, the probabilities for player 1 to win must be the probability for player 2 to win squared. Player 1 must defeat both player 2 and player 3.
So, for X=14, the probability of player 1 winning is now .0998 while the probability of not busting on a second spin is 0.3. Player 1 should spin again.
For X=15, the win probability is .149 while the probability of not busting is .25. Player 1 should spin again.
For X=16, the win probability is .217 while the probability of not busting is .20. Player 1 should stay.
This is equivalent to getting 80 or more on the wheel in the Price is Right.
2 thoughts on “The Price is Right Wheel: How to Spin it Optimally”
Nice! While I'm not particularly good at these riddles, I enjoy reading your analyses and solution attempts.
An attempt is all it was! My assumption was incorrect, which ultimately leads to an incorrect answer. 😦
Oh well. Live and learn.