Naranjas1
Joined: Nov 1, 2017
• Posts: 1
November 1st, 2017 at 9:41:19 AM permalink
This one is insidious. I'd love to see the math on it. In the following, I'm assuming readers have basic knowledge of the rules of the Price is Right Wheel.

It might have to modeled such that you assume every spinner attempts to spin a dollar. (Let's say you spin 0.95 on your first spin. Spinners should stay on 0.95, but maybe we can only model this scenario if we assume they will attempt again and try to hit 0.05). Or, it could be modeled such that we assume all spinners follow your optimal strategy outlined here /ask-the-wizard/tv-game-shows/

The harder part is the first dollar. There are 20 spaces on the wheel. You have a 1/20 shot on your first spin, and if you don't spin a dollar, you have an optional second 1/20 shot (to complete the dollar). Once all three contestants finish their spins, there are 8 outcomes that I can think of. 1) You went over \$1.00 and are eliminated. 2) You spun some value less than a dollar AND someone else beat you so you are eliminated. 3) You spun some value less than a dollar AND the two other contestants are eliminated, which ends the game. 4) You spun some value less than a dollar AND one other person tied you and the 3rd person is eliminated. 5) You spun some value less than a dollar AND both other contestants tied you and no one is eliminated. 6) You spun a dollar AND no one else spun a dollar. 7) You spun a dollar AND one other person spun a dollar. 8) You spun a dollar AND two others spun a dollar.

This is where my feeble mind cannot press on.

If you are involved in a NON-DOLLAR tie, you are given one additional 1/20 chance to spin the first dollar (tie-breaker spins count for the bonus). If a tie occurs again, you are given an additional chance IF the tie is NOT on the dollar. This scenario can theoretically happen to infinity. It might be a billion to one odds, but theoretically a spinner could have 10 chances to spin the first dollar (if their were 9 non-dollar ties prior to this).

The 2nd dollar is much easier. If you achieve \$1.00 through any of the above scenarios, you are given a single bonus spin. You have a 1/20 shot to spin a \$1.00 again and win \$25k. (ties occuring at this stage are not eligible for the \$25k, so we can discount them phew).

So considering all of this, if I'm a contestant on The Price is Right, what are the odds I win the \$25k bonus on the wheel?

If all of the above seems preposterous, I give you www.youtube.com/watch?v=DPGeJ8EfOkM

(A note: My question is what are the odds that a single person wins the \$25k bonus on the Wheel. Another fun one to tackle would be "What are the odds that at least one person wins 25k? At least two people? All three people?")
boymimbo
Joined: Nov 12, 2009
• Posts: 5994
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November 1st, 2017 at 10:07:33 AM permalink
You would need to assume the spin of the wheel is random because the person doing the spin really has no idea of where their first spin is going because they've never experienced the device before. You would also need to assume that in the excitement of the moment, the person has no idea to calculate (aka, how hard to pull) on the 2nd spin how to get to the \$1.00.

So, for you to get to a dollar, the odds are 1/20 (on the first spin) + 19/20 *1/20 = 9.75%.
The 2nd thing is that you would not try for the dollar if you have a reasonable expectation of going to the showdown which means that you would not try for the \$1 if (1) you have a reasonable expectation of winning or (2) you are the last spinner and you've already won. For example, if the first two people go "out" you would not try for the dollar if your first spin was a nickel because you would not take the .25% odds of winning \$25K over the 50/50 odds of winning around the same amount of merchandise in the showdown. That would be -EV.

That is, if your first spin is above the competitors and gives you a reasonable expectation (>50%) that you will go to the showdown you would not try the 2nd spin.

The odds of a tie without the dollar are hard to express because you have to assume an optimal betting assumption:
That is, what is optimal betting when you are batting first, second, and third.

That is over my head for the moment. However, when you are not batting first, you have to always spin if you are below the competitor's total.

Some kind of combinatoric analysis is required.
----- You want the truth! You can't handle the truth!
ams288
Joined: Sep 26, 2012
• Posts: 5461
November 1st, 2017 at 10:28:48 AM permalink
Did anyone see this?

I happened to be watching live as it happened and it was awesome. They were celebrating Drew's 10th anniversary hosting the show that week so 1.00 spins on the wheel were worth \$10,000.

Not sure if the embed works anymore so here's the link:

Ding Dong the Witch is Dead
TigerWu
Joined: May 23, 2016
• Posts: 4022
November 1st, 2017 at 10:31:47 AM permalink
I don't think you can calculate this for the real world because a major variable is that the contestant has control over the spin. If you know the location of the number you need, you can make an attempt to spin the wheel with a certain amount of force so that it stops close to where you want it. If it is your second or third spin, you would have more information as to the amount of strength required for this that you didn't have on your first spin. That means not all numbers would have the same odds of hitting.

On the other hand, I guess most people don't do this. I think they just grab the wheel and go, in which case if might be possible to calculate the odds since they are in effect just spinning blind.
Ayecarumba
Joined: Nov 17, 2009
• Posts: 6763
November 1st, 2017 at 11:06:29 AM permalink
Quote: TigerWu

I don't think you can calculate this for the real world because a major variable is that the contestant has control over the spin. If you know the location of the number you need, you can make an attempt to spin the wheel with a certain amount of force so that it stops close to where you want it. If it is your second or third spin, you would have more information as to the amount of strength required for this that you didn't have on your first spin. That means not all numbers would have the same odds of hitting.

On the other hand, I guess most people don't do this. I think they just grab the wheel and go, in which case if might be possible to calculate the odds since they are in effect just spinning blind.

Odds of hitting \$1 on the first spin: 1/20
Odds of not hitting \$1 on the first spin, then totaling \$1 on the second spin: 19/20 x 1/20
Odds of hitting \$1 on the bonus spin: 1/20

(1/20 + (19/20 x 1/20)) X 1/20 = (.05 + (.95 X .05)) X .05 = 0.004875

Assumptions:
Player always uses second spin if they don't hit the \$1 on the first spin.
Spins are random

*I recall that the host will set the wheel to a certain position (either on the \$1 or the slot before or after) before the first contestant's first spin, but not for the second. Does it get reset to the same position before each contestant's first spin? I think this makes the results less random. Also, how are the amount slots distributed around the wheel? Are they set so that amounts totaling \$1 are equally distant for all non-\$1 totals?
Simplicity is the ultimate sophistication - Leonardo da Vinci
TigerWu
Joined: May 23, 2016
• Posts: 4022
November 1st, 2017 at 11:18:52 AM permalink
Quote: Ayecarumba

Does it get reset to the same position before each contestant's first spin?

Yes, the wheel is reset to \$1.00 for each contestant. (<-- I was wrong about that part) It's also reset to \$.05 for each bonus spin.

Quote:

Also, how are the amount slots distributed around the wheel? Are they set so that amounts totaling \$1 are equally distant for all non-\$1 totals?

The numbers are in this order: 5, 1.00, 15, 80, 35, 60, 20, 40, 75, 55, 95, 50, 85, 30, 65, 10, 45, 70, 25, 90.
Last edited by: TigerWu on Nov 1, 2017
ThatDonGuy

Joined: Jun 22, 2011
• Posts: 4926
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November 1st, 2017 at 12:55:27 PM permalink
Quote: TigerWu

Yes, the wheel is reset to \$1.00 for each contestant. It's also reset to \$.05 for each bonus spin.

The second part is right, but the first is not. The wheel starts at 1.00 for the first contestant's first spin, but is never reset after any spin except for when a bonus spin is taken. (I think the reason it is set to the 5 on a bonus spin is, it prevents the possibility of it starting on the dollar, and then being spun to the 5, only for the host to point out that it doesn't count because it did not go all the way around once.)

Note that it is also not reset if tiebreaker spins are needed when two bonus spins end up in another tie.
Wizard

Joined: Oct 14, 2009
• Posts: 23301
November 1st, 2017 at 3:46:40 PM permalink
Let's assume the spin is fair and every outcome has a 5% chance.

I think we can also assume optimal player strategy.

The hardest part of the problem is calculating the probability of a tie.

I would suggest as a warm-up, calculate the optimal player strategy. I state the answer on my math problems site (problem 104).
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard

Joined: Oct 14, 2009
• Posts: 23301
November 1st, 2017 at 4:13:16 PM permalink
Let me suggest another two more manageable problems:

Question 3: What is the probability of at least one \$25,000 winner given a two-player tie?

0.102362204724409

Question 4: What is the probability of at least one \$25,000 winner given a three-player tie? (like in the video above)

0.156123746654209

In both cases, do not count the probability of winning the \$25,000 in the first round. Just the tie-breaker round(s).
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard