Price is Right, Big Wheel

Wizard,

In your Wizard of Odds page regarding the Price is Right Big Wheel, you suggest that, if the player spins a 65 on his first spin, he should stay.

I don't believe you factored in the fact that each other player gets two spins to try to beat you. That’s four 35% chances to beat you that you have to withstand. If my math is correct, those four shots will beat you 82% of the time.

Whereas if you spin again after getting 65, you will only go over 65% of the time. It’s less disastrous to spin again.

Even at 70, you go over 70% of the time if you spin again, but lose 76% of the time if you stay.

It seems to me like 75 is the correct number to stay on.

-- Bing

Quote: Bing987

I don't believe you factored in the fact that each other player gets two spins to try to beat you. That’s four 35% chances to beat you that you have to withstand. If my math is correct, those four shots will beat you 82% of the time.
-- Bing

Are you sure it's 4 chances? If the first of the two players spins a 60 on his first spin, he doesn't get a second chance and start over - he has 60 and spins again hoping to get 40 or less.

Quote: rdw4potus

Are you sure it's 4 chances? If the first of the two players spins a 60 on his first spin, he doesn't get a second chance and start over - he has 60 and spins again hoping to get 40 or less.

What Bing987 is saying is:
(a) The second player has a 35% chance of landing on between 70 and 100 on his first spin, and then stopping, guaranteeing that the first player will lose;
(b) If the second player needs a second spin, there will always be 7 numbers that will result in the total being between 70 and 100 (for example, if the first spin was 50, the second spin can be anywhere from 20 to 50) - again, a 35% chance of eliminating the first player;
(c) Assuming the second player did not beat the first player, cases (a) and (b) apply to the third player.

Each of the four cases has a 65% chance of not happening, and the first player wins only if none of them happen; there's only about a 17.85% chance of that happening.

Right, but also a successful improving spin doesn't guarantee victory either. So standing on 65 is stronger than it looks.

I'll have to think about this a little later unless someone gets to it first.

deleted

Bing,

Good question. I like it when people challenge something, rather than to take everything at face value.

I'm sure the Wizard factored that in... that Player 2 and Player 3 each get two spins to beat Player 1. That's part of the rules (puzzle) and the Wizards explanation clearly indicate he's aware of this.

Here's the link for those who haven't seen it:

https://wizardofodds.com/ask-the-wizard/tv-game-shows/

I wrote a computer simulation (since I don't know how to solve it otherwise) to check this. I THINK I've got it all working correctly. And if so, the Wizard is correct.

Based upon 10 million games, if Player 1 spins again on an initial total of 65 or less:

Player 1 wins 27.04% of the time
Player 2 wins 28.96% of the time
Player 3 wins 32.13% of the time
Player 1 and Player 2 tie for the lead 3.52% of the time
Player 1 and Player 3 tie for the lead 3.70% of the time
Player 2 and Player 3 tie for the lead 4.15% of the time
All three players tie for the lead .49% of the time

(As expected, other runs give nearly identical percentages.)


Based upon another 10 million games, if Player 1 spins again on an initial total of 70 or less:

Player 1 wins 26.89% of the time
Player 2 wins 29.00% of the time
Player 3 wins 32.37% of the time
Player 1 and Player 2 tie for the lead 3.44% of the time
Player 1 and Player 3 tie for the lead 3.62% of the time
Player 2 and Player 3 tie for the lead 4.19% of the time
All three players tie for the lead .47% of the time


The difference is very small but it's there. For all runs, Player 1 wins more often if he spins again with 65 or less ("stands" on 70) than he does if he spins again with 70 or less (stands on 75).

I'll continue to play with the program and look for errors.

Actually, I misread his answer. He says to spin again on 65 and I read it as to stay on 65. So, he is implying to stand on 70, which seems close. My back-of-the-envelope scribblings look like the accurate answer is about 72, which isn't a possible answer on the Price is Right Big Wheel.

What started my line of thinking, that ended with an internet search that dropped me onto this page, is, "What is the worst possible number to spin on your first spin, assuming that you were the first player to spin the Big Wheel?" The player this morning spun a 75 and I thought, "Well, that's a damned-if-you-do, damned-if-you-don't situation."

But then I thought, "Spinning a 5 is pretty bad too." Anything from 5 to 50 on your second spin is going to almost guarantee (> 90%) that you get knocked out.

No worries Bing. Until you brought this up, my memory recalled the same thing. I remember reading his answer to the question years ago. My long term memory thought the Wiz said to stay on 65. However, rereading his original answer, that was incorrect.

I'm ignoring ties for the moment because its 230am for me right now:


Stand on 70:
Win Percentage: (0.65)^4 = 17.85%

Spin on 70 and Improve:
75 and Win: (0.7)^4 = 24.01%
80 and Win: (0.75)^4 = 31.64%
85 and Win: (0.8)^4 = 40.96%
90 and Win: (0.85)^4 = 52.20%
95 and Win: (0.9)^4 = 65.61%
100 and Win: (0.95)^4 = 81.45%

Each one of these scenarios have a 5% chance of occurring giving a weighted average of:
0.05*(24.01%+31.64%+40.96%+52.20%+65.61%+81.45%) = 14.79%

Stand on 65:
Win Percentage: (0.60)^4 = 12.96%

Spin on 65 and Improve:
70 and win: (0.65)^4 = 17.85%
...see the numbers above.

Again each of these numbers have a 5% weight: 0.05*(17.85%+24.01%+31.64%+40.96%+52.20%+65.61%+81.45%) = 15.69%

So this helps demonstrate we should spin on 65, and NOT spin on 70. I see many people stand on 65 when I watch and spin on 60.

Quote: Bing987

I don't believe you factored in the fact that each other player gets two spins to try to beat you.

Of course I factored that in! You must have a very low opinion of me.

You have to consider that if you spin again, even if you don't go over you still may not win.

This is also problem 104 on my math problems site. There I state that assuming everyone follows optimal strategy the odds of winning are:

Player 1: 30.82%
Player 2: 32.96%
Player 3: 36.22%

deleted