Optimal "Price is Right" strategy

[BadPenguin]

I was sitting in a deli the other day waiting for my sandwich to be made and "The Price is Right" was playing on the TV. It was the part at the end of the first half where they see which one of several people gets to go on to the ... big thing at the end, I forget what it's called. Anyway, the game they play at this stage is the following:

There's a big wheel marked from $.05 to $1.00 in increments of 5 cents. There are 3 players (maybe sometimes 2 or 4, I'm not sure) who play in some predetermined order. The first player spins the wheel and sees where it stops - he can stop and take the result of that spin as his score or he may immediately spin again, in which case the result of the second spin is added to the first to give his score. The object is to get as high a score as possible without exceeding $1.00 . If you go over $1 you lose.

Then the second player does the same thing, again with the option to spin either once or twice. He needs to score higher than the first player (if the first player didn't go over $1, that is) in order to take the lead. The third player then follows in the same way.

Obviously it's a big disadvantage to be "under the gun" in this game. The question is, what's the optimal strategy to follow if you're spinning first? (Or, more generally, what's the optimal strategy if there are n players yet to spin behind you and everyone in front of you went over $1 and is out.) Clearly if your first spin is .05 you're going to spin again, and if it's $.95 or $1.00 you're not. Where is the crossover point? In other words, what's the lowest first spin that will cause you to stand on that score?

I tried to solve this both mathematically and computationally and got different answers, so either my math or my code is wrong - probably my math, I'm guessing. (It might be easier to approach the problem - at least on the mathematical side - if you assume the wheel can return any real number between 0 and 1.00 rather than only multiples of .05.)

[TheHammer24]

First post on the probability forum, I apologize that my answer is probably incomplete let alone wrong.

Let's say your the first person and you spin a value x from (1 to 20) x 5. To calculate your possibility of winning, you must figure out the number of values on the wheel that beat you (i.e. if you spin 90, 95 and 1.00 beat you) = y.

Your opponent will have that oppurtunity twice since if he does not beat you on the first spin he will have y chances of winning on the second spin. So your chances of winning are ((20-y)/20)((20-y)/20) / (number of opponents to spin).

So if you spin .90 with two opponents behind you, two values beat you. (18/20)(18/20)/(2) = 62% chance of winning.


However, you would have to figure out your chance of winning if you improve minus the likely hood you would go over since thats an automatic dq. If your winning chances were better spinning again you would spin again.


y1 = (1 - x+1 five cent increment)

< {[(20-y1)/20)*((20-y1)/20)] / (number of opponents to spin)+[(20-y2)/20)*((20-y2)/20)] / (number of opponents to spin)+ ....} / y > - <(1-y)/20>


I hope this at least seems logical

[PokerGoblin]

Just play Plinko. That game rulez supreme.

PG

[Tom1975]

http://www.nd.edu/~ndmag/ilm2au97.html

[d10]

Yes, but unfortunately you don't get to choose which game you want to play. If you could I'm sure everyone would choose Plinko.

[Eegs]

I heard it's hard to get on the show. You have to be really enthusiastic and stand out in the crowd and wear a bob barker t-shirt or something. Cheerleaders have a huge edge.

[BadPenguin]

Thanks - I should have known somebody had already looked at it.

I wrote a Java program to simulate this and (generalizing that the wheel could stop anywhere from 0 to 1.00, not just on multiples of .05) came up with these, depending on the number of people yet to spin after you:

1 opponent = stop on .54 or higher, spin again on .53 or lower

2 opponents = stop on .66 …, spin again on .65 …

3 opponents = stop on .72 …, spin again on .71 … (this never happens on the actual show, since there are only three contestants maximum, I believe)

(So for the actual game, you’d stop on .55, .70, or .75 respectively.)

The 1 opponent result is one step off from the one in the article you linked to; the 2 opponent number is the same. I simplified by assuming that a later player had to actually beat you, rather than tie and go to a spin-off as I think happens on the show. That might explain the slight difference.

In the case of 2 people behind you, this also assumes that player 2 follows a simple-minded strategy and stops if his first spin beats your score, even if it's a low result. (You spin .05 and .10 for a score of .15; he spins .20 on his first spin. He should obviously spin again with one person yet to play behind him.) If you assume he follows an optimal strategy as well it might change things slightly - I haven't tried looking at that.

For the math geeks out there you can use freshman-level calculus to get a polynomial, one of whose roots is the answer to the problem for a given player position. However, you still need to solve the polynomial computationally since it doesn't factor nicely. (Anyone who wants the gory details can PM me.)

[Chris Dow]

Don't forget that if you can reach exactly 1 dollar you get I think $1000 and a spin where if you get green (2 spots) you get $5000 more and $10000 for hitting the dollar, that probably changes the ev fairly significantly.

[Snoogins47]

[ QUOTE ]
Don't forget that if you can reach exactly 1 dollar you get I think $1000 and a spin where if you get green (2 spots) you get $5000 more and $10000 for hitting the dollar, that probably changes the ev fairly significantly.

[/ QUOTE ]

There's some other factors too. If you're spinning the wheel the SECOND time and the person who won the first time is a complete feeb, you get a lot more value out of just GETTING to the showcase showdown than if he's a PhD of Consumer Economics with an emphasis in Automobile, Recreational Vehicle, Furniture, and Travel-Package pricing.

Don't forget to have your pets spayed or neutered.

[TheHammer24]

[ QUOTE ]
[ QUOTE ]
Don't forget that if you can reach exactly 1 dollar you get I think $1000 and a spin where if you get green (2 spots) you get $5000 more and $10000 for hitting the dollar, that probably changes the ev fairly significantly.

[/ QUOTE ]

lmao, nh


There's some other factors too. If you're spinning the wheel the SECOND time and the person who won the first time is a complete feeb, you get a lot more value out of just GETTING to the showcase showdown than if he's a PhD of Consumer Economics with an emphasis in Automobile, Recreational Vehicle, Furniture, and Travel-Package pricing.

Don't forget to have your pets spayed or neutered.

[/ QUOTE ]