The Price is Right - Should you stay or spin on 65 cents?

I was watching The Price is Right yesterday, and not being an expert on probability, I wondered what the best option that a contestant should take when spinning the wheel.

So the situation is as follows:

You are the first contestant to spin the wheel, meaning there are two more people about to spin after you. So on your first spin, you get 65 cents.

Do you have a higher probability of making it to the showcase if you stay on 65 cents or spin the wheel again and risk going over? If the wisest option is to spin again, at what point would it be wiser to stay? 70 cents? 75 cents?

I have seen people stay as low as 55 cents, and that has always bugged me.

Need more info, such as all of the values on the wheel.

I guess you never watch the show? The wheel has values from 5 cents to one dollar in five cent increments, thus there are twenty unique values on the wheel, more or less randomly distributed around it.

The point of the game is to get closest to exactly $1.00 without going over. You get up to two spins of the wheel.

The strategy is easy for the second person, they only spin again if their first spin is less than the total of the first person.

The OP asks, how high should the first spin be for the first person to hold?

Haj

At 65 cents each of the following people have 7 chances out of 20 in beating you on their first spin; combined that’s a 70% chance of you losing. Given that, I’d say 65 cents is too low and you should only stick at 70 cents or above. Not sure how to reason about the second spin.

Yes, stay on 65.

Assume that you stay on 65 and that the next contestant will spin again if he ties you on the first spin. Then your chance of losing if you stay is

7/20 (if the next contestant spins 70, 75, …, 100 on the first spin)

  • 7/400 (5+65, 5+70, …, 5+95 on two spins)
  • 7/400 (10+60, 10+65, …, 10+90 on two spins)
  • 7/400 (65+6, …, 65+35 on two spins)

for a total of 7/20+13*7/400 = 0.5775.

On the other hand, your chance of losing if you spin again is at least 13/20=0.65, because of the risk of going over 100. So staying has to be the better bet.

What if your first spin is 55? That’s harder. The chance of losing if you stay on 5n is exactly (20-n)/20 + n(20-n)/400. So for 55, the chance of losing if you stay is 0.6975. Your chance of going over on a second spin is only 0.55, but you can still lose even if you don’t go over, so your total chance of losing if you spin again is

0.55 + (0.64 + 0.5575 + … + 0.0975)/20 = 0.7035

So yes, stay on 55 as well.

You should also factor in the odds of going over $1.00 on that second spin…

Argh, sorry Orbifold, you replied at the same time I did. Disregard the last post.

Also factor in the fact that there are TWO people who would spin after you, not one.

There are 20 possible first spins. Of those 20, 7 lie above 65; 70, 75, 80, 85, 90, 95, and 100. Therefore, if you hold first at 65, there’s a 57.75% chance you’ll be beaten by one of the subsequent competitors just on her FIRST spin - holding at 65 essentially means you’re betting on yourself to lose even if your opponents weren’t allowed to spin twice.

I’m not willing to do enough number crunching to figure out the odds you’ll be beaten by a two-spin turn, but it’s got to be 70% likely or so if you hold at 65. My quick and dirty estimate is that by holding at 65 as the first player, you will be beaten by a first spin 57.75% of the time (that number is dead certain) and by a second spin about 9% of the time (that’s a rough guess based on a few quick calculations - figuring the average first spin below 65, the average expected second spin, figuring the likelihood the second spin will beat you, still not beat you or break 100, and then adjusting for the number of trials so you aren’t beaten two different ways.)

I am assuming here no player will be retarded enough to beat you on the first spin and then spin again; granting that The Price Is Right does not require a Ph.D. to compete, I’ve seen a thousand episodes and I’ve never seen anyone do that. If someone beats 65 on the second or third turn they always play conservatively and hold, so you can count on someone who follows you with a 70 to take their chances with it and screw you.

So by holding at 65, you should lose about 65-70 percent of the time.

However, if you spin on 65, you have a 65% chance of immediately losing anyway, since that’s your odds of going over 100, which means you lose no matter what your opponents do afterwards. Plus there’s a chance you can spin twice, not go over, but still lose. If you spin at 65, your average expected score the 35% of the time you don’t go over is about 82.5 - so if anyone nails 85, 90, 95, or 100, you’re history anyway. The odds of one of the two subsequent opponents getting 85 or better is a lot higher than you might think - 36% of the first spin alone, probably 45-50% overall.

So while not spinning as the first contestant at 65 will beat you around 65-70% of the time, spinning will beat you even more often, probably 80% of the time. You should hold at 65.

Clearly, of course, spinning first is a huge disadvantage. Since the second and third competitors can modify their strategy based on their spin - the third person never has to guess at all, and I’ve seen people win with 15 cents on the third spin because the first two people had to spin twice, busted, and gave the third competitor a free ticket. You can’t do much as the first competitor.

Wait, if there are two contestants, then wouldn’t the probability that both would lose be (1 - .4225) ^ 2 = 17.85%, meaning the probability that at least one of them would beat the first player be 82.15% > 65%? So it would actually be advantageous to spin again?

Or… actually, that would be (1 - .5775) ^2 = 17.85%.

Assume Beavis and Butthead are contestants two and three respectively.

Beavis tells Bob he wants to pass but he is going to pull one of two
cards out his back side after Butthead spins. Beavis will spin one or
two times, without regard to what he gets or the number he must beat, based
solely on the card drawn.

Butthead gets your 65 to beat. Even Butthead knows to stop if he exceeds
65 or to spin a second time if under. Hopefully if Butthead gets exactly
65 Bob will whisper to him that 7/20 is smaller than 1/2, the odds of
winning in a playoff spin.

So Butthead’s odds:
35% he will win on the first spin
5% he will hit 65 and stop for the playoff spin
60% he’s under and will spin again:
35% of these he will win
5% he will hit 65

The total then is:
35% + 35%(60%) + 1/2(5%) + 1/2(60% times 5%)=60%

Now Beavis steps to the spinner and he’s going to spin one or
two times completely independent of what he gets or what is needed.
Beavis may beat BH, but he can’t increase your chances
above the 40% left after BH.

You’re right…my analysis assumed that there was only one contestant following you. If there is only contestant following you then you should stay, but if there are two contestants following you than you actually should switch:

Your odds of beating one other contestant if you stay on 5n is f(n)=1-((20-n)/20+n(20-n))/400). Your odds of beating two contestants is therefore f(n)^2, if you stay. But if you spin again, your odds of beating one contestant are

(f(n+1)+f(n+2)+…+f(20))/20

while your odds of beating two contestants if you spin again is

(f(n+1)^2+f(n+2)^2+…+f(20)^2)/20

For n=13 (i.e., if you get 65 on the first spin) your odds of beating one contestant is 0.4225 if you stay and about 0.2564 if you spin again. But if there are two contestants following you, your odds of winning are about 0.1785 if you stay and 0.1979 if you spin again.

So to sum up: if there is one contestant following you, then spin again at 50 and stay at 55. If there are two contestants following you, then spin again at (pause for calculations)…it looks like you should spin again at 65 and stay at 70.

The odds of you spinning once, holding, then losing are (1-s^2) if you spin $.s. So the odds of doing so on a spin of .55 are .6975.

The odds of losing given a spin of $.s*5 and a second spin are

(odds of busting) + (odds of not busting) * sum from i = 1 to (1-s)/5 of [ (5/(1-s) * the odds of losing when standing with $.(s+i)*5 ].

That is, for .60,

.6 + .4*(1/8)(1-.65^2) + (1/8)(1-.70^2) + … + (1/8)*(1-1^2) = .7225

For .65,

.65 + .35*(1/7)(1-.70^2) + … + (1/7)(1-1^2) = .743625 vs. .5775 for not spinning

For .55,

.55 + .45*(1/9) well, you know… = .7045 vs. .6975 for not spinning

For .5 = .6704375 vs. .75 for not spinning

So you should hold exactly when you spin at least 55 cents.