Statistically, on The Price Is Right, when should you spin again on the big wheel?

Question

You have to get as close to $1 with the sum of 1 or 2 spins without going over $1. The wheel has values from 5 cents to $1 in increments of 5 cents. Three contestants compete for who can get closest to a dollar without going over. If you spin a low value and decide to stay, you risk the person(s) after you beating you. If you spin a high value and decide to spin again, you risk going over. So what value is the cut off of when you should stop or spin again? This should differ if you are the first of the three to spin or if you are the second. If you are the third, it is easy because you know what you have to beat.

So far no one has got it. It is much more complicated than it seems.

I believe a good clue would be that your chances of going over are less than your chances of either person spinning higher than you.

Answer 1

Hmm, interesting question, actually, and a practical application of math. Hey, all you folks who ask what this stuff is good for, here is a real-life situation where you can use it!

Getting to the problem. Obviously, the contestant to go last spins again if his score is lower than the best score. If tied, there is a spin-off where his winning chance is 50%; so, if there is a tie, he should respin if he has 45 or less (giving him at least 11 out of 20 good numbers), not spin if he has 55 or more (giving him at most 9 out of 20 good numbers - better to go to the spin-off, where he has a 50% chance), while 50 is a "push".

Now, for the contestant to go 2nd to last: First, let's assume the first player busts (of course, the second player respins if his score is lower than the first player), which essentially makes this a two-player game. Based on what number the second player stops on, this is the winnning percentage I calculate for the last player:

100 - 4.875%
95 - 14.375%
90 - 23.375%
85 - 31.875%
80 - 39.875%
75 - 47.375%
70 - 54.375%
65 - 60.875%
60 - 66.875%
55 - 72.375%
50 - 77.375%

So, you may think the second player should respin whenever his odds are under 50%, but that is actually incorrect. For example, if the second player has 70, giving him a 45.625% chance of winning, respinning would give him a 70% chance of going bust - he is better off stopping with the 45.625% chance. To determine whether or not respinning improves the second player's chances, a second calculation is needed that considers each of the 20 scores after the second spin - some of them bust the player but the others improve his chance of winning. Based on that, here is a new table, giving the last player's chance of winning if the second player respins with the score given:

95 - 95.24375%
90 - 90.9625%
85 - 87.13125%
80 - 83.725%
75 - 80.71875%
70 - 78.0875%
65 - 75.80625%
60 - 73.85%
55 - 72.19375%
50 - 70.3125%
45 - 69.68125%

So, comparing the two tables, the critical points are 55 and 60. If the second player has 60, he should stop - the last player's chance of winning improves if the second player spins. At 55, however, the second player should spin, improving his odds by about 0.2%.


Moving on to the three player game... now it gets really messy. The player to go first needs a score that can beat the other two players. Does he stop on 55 again, hoping the other two players both bust? He'll need a better score than that to have a 50% chance of winning, but, as was noted in the two-player analysis, stopping with a 30% chance can be better than making a second spin. As for how to do a complete analysis, uh, beats me. This may require a computer to do all the math.

* * * * * *

Thinking about the three-player game, maybe it isn't as messy as I initially thought - the only additional consideration for the first player, as opposed to the second player, is that either player will stop if the first player's score is beaten. This can be calculated from the first table. For this table, listed is the chance of the first player losing (i.e. either player beats him) if he stands on the score given:

100 - 9.51%
95 - 26.68%
90 - 41.29%
85 - 53.59%
80 - 63.85%
75 - 72.31%
70 - 79.18%
65 - 84.69%

Similar to the two-player game, I can now calculate the odds if the first player decides to respin. Since I’m getting a little tired crunching these numbers, for this last table I’ll only list the critical ones:

70 - 83.4%
65 - 82.3%

In other words, the first contestant should respin on 65, stop on 70.

In conclusion, I think the optimum strategies are.....

First player: spin on 65, stop on 70

Second player: spin on 55, stop on 60 (assuming his score at least matches the first player)

Answer 2

I believe the first Answerer is mostly correct, although it's probably wise to point out that the $1.00 space on the wheel skews the probabilities somewhat. After all, there are 9 spaces less than 50, a single 50 space, and 10 spaces with values more than 50.

This feature of the wheel reminds me of the 0 and 00 spaces on some roulette wheels, which go in favor of "the house" and against "the player."

No Answer yet addresses what the Asker wanted: A discussion of strategies based on the player's position in line (1st, 2nd or 3rd). Obviously, the last player need only spin a second time if a previous player spun a higher result.

Hope this helps!

Answer 3

Statistically, the odds are in your favour of not going over $1 if you're at 45 cents or less, evens if you're at 50 cents and against you if you have over 50 cents.

Answer 4

60 cents or less and two more to spin 50 cents or less 1 more to spin