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The following situation happened on the Price is Right and I was curious about the optimal response. The rules are: A contestant rolls a wheel with 5 cent increments from 5 - 100 (20 numbers total). A contestant can choose to spin the wheel once and accept the number the wheel landed on (stay) or spin again and have this new number added to the previous number. If the contestant has a number that is over 100 cents they automatically lose. The object of the game is to roll the highest number out of a set of contestants.

GirlA rolled once and rolled 60 cents. GirlA knows GirlB will play afterwards. Should GirlA roll again? At what number is the expected value neutral / what range of numbers should GirlA stay on? If there is more than one player behind, what range of numbers is best to stay on?

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  • $\begingroup$ If I understand the rules, $A$ wins if two conditions hold: (1) Her combined results add up to no more than 100; and (2) if her total is valid, it still must be higher than the various totals of all the following contestants (provided their totals are also valid). Is that right? $\endgroup$ – Théophile Jul 11 '14 at 19:56
  • $\begingroup$ Yes. Also assume that if there is a tie then the first person to have the number that is tied wins. $\endgroup$ – mathewbruens Jul 11 '14 at 20:04
  • $\begingroup$ I remember solving this problem on rec.puzzles around 1993. That seems more recent to me than it is. :-o $\endgroup$ – Brian Tung Feb 14 '17 at 7:45
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I'll look at the two player version, although IIRC the TV show was for three players. Also, instead of the values being 5 to 100 in steps of 5, WLOG just consider the wheel to be 1 to 20.

Well I believe the first thing to figure out is, if player one retires with value $x$, what is probably that player two wins? He can get either more than $x$ on his first roll, or with two rolls:

  • hit $y$ on the first roll
  • hit more than $x - y$ but no more than $20 - y$ on the second roll

The chance of second player Winning in one roll is then

$$\sum_{y=x+1}^{20}\frac{1}{20} = \frac{20 - x}{20}$$

and winning in two rolls is

$$\sum_{y=1}^x\overbrace{\frac{1}{20}}^\text{Chance of hitting y} \times \underbrace{\sum_{z=x - y + 1}^{20-y}\frac{1}{20}}_\text{Chance of hitting a winning second roll}$$

$$=\frac{1}{20\times 20}\sum_{y=1}^x \sum_{z=x - y + 1}^{20-y} 1$$

$$=\frac{1}{20\times 20}\sum_{y=1}^x 20 - x$$

$$=\frac{20x - x^2}{20\times 20}$$

So if player 1 retires with $x$, then the chance that player two wins is $p_2(x) = \frac{20 - x}{20} + \frac{20x - x^2}{20\times 20} = \boxed{1 - \frac{x^2}{400}}$

Ok so now we have to answer the second part: if you are player 1, and you roll $x$, should you reroll? For this you calculate the chance of winning if you reroll, and the chance of winning if you won't reroll, and choose which one is larger.

If you reroll, in order to win:

  • roll value $z$ between $1$ and $20 - x$ inclusive
  • player 2 must lose with player 1 retiring on $x + z$

$$\begin{align} \text{reroll win chance} &= \sum_{z = 1}^{20 - x} \frac{1}{20} (1 - p_2(x + z)) \\ &= \frac{1}{20} \sum_{z = 1}^{20 - x} \frac{(x+z)^2}{400} \\ &= \frac{1}{20^3} \sum_{z = 1}^{20 - x} x^2+2xz +z^2 \\ &= \frac{1}{20^3} \left((20-x)x^2+2x\frac{(20 - x)^2 + (20 - x)}{2} +\frac{(20-x)((20-x)+1)(2(20-x)+1)}{6} \right)\\ &= \frac{17220 - x - 3x^2 -2x^3}{48000} \end{align}$$

If you don't reroll, the chance of your winning is $1 - p_2(x) =\frac{x^2}{400}$.

So summarized:

$$ \newcommand{\cent}{{\mathrm{c}\mkern-6.5mu{\mid}}} \begin{array} {|c|c|c|} \text{Player 1 first roll} & \text{Winning chance if reroll} & \text{Winning chance if stay} \\ 5 \cent & 0.358625 & 0.0025 \\ 10 \cent & 0.358125& 0.01 \\ 15 \cent & 0.357 & 0.0225 \\ 20 \cent & 0.355 & 0.04 \\ 25 \cent & 0.351875& 0.0625 \\ 30 \cent & 0.347375& 0.09 \\ 35 \cent & 0.34125 & 0.1225 \\ 40 \cent & 0.33325 & 0.16 \\ 45 \cent & 0.323125& 0.2025 \\ 50 \cent & 0.310625& 0.25 \\ \hline 55 \cent & 0.2955 & 0.3025 \\ 60 \cent & 0.2775 & 0.36 \\ 65 \cent & 0.256375& 0.4225 \\ 70 \cent & 0.231875& 0.49 \\ \hline 75 \cent & 0.20375 & 0.5625 \\ 80 \cent & 0.17175 & 0.64 \\ 85 \cent & 0.135625& 0.7225 \\ 90 \cent & 0.095125& 0.81 \\ 95 \cent & 0.05 & 0.9025 \\ 100 \cent & 0 & 1 \end{array}$$

So player 1 should reroll if he gets anything less than $x=11$ or 55 cents. But unless he gets at least 75 cents, he's probably going to lose anyway. That's a pretty unfair game, player 1 only has a ${6 \over 20}$ chance of winning overall.

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Long story short: She should not roll again.

Supposing Girl A does not spin again, we have the following possibilities.

Case 1: It is clear that there is an $\frac{8}{20}=\frac{2}{5}=40$% chance that Girl B rolls (strictly) higher than 60 (in which case, Girl B wins). In this case, Girl B holds (and presumably wins).

Case 2: Girl B rolls less than or equal to 60. This outcome is $60$% likely. Then, Girl B must roll again to beat girl A. Note that if Girl B rolls a $5n$ on her first attempt, where $n\in \{1,...,12\}$, the probability of her rolling "over" is $5n$%. What Girl B needs to roll enough so that her score is greater than 60 (and not exceeding 100) is $\{60-5(n-1),...,100-5n\}$. This occurs $\frac{8}{20}=40$% of the time.

Putting all of this together, if Girl A remains at 60, then Girl B has a $.4+.6\cdot.4 =.4+.24=.64=64$% chance of getting a higher score than Girl A.

Now, if Girl A decides to roll again, she has a $60$% of rolling over (or losing right away). Given this new score, say $5m$ (that is not a roll over), we have that Girl B can either win on her first roll or roll again. Note that $m\in \{13,...,20\}$.

Case 1: Girl B rolls greater than Girl A (in the case where $m\neq 20$) we have this is $100-5m$%.

Case 2: Girl B rolls less than or equal to than Girl A. This has probability $5m$%. Suppose that Girl B rolls $5t$. Then, the probability that Girl B, on her second turn, gets higher than Girl A, but not rolling over, we have $\{5m-5(t-1),...,100-5t\}$.

Now, we need to put these together to arrive at our final answer: The probability of Girl A not rolling over and Girl B beating her is $(.4)(\sum_{i=1}^8(.1)(\frac{8-i}{20}))=(.4)(\frac{7}{50})=(.4).14=0.056$. Therefore, the probability of Girl A losing is $.6+0.056=.656=65.6$%.

In conclusion, Girl A is in a very tough position. There is a $65.6$% chance of her losing if she rolls again and only a $64$% chance of losing if she chooses to not roll. Therefore, it's very close, but she shouldn't roll again.

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I made a python function to test this problem.I will put the code at the bottom and you may test it at this site.

Please note that I used a 1 to 100 system rather than a 1 to 20 system.

Scroll to bottom for conclusion

I found that when the first number that is spun gets to 50 the odds of you winning and losing are approximately equal.

However, as the number you first spin gets higher, the average amount you gain drops.

Here are 3 test each with 100,000 spins.(Please note these are just approximations)

When the number you have spun on your first spin is 40.

Wins:59473
Losses:40527
Average gain:30.46

When the number you have spun on your first spin is 50.

Wins:49593
Losses:50407
Average gain:25.505

When the number you have spun on your first spin is 60.

Wins:39597
Losses:60403
Average gain:20.466727277318988

As you can see, every time the number we spun at first is increased by 10, the average amount gained goes down by 5.

The obvious conclusion is that if you rolled a 60, you should definitely not spin again. The highest roll were the odds would be in your favor would be a 45.

from random import randint

def test(spin):
    w=0
    l=0
    gain=0
    wins=0
    tests=1000
    for i in range(tests):
        new=spin+randint(1,101)
        if new>100:
            l+=1
        else:
            w+=1
            gain+=new-spin
            wins+=1

    return ("Wins:"+str(w)+"\nLosses:"+str(l)+"\nAverage gain:"+str(gain/wins))



#Set This equal to the number the person spun first
first_spin=50

print(test(first_spin))
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There is actually a published paper about that question:

"To Spin or Not To Spin? Natural and Laboratory Experiments from The Price is Right", Rafael Tenorio and Timothy N. Cason, The Economic Journal, 2002.

Your question is answered in proposition 1:

Should girlA roll again?

No.

At what number is the expected value neutral / what range of numbers should girlA stay on?

At 50 she is indifferent and for >50 she should stay.

If there is more than one player behind, what range of numbers is best to stay on?

For 3 they show that 65 is the number to stay on, for a higher number of players? No idea (and it is not really easy to derive as you have to recalculate the strategies of everyone else).

Btw: They state that Contestant 1 wins 30.82% of the time, Contestant 2 wins 32.96% of the time, and Contestant 3 wins 36.22% of the time.

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More on this problem in the paper P.Coe and W. Butterworth, Optimal stopping in the showcase showdown, American Statistician, 49, 1995, p. 271-275. The problem and variants are also extensively discussed in the book Understanding Probability by Henk Tijms

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This is a great question and I record the Price is Right daily to watch only the big wheel. I looked at the proposed answers above and they are brilliant answers to the wrong question.

The reason the calculation of the probabilities is so difficult is that there are three people involved, each with a choice of whether to spin the wheel once or twice, except in the instance where the first two contestants go over a dollar in which case the third contestant automatically wins, but only gets to spin the wheel once for an initial score. The further complication is that there can be ties which does not appear to be accounted for in the calculations above. If there is a tie and the tie is the highest score without going over, there is a spin-off where each contestant gets one additional spin and the dollar bonus is still in play. Further, if any contestant's spins total one dollar, they are awarded $1,000 and an additional spin. If on an additional spin, they land on 5 or 15 cents, they win an additional $10,000 and if they land on a dollar, they win an additional $25,000. One more thing to consider is that the numbers are in a particular non-sequitur order. I would say maybe that doesn't matter except that a contestant can't just spin the wheel targeting a particular range (other than the 5 cent, 1 dollar, 15 cent range on the additional spin) even though many try to do so.

The winning percentages pointed out in the article seem about right, intuitively, and while the calculations above suggest that the first person to spin is severely disadvantaged, by observation, it has not been the case. (I have been recording the show and watching who wins on the big wheel for nearly a year. That's about 420 actual big wheel contests.) The third person already know the scores of the first two contestants and has no incentive to quit spinning with a losing number. The second contestant knows the score of the first and again has no incentive to quit unless they either beat or tie the first contestant. The first contestant knows that there will be two contestants spinning after them and has an incentive to try to get a higher number

But, each contestant, at least on their first spin has no "feel" for the wheel, so it might be fair to say that the first spin of each contestant is completely random. And, the second spin may only be slightly influenced by the "feel" gained in the first spin. Again, the order of numbers on the wheel is not random, but I don't have it to give. So, at least for the initial two spins, I would say you could ignore "feel," but it is the whole idea of the dollar bonus spin, so don't think contestants are unaware of this factor.

All of that said, on the very first spin of each contestant, the probability for any score is going to be 1/20. To determine if the first contestant should take a second spin, the risks are:

A. What is the probability that the score on the first spin, when added to the score of the second spin will exceed one dollar and cause the contestant to automatically be disqualified?

B. What is the probability that on one spin, the second contestant will a)tie or b) surpass the total of the first spin.

C. What is the probability that on two spins, the second contestant will a)tie or b) surpass the total of the first spin without going over a dollar.

D. What is the probability that on one spin, the third contestant will a) tie or b) surpass the total of the first spin.

E. What is the probability that on two spins, the third contestant will a) tie or b) surpass the total of the first spin.

F. What is the probability that the first contestant on a second spin, would score a total of one dollar or less. (This should have been first or second in my list.)

So, what I get from this list is that if:

A + F >= Ba + Bb + Ca + Cb

the first contestant should not spin.

If the first spin is 50 cents, then A + F would be 1. Ba would be .5. Bb would be .25. Ca would be .5 and Cb would be .25. The total for the two following contestants is 1.5, the first contestant should spin again.

If the first spin is 55 cents, then A + F would be 1, so there is a flaw. A should be the probability that the second spin will cause the total to be less than or equal to a dollar so that the term totaling to spin again number goes down as the score in the first spin goes up. Recalculating that way, A+F equals .45 + .45, .90. I'm missing something but will push through to see if it shakes out. Ba would be .45, Bb would be (Okay, well first, tie would only be 1/20 on the first spin always, so that part is wrong. Wow, I am crumbling, I totally forgot D and E above. I guess I'm too sleepy.) Help!

I broke down most of the elements of the probability problem, although something is telling me I still missed two of them (one I suspect is ties on the first contestant's side of the equation). But, I can't think through how to set up the equation. The rules are there. Holy smoke, its 2:30 am.

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First, we have to lay out some assumptions. 1) The players have no "skill" in that they can get a "feel" for the wheel and improve their odds of hitting a specific number (or range) on subsequent spins. So, each spin is independent and random. 2) The goal for the players is only to beat the other two contestants, and they will not try to get to $1.00 if it jeopardizes their odds of beating the other contestants. The $1.00 bonus spin is just that, a separate bonus not calculated into your spin strategy. 3) All the players will spin optimally, as we don't have any prior information about their computational abilities. I suppose if you wanted to solve this more for "real world" application you could allow for some degree of sub-optimal play (such as in blackjack when people hit on 13 vs. 5 because "they have a gut feeling"). So, for the sake of this solution, we are only looking at the pure math, and aren't factoring in skill, greed, and other's play ability.

To solve the three person problem, it's easiest to work backwards. The last person to spin (Player C) doesn't have any choice other than if they tie the best score. If their first spin is too low, they must spin again. If higher, they win unless they spin > $1.00. If they tie, they can choose to go to a one spin spin-off, or spin again. Since the spin-off has a 50% chance, we can assume Player C will choose to spin again if they tie the best score with 0.50 or less (since they would have a better chance of beating the tie with a second spin). So, we essentially know the "rules" for the Player C.

We can then build out the conditional probabilities for each scenario of Player B's first spin to determine their optimal strategy vs. Player C.

1st Roll    Stay      Spin       Max
 0.05    0.00250     0.33988     0.33988 
 0.10    0.00875     0.33944     0.33944 
 0.15    0.02000     0.33844     0.33844 
 0.20    0.03625     0.33663     0.33663 
 0.25    0.05750     0.33375     0.33375 
 0.30    0.08375     0.32956     0.32956 
 0.35    0.11500     0.32381     0.32381 
 0.40    0.15125     0.31625     0.31625 
 0.45    0.19250     0.30663     0.30663 
 0.50    0.23875     0.29469     0.29469 
 0.55    0.28750     0.28031     0.28750 
 0.60    0.34125     0.26325     0.34125 
 0.65    0.40000     0.24325     0.40000 
 0.70    0.46375     0.22006     0.46375 
 0.75    0.53250     0.19344     0.53250 
 0.80    0.60625     0.16313     0.60625 
 0.85    0.68500     0.12888     0.68500 
 0.90    0.76875     0.09044     0.76875 
 0.95    0.85750     0.04756     0.85750 
 1.00    0.95125     0.00000     0.95125 

Assuming Player B beat Player A's score on their first spin, and had a choice whether or not to spin again, they should spin again on 0.50 or lower, and stay on 0.55 or higher. (Note that 0.55 gives the worst odds to win, similar to getting a 16 vs. 2 in blackjack. It's probably not enough to hold up, but if you take another card, you'll still lose.) So, here we have essentially solved the two-person problem.

NOTE: Player B has a disadvantage over Player C (45.76% chance of winning), because they are forced to choose first. This is very similar to blackjack, where the house has the edge by being the last player and just follows a strict set of rules.

Now that we know what Player B will do on their first spin (spin again if less than Player A or 0.55), we can calculate a similar matrix of odds for Player A's choices:

1st Roll    Stay      Spin       Max
 0.05    0.13750     0.37438     0.37438 
 0.10    0.13875     0.36744     0.36744 
 0.15    0.14250     0.36031     0.36031 
 0.20    0.14875     0.35288     0.35288 
 0.25    0.15750     0.34500     0.34500 
 0.30    0.16875     0.33656     0.33656 
 0.35    0.18250     0.32744     0.32744 
 0.40    0.19875     0.31750     0.31750 
 0.45    0.21750     0.30663     0.30663 
 0.50    0.23875     0.29469     0.29469 
 0.55    0.28750     0.28031     0.28750 
 0.60    0.34125     0.26325     0.34125 
 0.65    0.40000     0.24325     0.40000 
 0.70    0.46375     0.22006     0.46375 
 0.75    0.53250     0.19344     0.53250 
 0.80    0.60625     0.16313     0.60625 
 0.85    0.68500     0.12888     0.68500 
 0.90    0.76875     0.09044     0.76875 
 0.95    0.85750     0.04756     0.85750 
 1.00    0.95125     0.00000     0.95125 

This comes out very similar to the previous matrix, and 0.55 or higher is still the optimal choice for Player A to stop.

NOTE: Player A has a 46.38% chance to beat Player B, so again A is at the worst disadvantage since he has to go first.

Based on these given decisions, it is easier to now build out a 3-way probability matrix to determine each player's overall odds:

Player A   30.26%
Player B   33.01%
Player C   36.74%
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