The Price is Right optimal play 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?
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 $x + z$


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

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

So summarized:

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

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.
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%