(Scenario with 1 dice):
Denote $P(r,s)$ probability to get sum $s$ by $r$ rolls.
Denote $Q(r,s) = 1-\sum\limits_{q=1}^{s-1}P(r,q)$ $~$ probability to get sum $\ge s$ by $r$ rolls.
Obviously:
$$
P(1,1)=P(1,2)=\ldots=P(1,10)=0.1;
$$
then
$$
P(r,s) = \sum\limits_{q=s-10}^{s-1} P(r-1,q)\cdot 0.1
$$
(i.e. sum $s$ we can get of "previous" sums $s-10$, or $s-9$, ..., or $s-1$).
Step-by-step, we can fill table of probabilities $P(r,s)$.
A few conclusions:
$Q(16,100) = 1-\sum\limits_{q=1}^{99}P(16,q) \approx 0.159924$;
$Q(17,100) \approx 0.307628$;
$Q(18,100) \approx 0.483771$;
$Q(19,100) \approx 0.654096$;
$Q(20,100) \approx 0.791809$;
$Q(21,100) \approx 0.887108$;
$Q(22,100) \approx 0.944590$;
$Q(23,100) \approx 0.975252$.
Now you can choose closest values:
$25 \%$: ~ 17 rolls;
$50 \%$: ~ 18 rolls;
$75 \%$: ~ 20 rolls;
$90 \%$: ~ 21 rolls.
(Scenario with 2 dices):
As dices are independent, then all that we need is to consider even number of single rolls, and divide them by $2$.
$Q_2(8,100) =Q(16,100) \approx 0.159924 (\approx 15.99 \%);$
$Q_2(9,100) =Q(18,100) \approx 0.483771 (\approx 48.38 \%);$
$Q_2(10,100)=Q(20,100) \approx 0.791809 (\approx 79.18 \%);$
$Q_2(11,100)=Q(22,100) \approx 0.944590 (\approx 94.46 \%).$
In the following I shall treat the continuous version of the game: The rolling of the die is modeled by the drawing of a real number uniformly distributed in $[0,1]$.
Only the bettor can have a strategy, and this strategy is completely characterized by some number $\xi\in\ ]0,1[\> $. It reads as follows: Draw once more if the current sum $x$ is $<\xi$, and stay if the current sum is $>\xi$. The problem is to find the optimal $\xi$.
Claim: When the bettor stays at some $x\in[0,1]$ his chance $q(x)$ of losing is $=(1-x)e^x$.
Proof. The bettor loses if the successive drawings of the dealer produce a partial sum $s_{n+1}$ in the interval $[x,1]$. This means that there is an $n\geq0$ with $s_n\leq x$ and $x\leq s_{n+1}\leq1$. The probability for $s_n\leq x$ is easily seen to be ${x^n\over n!}$, and the probability that the next drawing causes $s_{n+1}$ to lie in the given interval of length $1-x$ is $1-x$. It follows that
$$q(x)=\sum_{n=0}^\infty{x^n\over n!}\>(1-x)=(1-x)e^x\ .$$
From this we draw the following conclusion: If the bettor has accumulated $x$ and decides to make one additional draw then his chances of losing are
$$\tilde q(x)=\int_x^1 q(t)\>dt\>+x =(2-x)e^x+x\ ,$$
whereby the $+x$ at the end is the probability of busting.
The function $x\mapsto q(x)$ is decreasing in $[0,1]$, and $\tilde q$ is increasing in this interval, whereby $q(x)=\tilde q(x)$ when $x=\xi$ with $\xi\doteq0.570556$. When $x<\xi$ then $\tilde q(x)<q(x)$. This means that an additional drawing at $x$ decreases the probability of losing. When $x>\xi$ then the converse is true: An additional drawing would increase the probability of losing.
Best Answer
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:
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:
$$\begin{align} \text{reroll win chance} &= \sum_{z = 1}^{20 - x} \frac{1}{20} (1 - p_2(x + z)) \\ &= \frac{1}{20} \sum_{z = 1}^{20 - x} \frac{(x+z)^2}{400} \\ &= \frac{1}{20^3} \sum_{z = 1}^{20 - x} x^2+2xz +z^2 \\ &= \frac{1}{20^3} \left((20-x)x^2+2x\frac{(20 - x)^2 + (20 - x)}{2} +\frac{(20-x)((20-x)+1)(2(20-x)+1)}{6} \right)\\ &= \frac{17220 - x - 3x^2 -2x^3}{48000} \end{align}$$
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.