In the show The Price Is Right, contestants spin the Big Wheel, which has 20 sections randomly distributed, from 5 cents to $1.00 in 5-cent increments. The objective is to get the highest amount without going over $1, with one initial spin and an optional second spin. Clearly the first contestant is at a disadvantage, as the later contestants simply opt for the second optional spin if the first spin is not enough. For the first contestant, there needs to be a strategy consisting of a critical value for the first spin, at or above which the second spin is not preferred, as the probability of going over is too high. It's also a discrete case, as the critical value can only be from 5 to 100, in increments of 5.
In the actual show, there are three contestants. For the sake for simplicity, only two contestants are assumed in this simulation. In the case of tie, rerun the trial. For each possible critical value from 5 to 100 cents, 1 million trials are run in this Monte Carlo simulation to determine the winning percentage of player 1 under that strategy.
Here's the graph with the critical point on the x-axis and the winning percentage on the y-axis:
The maximum occurs at critical value of 55, corresponding with a winning percentage of 45.55%. At a value of 50, the winning percentage is up there at 45.22%. Even closer is 45.52% at a value of 60. The winning percentage gradually tapers off. Even at a value of 70, the winning percentage is still at 44.25%. Here's the MATLAB script to generate the data and graph:
x = zeros(1,20);
y = zeros(1,20);
for crit = 1:20
x(crit) = 5*crit;
n = 1000000;
win = zeros(1,n);
for i = 1:n
done = false;
while ~done
firstSpin = ceil(20*rand)*5;
if firstSpin >= x(crit)
playerOne = firstSpin;
else
playerOne = firstSpin + ceil(20*rand)*5;
end
if playerOne > 100
playerOne = 0;
end
playerTwo = ceil(20*rand)*5;
if playerTwo < playerOne
playerTwo = playerTwo + ceil(20*rand)*5;
end
if playerTwo > 100
playerTwo = 0;
end
if playerOne > playerTwo
win(i) = 1;
done = true;
elseif playerOne < playerTwo
done = true;
end
end
end
y(crit) = sum(win)/n;
end
plot(x,y); grid on; xlabel('player 1 critical point'); ylabel('player 1 winning percentage')
In the code above, simply vary the value of n to change the number of trials at each strategy. Of course, this optimal strategy will be different if the third contestant is taken into consideration in the actual game.
