Analyzing Double Plus One
Steven J Miller
sjmiller@math.brown.edu

In this note we analyze the strategy of double plus one, in particular, calculating the probability of getting five consecutive blacks in n spins (red and black each occur 50% of the time).

We first solve the difference equation. The behavior is governed by the root that is largest in absolute value, which here is a bit less than 1. This is to compute a[n], the probability of not having 5 consecutive blacks in n spins.

Solve[r^5 - (r^4/2) - (r^3/4) - (r^2/8) - (r/16) - 1/32 == 0., r]

We now use RSolve to solve the recurrence relation. As this involves a quintic, we put in a 16. inside so it approximates the solution and doesn't attempt to symbolically manipulate. Note how easy it is to incorporate initial conditions. As the probability of not having 5 consecutive blacks is always 1 if we have 4 or fewer tosses, the initial conditions are easy.

?RSolve

RSolve[{a[n+5] == a[n+4]/2 + a[n+3]/4 + a[n+2]/8 + a[n+1]/16. + a[n]/32, a[0] == 1, a[1] ==1, a[2] == 1, a[3] == 1, a[4] ==1.}, a[n],n]

We now have a formula for a[n]. What we want is b[n], the probability we do get 5 consecutive blacks in n spins.

b[n_] := 1 -((-0.0780088137387845`-0.06154991357128762` I) (((-0.33917535648499836`-0.22926809363657225` I))^n)-(0.0780088137387845`-0.06154991357128762` I) (((-0.33917535648499836`+0.22926809363657225` I))^n)+(0.04924136333569412`-0.08699322803629538` I) (((0.09768829732362702`-0.4244268202731228` I))^n)+(0.04924136333569412`+0.08699322803629538` I) (((0.09768829732362702`+0.4244268202731228` I))^n)+(1.0575349008061807`+2.922587190776818`*^-40 I) ((0.9829741183227426`)^n))

b[100.]
b[200.]
b[300.]
b[400.]

f[x_] := Abs[b[Floor[x]]]

Plot[{0,1,f[x]},{x,4,400}, PlotLabel -> "Probability get 5 consecutive blacks in n spins (fair roulette wheel)"]