An almost straightforward riddle from The Riddler involving a triple gambler’s ruin: Dawn competes against three players Alessandra, Berenike, and Chinue, with probabilities of winning one round ¾, ½, and ¼, respectively, until the cumulated score reaches ±15, ±30, and ±45, for the first, second, and third games. What is Dawn’s optimal sequence of adversaries?
First, a brute force R simulation shows that the optimal ordering is to play the three adversaries first weakest, third strongest and middle fair:
ord=function(p){
z=2*(runif(1)<p[1])-1
while(abs(z)<15)z=z+2*(runif(1)<p[1])-1
y=2*(runif(1)<p[2])-1
while(abs(z+y)<30)y=y+2*(runif(1)<p[2])-1
x=2*(runif(1)<p[3])-1
while(abs(z+y+x)<45)x=x+2*(runif(1)<p[3])-1
return(x+y+z>0)}
mcord=function(p,T=1e2){
for(t in 1:T)F=F+ord(p)
return(F/T)}
comp=function(T=1e2){
return(c(mcord(c(.5,.55,.45),t),
#mcord(c(.5,.45,.55),t),#1-above
mcord(c(.55,.5,.45),t),
#mcord(c(.45,.5,.55),t),#1-above
mcord(c(.55,.45,.5),t)
#mcord(c(.45,.55,.5),t)))#1-above
))}
where I used probabilities closer to ½ to avoid estimated probabilities equal to one.
> comp(1e3) [1] 0.051 0.038 0.183
(and I eliminated the three other probabilities by sheer symmetry). Second, checking in Feller’s bible (Vol. 1, XIV.3) for the gambler’s ruin probability, a simple comparison of the six orderings confirms this simulation.
Considering a binary random walk, starting at zero, what is the probability of being almost sure of winning at some point only to lose at the end? This is the question set by 
