fast track & slow lane

A riddle from the Riddler where N cars going at random (iid) speeds drive a road with a slow and a fast lane, each car choosing the fast lane iff any of the cars ahead in the slow lane is slowerthan them. With the question of the average number of car convoys.

If there were one single lane, the problem would be to determine how many times a smaller realisation and it has been solved in a much older Riddler. Namely

In the two-lane system, the slow one only gathers cars with speeds lowest than the current speed, i.e. a decreasing sequence of speeds, creating single car convoys.  The fast lane thus captures cars that are above the current minimum in the sequence, which, as it converges to the global minimum at some points, means that all following cars are found in the fast lane. i thought this would bring the number of convoys close to twice the above logarithmic sum (which sealed my destiny during an entrance oral exam 40 years ago!), but there are actually more of them for N large enough , which may be due to the possibility of the slow lane to capture more moderate speed cars in the beginning… The above compares the number of convoys for one lane (in red) and two (in gold), as well as the remarkable fit when regressing these numbers against log(N).

Here is the R code I used for that simulation

convoy=function(N,T=1e5){
  for(t in 1:T){
    speed=runif(N)
    slow=fast=NULL
    slow=speed[1]
    for(i in 2:N){
      if(speed[i]<min(slow))slow=c(slow,speed[i]) 
      else fast=c(fast,speed[i])} 
    F=F+length(slow) 
    if(length(fast)>0)F=F+1+sum(fast[-1]<cummin(fast)[-length(fast)])}
  return(F/T)}

the Kelly criterion and then some

The Kelly criterion is a way to optimise an unlimited sequence of bets under the following circumstances: a probability p of winning each bet, a loss of a fraction a of the sum bet, a gain of a fraction b of the sum bet, and a fraction f of the current fortune as the sum bet. Then

is the fraction optimising the growth

Here is a rendering of the empirical probability of reaching 250 before ruin, when starting with a fortune of 100, when a=1, p=0.3 and f and b vary (on a small grid). With on top Kelly’s solutions, meaning that they achieve a high probability of avoiding ruin. Until they cannot.

The Ridder is asking for a variant of this betting scheme, when the probability p to win the bet is proportional to 1/(1+b), namely .9/(1+b). In that case, the probabilities of reaching 250 (using the same R code as above) before ruin are approximated as followswith a maximal probability that does not exceed 0.36, the probability to win in one go, betting 100 with a goal of 250. It thus may be that the optimal choice, probabilitiwise is that one. Note that in that case, whatever the value of b, the Kelly criterion returns a negative fraction. Actually, the solution posted by the Riddler the week after is slightly above, 0.3686 or 1−(3/5)^9/10. Which I never reached by the sequential bet of a fixed fraction of the current fortune, eps. when not accounting for the fact that aiming at 250 rather than a smaller target was removing a .9 factor.

the other side of the dice

A simple riddle from the Riddler: if a standard fair six-faced dice is falling on an edge rather than a face, what is the expectation of the sum of the faces sharing this edge?

The solution proposed there is however somewhat convoluted, when the average is simply

since the only face that does not share an edge with face i is 7-i…

shelled and riddled

Consider a shell game with three shells and a ball with The Riddler constraint that the location of the shell with the ball is always exchanged with the location of an empty shell, randomly chosen. If one starts with the ball as rightmost, what is the distribution of the location of the ball after N steps?

Running an exploratory R code like

o=rep(0,3)
for(n in 1:1e6){
  b=c(0,0,1)
  for(t in 1:N){
    i=sample((1:3)[!b],1);b=0*b;b[i]=1}
  o=o+b}

shows that the difference in probability is between the rightmost position and both others, starting at zero, and evolving as p⁺=(1-p⁻)/2, with the successive values 0,1/2,1/4,3/8,5/15,11/32,… Very quickly converging to 1/3.

Goats do room

The riddle of the week is about 10 goats sequentially moving to their room, which they have chosen at random and independently (among ten rooms), unless another goat already occupies the room, in which case they move to the first free room with a higher number or fail. What is the probability that all goats end up in a room?

Coding the experiment is straightforward:

g=sample(1:N,N,rep=TRUE)
o=0*g
for(i in 1:N){
    if(min(o[g[i]:N])){f=f+1;break()
    }else{
      o[min(which(!o[g[i]:N]))+g[i]-1]=1
    }}}

returning an estimated probability of approximately 0.764.

As I had some free time during the early mornings at ISBA 2022, I tried to reformulate the question as a continuous event on uniform order statistics, turning to be at most one uniform larger than (N-1)/N, at most two larger than (N-2)/N, and so on… Asking the question on math.stackexchange quickly produced an answer that reversed engineered my formulation back to the goats (or parking lot), with a generic probability of

which of course coincides with the Monte Carlo approximation!

As an aside, I once drank South-African wines named Goats-do-Roam and Goat-Roti at my friends Jim and Maria’s place,  and they were quite enjoyable!