Cross Validated Asked by EngrStudent on August 30, 2020

**Question:**

Is it incorrect to think there is a “sweet spot” where more samples slightly decreases the likelihood of a “Best pick” in the Secretary Problem?

**Details:**

The “Secretary Problem” from “optimal stopping” is a classic in decision theory. It also relates to the game “deal or no deal”, and selection of candidate employees.

It seems here are two “forces” working here: the more samples I have, the more likely that the first ones are far from the max, and the more samples I have the closer the max drawn is to the absolute max.

Here is a single run

- I uniformly randomly generated N candidates
- let k = 2 candidates be evaluated but not selected
- pick the next candidate who is better than the first 2
- compute the distance between picked candidate and the best drawn candidate and store that.

I wrapped this into a simulation (code below):

- sweep the total candidates from N=5 to N=30
- repeat the process many (n=1e5) times at each N
- (key) I then estimate the quantile where gap is equal to half the domain. This is the place where the results are a wash, where the approach yields a candidate that is half the domain away from the best. (I define this as a “dud”.)

When I run this I get that at N = 10 is a substantial transition in rate of duds, and I am tempted to think it is a plateau. The likelihood of a “dud” here is 13.671%. For all N greater than 10 the percentile is of the form 13.x%. however…

When I:

- plot the quantiles for N >= 10, and use a smoothing spline fit, there is

a local minimum at 20 - eyeball the numbers in the table there seems to be an interior minimum
- plot the values, and use a robust kernel fit, there are interior minima.

Is it legitimate to say there is an interior minimum? Is there an analytic, or other, way to dig into this? Did I miss something obvious?

Code:

```
library(pracma) #for fzero
N_loops <- 100000
N_trials <- 25
xz <- numeric()
# set.seed(1)
for (k in 3:30){
store <- numeric()
for (i in 1:N_loops){
y <- runif(n = k)
best_idx <- which(y==max(y),arr.ind=TRUE)
do_keep <- FALSE
my_keep <- 0
for (j in 1:k){
#draw one
yi <- y[j]
#find in rank
all_ranks <- rank(y[1:j])
this_rank <- tail(all_ranks,1)
#compare with acceptance value
if (j/exp(1) <1){
can_keep = FALSE
} else {
can_keep = TRUE
}
#do you keep
if (can_keep ==TRUE & do_keep == FALSE){
if (this_rank >3){
do_keep <- TRUE
my_keep <- yi
}
}
}
store[i] <- my_keep
}
store <<- store
myfit <- function(phi){
if (phi > 1){
phi <- 1
} else if (phi < 0){
phi <- 0
}
err <- quantile(x = store,probs = phi)-0.5
return(err)
}
temp <- fzero(myfit,0.1)
xz[k] <- unlist(temp)
print(xz[k])
}
```

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