Generate a random number, excluding a single number

Question

I'm writing a Monty Hall simulator, and found the need to generate a number within a range, excluding a single number.

This seemed easy, so I naively wrote up:

(The g/... functions are part of my personal library. Their use should be fairly clear):

(defn random-int-excluding
  "Generates a random number between min-n and max-n; excluding excluding-n.
  min-n is inclusive, while max-n is exclusive."
  [min-n max-n excluding-n rand-gen]
  (let [rand-n (g/random-int min-n max-n rand-gen)
        rand-n' (if (= rand-n excluding-n) (inc rand-n) rand-n)]
    (g/wrap rand-n' min-n (inc max-n))))

This generates a random number within the range, and if it equals the excluded number, adds one; wrapping if necessary. Of course this ended up giving the number after the excluded number twice the chance of being picked since it would be picked either if it or the excluded number are chosen. Sample output frequencies for a range of 0 to 10 (max exclusive), excluding 2:

([0 0.099882]
 [1 0.100355]
 [3 0.200025]
 [4 0.099912]
 [5 0.099672]
 [6 0.099976]
 [7 0.099539]
 [8 0.100222]
 [9 0.100417])

Then I read this answer, which seemed much simpler, and based on it, wrote up:

(defn random-int-excluding
  "Generates a random number between min-n and max-n; excluding excluding-n.
  min-n is inclusive, while max-n is exclusive."
  [min-n max-n excluding-n rand-gen]
  (let [r1 (g/random-int min-n excluding-n rand-gen)
        r2 (g/random-int (inc excluding-n) max-n rand-gen)]
    (if (g/random-boolean rand-gen) r1 r2)))

Basically, it splits the range into 2 smaller ranges: from the min to the excluded number, and from excluded number + 1 to the max. It generates random number from these ranges, then randomly chooses one of them. Unfortunately though, as I noted under the answer, this gives skewed results unless both the partitions are of equal size. Sample output frequencies; same conditions as above:

([0 0.2499497]
 [1 0.2500795]
 [3 0.0715849]
 [4 0.071297]
 [5 0.0714366]
 [6 0.0714362]
 [7 0.0712715]
 [8 0.0715285]
 [9 0.0714161])

Note the numbers part of the smaller range before the excluded number are much more likely. To fix this, I'd have to skew it to pick numbers from the larger range more frequently, and really, I'm not proficient enough in maths in general to understand how to do that.

I looked at the accepted answer from the linked question, but to me, it seems like a version of my first attempt that accepts more than 1 number to exclude. I'd expect, against what the answerer claimed, that the numbers at the end of the exclusion range would be favored, since if a number is chosen that's within the excluded range, it just advances the number past the range.

Since this is going to be one of the most called functions in the simulation, I'd really like to avoid the "brute-force" method of looping while the generated number is excluded since the range will only have 3 numbers, so there's a 1/3 chance that it will need to try again each attempt.

Does anyone know of a simple algorithm to chose a random number from a continuous range, but exclude a single number?

Answer 1

To generate a number in the range [a, b] excluding c, simply generate a number in the range [a, b-1], and if the result is c then output b instead.

Answer 2

Just generate a lazy sequence and filter out items you don't want:

(let [ignore #{4 2}]
  (frequencies
    (take 2000
          (remove ignore (repeatedly #(rand-int 5))))))

Advantage to the other approach of mapping to different new values: This function will also work with different discrete random number distributions.

Answer 3

If the size of the collection of acceptable answers is small, just put all values into a vector and use rand-nth:

http://clojuredocs.org/clojure.core/rand-nth

(def primes [ 2 3 5 7 11 13 17 19] )
(println (rand-nth primes))
(println (rand-nth primes))
(println (rand-nth primes))

~/clj > lein run
19
13
11

Update

If some of the values should include more than the others, just put them in the array of values more than once. The number of occurrances of each value determines its relative weight:

(def samples [ 1 2 2 3 3 3 4 4 4 4 ] )
(def weighted-samples 
  (repeatedly #(rand-nth samples)))

(println (take 22 weighted-samples))

;=> (3 4 2 4 3 2 2 1 4 4 3 3 3 2 3 4 4 4 2 4 4 4)

If we wanted any number from 1 to 5, but never 3, just do this:

(def samples [ 1 2   4 5 ] )
(def weighted-samples 
  (repeatedly #(rand-nth samples)))

(println (take 22 weighted-samples))

(1 5 5 5 5 2 2 4 2 5 4 4 5 2 4 4 4 2 1 2 4 1)