The 16 Game
October 29, 2013
A local store has a promotional game in which scratch-off cards have sixteen spaces that cover a random permutation of the numbers 1 through 16. Customers scratch off the spaces at random. If the scratch-off reveals the number 3, the card loses. If the scratch-offs reveal the numbers 1 and 2, in either order, the card wins. Winning cards receive a discount on store merchandise.
Your task is to write a program that determines the average winning percentage. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
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The only thing that matters is the relative order of 1, 2, and 3 in a permutation. Since 1/3 of the permutations have 1 and 2 before 3, the classical probability of winning is 1/3 and the expected percentage of winning cards in a pack is a little above 33. I didn’t write a program.
Solution in Racket.
#lang racket (define (shuffle! v) (define len (vector-length v)) (for ([i (in-range len)]) (define j (+ i (random (- len i)))) (define tmp (vector-ref v i)) (vector-set! v i (vector-ref v j)) (vector-set! v j tmp)) v) (define slots-per-card 16) (define (scratch-card card) (for*/fold ([x 0]) ([i (shuffle (range slots-per-card))] [c (in-value (vector-ref card i))] #:break (or (= c 3) (= x 2)) #:when (or (= c 1) (= c 2))) (add1 x))) (define (winning-% [tries 10000]) (define card (build-vector slots-per-card add1)) (define n (for*/sum ([_ (in-range tries)] [_ (shuffle! card)] [x (in-value (scratch-card card))] #:when (= 2 x)) 1)) (exact->inexact (/ n tries)))Oops! The winning-% function had a bug in the second for*/sum clause.
(define (winning-% [tries 10000]) (define card (build-vector slots-per-card add1)) (define n (for*/sum ([_ (in-range tries)] [_ (in-value (shuffle! card))] ;; <== [x (in-value (scratch-card card))] #:when (= 2 x)) 1)) (exact->inexact (/ n tries)))Trying Nimrod:
import math var numTrials = 1000000 j, tmp: int a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16] tot = 0 for n in 1..numTrials: for i in countdown(15,1): j = random(i+1) tmp = a[i] a[i] = a[j] a[j] = tmp if a.find(3) > a.find(2) and a.find(3) > a.find(1): tot += 1 echo(tot / numTrials)A slightly different approach, which recursively calculate the probability. It is “generalized” to take any N >= 3, but that doesn’t matter :-)
module TheSixteenGame(
solve
) where
solve :: Int -> Double
solve n
| n <= 2 = error "Input integer should be >= 3"
| otherwise = twoLeft n
twoLeft :: Int -> Double
twoLeft n
| n == 2 = 1.0
| otherwise = pickOne * oneLeft (n – 1) + pickNone * twoLeft (n – 1)
where
pickOne = 2.0 / fromIntegral n
pickNone = fromIntegral (n – 3) / fromIntegral n
oneLeft :: Int -> Double
oneLeft n
| n == 1 = 1.0
| otherwise = pickOne + pickNone * oneLeft (n – 1)
where
pickOne = 1.0 / fromIntegral n
pickNone = fromIntegral (n – 2) / fromIntegral n