The question I'm solving is if I receive 1/5 coupons at random a day, how many days will it take to collect 2 copies of each of the 5 coupons?
My code is only checking for 1 of each coupon instead of two but I'm not sure where I can make revisions. I edited the while True statement to include a list to my cards_own statement like this cards_own != [1,1,2,2,3,3,4,4,5,5] but then my output didn't print at all. I think maybe using the random.choice function would be a better option. Any recommendations?
import numpy as np
# number of simulations
n =10000
# defining function to check if we got all 5 cards indexed from 0 to 4
def all_cards(arr2,arr1=np.array(range(5))):
return np.array_equal(arr1,arr2)
days_taken = 0
# performing simulation
for i in range(n):
# initializing empty list to store card obtained in each day
cards_own =[]
# days taken in each simulation
days =1
# looping until i get 5 cards and count the days taken for finding average
while True:
value = np.random.randint(0,5)
cards_own.append(value)
if len(cards_own)>=5:
if all_cards(arr2=np.array(list(set(cards_own)))):
days_taken+=days
break
days+=1
average = days_taken/n
# recording the result in result with given precision
result = round(average,2)
print("Average days taken = ",result," days")
So there are a few things that I think you could change here. First off you can change your if statement to read if len(cards_own) >= 10 since you need at least ten coupons to have 2 of everything. Additionally, in your cards_all function, you can loop through all of the possible coupons (in this case [0, 4] and check if there are at least two occurences for each of them in the array (so you don't need arr2). Something like:
for i in range(5):
if np.sum(arr1 == i) < 2:
return False
return True
the if statement is just counting how many times the element i appears in the array. Also you should not be converting arr1 to a set in the function call since sets cannot have duplicate values so checking for 2 of each coupon would never work like that. Hopefully this helps!