Apr 25, 2016
Two people join a dating site independently. Upon signing up they pick 5 words to describe themselves out of 25 given words. If 4 out of 5 of the words are the same the dating site declares they're a match. What's the probability those couples match?
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1 of 20 problems on a one hour test
This question was a 1st round automated interview test.
Given problems like this it's safe to assume the picks are independent both between the people, and even between the words themselves. In real life if the adjectives "vegetarian" and "steak eater" were in the same list it's unlikely the same person would pick both and the choices. Would not entirely be independent. Likewise if the two people were already dating it that also would not be independent. But this is a probability multiple choice test so assume independence.
Another trick which greatly simplifies the problem is to assume one person already picked 5 words so find the probability the other person picks 4 out of the 5 words. Mathematically it's the same but conceptually it helps me work with the problem more easily.
If you're unlucky there's a chance you'll blank on the elegant
way to solve this.
However if you're partially lucky you'll be sitting in front of a computer with no one around.
As it turns out I was partially lucky.
#Python Monte Carlo Simulation import numpy as np import pandas as pd s = pd.Series(list(range(24))) number_of_trials, matches = (0,0) while number_of_trials < 1000000: c1 = s[np.random.choice(24,5, replace = False)] c2 = s[np.random.choice(24,5, replace = False)] if c2.isin(c1).sum() >= 4: matches +=1 number_of_trials +=1 print(matches/number_of_trials)
The above code is as crude as it gets. Python randomly picks 5 integers out of 24 with no replacement twice. It turns compares it's picks. If 4 match the numerator increments by one. For each pair of random picks the denominator increments by one.
If we do this a million times we get a number really close to the closed form solution. It actually turns out that this simulation gets close to the real answer really quickly in close to 1000 iterations which is less than 2 seconds on my laptop.
This type of solution is called the Monte Carlo method. It's actually very useful for a wide range of problems that are too complicated for closed form solutions. But for this problem there are better solutions.
But before we get to the better solution let's say you're really unlucky. You both forgot the elegant method and don't have a computer. This problem can be brute forced.
To restate the problem this individual needs to pick 4 specific words out of 25 with no replacement with 5 picks. Let's call this individual Sam.
Limiting this to two picks on the first pick there's a 5 out of 24 chance that Sam will pick a "match" word and a 19 out of 24 that the person will pick a "non match" word.
However on the second pick the probability is dependent on the first pick. If the first pick was a "match" word there is now a 4 out of 23 chance that Sam will pick a "match" word. Likewise there is a 19/23 chance Sam will pick a non match word.
Since each pick is independent we can multiply the probabilities to get the chance of a sequence of picks. We also can add each way the picks could play out to get the total chance at least 4 match. To save yourself a lot of multiplication and division here's the math in Python.
print( #All five match (5/24)*(4/23)*(3/22)*(2/21)*(1/20) + #First four are match words, last one isn't (5/24)*(4/23)*(3/22)*(2/21)*(19/20) + #First 3 are match words, 4th doesn't, last does (5/24)*(4/23)*(3/22)*(19/21)*(2/20) + #First 2 are match words, 3rd doesn't, last 2 do (5/24)*(4/23)*(19/22)*(3/21)*(2/20) + #First matches words, 2nd doesn't, last 3 do (5/24)*(19/23)*(4/22)*(3/21)*(2/20) + #First doesn't match, last 4 do (19/24)*(5/23)*(4/22)*(3/21)*(2/20) )
If you multiplied and divided by hand it would be a pain but you would get the exact answer.
If you typed the above by hand into Python it would also be a pain and you'd get almost the right answer. (You don't because of floating point error)
If you were really on your game though you'd realize that this is a combinatorics problem. Essentially we need to pick X winning combinations out of Y possible combinations.
In combination notation it looks like this.
In python we can use the Combinations function to get the answer
from itertools import combinations def cl(n,k): ''' k combination of out length n set ''' return len(list(combinations(range(n),k))) numerator = cl(5,4) * cl(19,1) + cl(5,5)*cl(19,0) denominator = cl(24, 5) numerator/denominator
There are some additional tricks that would allow you to simplify the calculation even further but I leave it up to you to explore combinatorics further