Choose a Number

Probability Level 4

Six people choose a whole number from the range 1 to 10 (inclusive) independently and randomly. What is the probability that there will be two matching numbers and four non-matching numbers?

Example: 6, 2, 7, 4, 6, 10 works

Example: 2, 7, 1, 4, 4, 1 doesn't work

Example: 6, 2, 7, 3, 7, 7 doesn't work

Write your answer as a percentage.


The answer is 45.36.

Jesse Li by Jesse Li , 16, USA

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3 solutions

Jesse Li
Jun 19, 2019

Let's suppose that the second and first person choose the same number and the others choose different numbers.

The probability of that happening is 1 10 × 9 10 × 8 10 × 7 10 × 6 10 \frac{1}{10}\times\frac{9}{10}\times\frac{8}{10}\times\frac{7}{10}\times\frac{6}{10} because the second person has a 1 10 \frac{1}{10} chance of choosing the same number as the first person, the third person has a 9 10 \frac{9}{10} chance of choosing a number different from the first two people, and so on.

However, there are ( 6 2 ) = 15 {6 \choose 2} =15 ways of choosing the two people whose numbers match, so we must multiply the answer we got, 0.03024, by 15, to get 0.4536, which is 45.36 \boxed{45.36} as a percentage.

4 Helpful 0 Interesting 2 Brilliant 0 Confused

The "counting" version of this (if yours is the probability version) is to say there are 1 0 6 10^6 possible choices altogether, of which ( 6 2 ) × ( 10 1 ) × ( 9 4 ) × 4 ! = 15 × 10 × 126 × 24 = 453600 {6 \choose 2} \times {10 \choose 1} \times {9 \choose 4} \times 4! = 15 \times 10 \times 126 \times 24 = 453600 meet the criterion. This comes from ( 6 2 ) {6 \choose 2} ways to pick two people with the same number, ( 10 1 ) {10 \choose 1} choices of what that number is, ( 9 4 ) {9 \choose 4} ways to pick four other numbers, and 4 ! 4! ways to assign them to the remaining four people. (I'm not posting this as a separate solution because it isn't, really, but it gives a slightly different way of looking at it.)

Chris Lewis - 1 year, 11 months ago
David Stiff
Jun 19, 2019

Or in Python! 

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import random

def run_test(iterations):
    print('Running', iterations, 'tests...')
    test_results = []

    for test in range(iterations):
        choices = [random.randint(1, 10) for i in range(6)]
        numbers = set(choices)

        count_list = []
        for number in numbers:
            number_count = len([True for choice in choices if choice == number])
            count_list.append(number_count)

        pair_count = len([count for count in count_list if count == 2])

        if pair_count == 1:
            test_results.append(True)
        else:
            test_results.append(False)

    return test_results

# Modify the number here in run_test() to try a greater number of trials
# I wouldn't recommend going too far over 1,000,000 or else the interpreter will run out of execution time :)
test_results = run_test(1000000)

true_count = len([i for i in test_results if i == True])
false_count = len([i for i in test_results if i == False])
total = true_count + false_count

probability = true_count/total

print('Permissable:', true_count)
print('Not permissable:', false_count)
print('The probability then of obtaining a permissable set is:', probability)

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Kyle T
Jun 19, 2019

More of a brute force solution, but the clever thing here is we're looking for cases where we have 5 distinct numbers out of the 6 that we've chosen 

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<?php
$s = 0;
$t=0;
for($a=1;$a<=10;$a++){
    for($b=1;$b<=10;$b++){
        for($c=1;$c<=10;$c++){
            for($d=1;$d<=10;$d++){
                for($e=1;$e<=10;$e++){
                    for($f=1;$f<=10;$f++){
                        if(count(array_unique(array($a,$b,$c,$d,$e,$f)))==5){
                            $s++;
                        }
                        $t++;
                    }
                }
            }
        }
    }
}
echo 100*($s/$t); //45.36
?>

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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