STATEMENT
4 friends (Alex, Blake, Chris and Dusty) each choose a random number between 1 and 5. What is the chance that any of them chose the same number?
(original problem here: https://www.mathsisfun.com/data/probability-events-conditional.html)
SOLUTION
Billy compares his number to Alex's number. There is a 1 in 5 chance of a match. Hence, probability of $1/5$ for match and $4/5$ for no match.
But there are now two cases to consider:
- If Alex and Billy did match, then Chris has only one number to compare to.
- If Alex and Billy did not match then Chris has two numbers to compare to. So, there is a $2/5$ chance of Chris matching (against both Alex and Billy). And a $3/5$ chance of not matching.
I don't understand the second point. If Alex and Billy did not match then Chris has three people to compare to: Alex, Blake and Chris. Let's take an example:
Alex = 1
Blake = 2
Chris = 1
Dusty = 3
Now, since Alex and Blake do not match, Chris still has to compare his number against all three to check for a match and there is only one match. So chance of getting a match is $1/5$
I found a way to solve it. To understand this, we have to understand two assumptions.
Assumption #1: There is one or more than one match.
Tis assumption is direct result of the problem statement: What is the chance that any of them chose the same number?
Now Alex compares his number to Billy. Two people are comparing if they got the matching number. So probability of match and no match are $1/5$ and $4/5$ respectively. It's all fine till here.
Now we talk about Chris and before we go ahead, here comes the second assumption.
Assumption #2: Chris is the the third person to come into the picture. Dusty is here too but we are not taking him into the account yet. We are working right now with three people, Alex, Blake and Chris. Let's take an example:
Alex = 1
Blake = 2
Chris = 1
Dusty = 3
Since there is no match between Alex and Blake and we are limiting ourselves to three people, third person is Chris. So Chris gotta match his number with Alex and Blake, two people. What is the probability of matching?
$2$ people = $2/5$ probability of matching the number, and hence $3/5$ of not matching.
Now we assume, Chris matches with Alex in our example. So it is all fine but we have to compute overall conditional probability. Hence we will go with "no match" case and bring Dusty. Who now has three people to match his number with: Alex, Blake and Chris.
$3$ people = $3/5$ probability of matching the number, and hence $2/5$ of not matching.
Hence, probability of What is the chance that none of them chose the same number? is $(4/5 x 3/5 x 2/5) = 24/125$ and hence the complement of it is What is the chance that one of them chose the same number? and that is:
$1 - \frac {24}{125}$ $= \frac {125 - 24}{125} = \frac {101}{125} = 80.8$%