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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 $1/5$

ArnuldOnData
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  • If Alex chose $A$ and Blake chose $B$ with $A\neq B$ then, to avoid a match, Chris must have chosen one of the $3$ numbers other than $A,B$. There is a $\frac 35$ chance of that. – lulu May 06 '20 at 14:59
  • Note: this is the long way round to solve the stated problem. It is $\textit {much}$ easier to work out the probability that they all chose different numbers. – lulu May 06 '20 at 15:00
  • @lulu no they didn't. There is exactly one match as per the "condition" of conditional probability here in the problem statement. Regarding "avoiding a match", look at my example, Chris and Alex have matching number – ArnuldOnData May 06 '20 at 15:36
  • I don't understand what "no they didn't" means here. Who is they and what didn't they do? – lulu May 06 '20 at 15:37
  • I meant the same "they" you mentioned in your Note in comment – ArnuldOnData May 06 '20 at 15:38
  • Still not following. My point is that it is a lot easier to compute the probability that the four chose different numbers. Then you can just subtract that value from $1$ to get the answer you seek. Trying to write out all the branches for the problem as stated is inefficient and error prone (but perfectly possible). – lulu May 06 '20 at 15:40
  • Problem is about What is the chance that any of them chose the same number? You are actually proposing to discard what we need to find? – ArnuldOnData May 06 '20 at 15:44
  • Once again: it is often easier, as it is here, to compute the probability of the complementary event. Then you can get what you want by subtracting from $1$. – lulu May 06 '20 at 15:48
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    Here the probability that they chose different numbers is $\frac {5\times 4\times 3\times 2}{5^4}=\frac {24}{125}$. Hence the probability that there is a match amongst the four is $1-\frac {24}{125}=\frac {101}{125}$. Very simple. – lulu May 06 '20 at 15:50
  • That looks like you used Permutations, I found same yesterday in Casella and Berger's book. Looks like I should read more on that. It is much simpler and straightforward solution from there. – ArnuldOnData May 06 '20 at 16:31

1 Answers1

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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:

$1 - \frac {24}{125}$ $= \frac {125 - 24}{125} = \frac {101}{125} = 80.8$%

ArnuldOnData
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