# Tricky conditional probability question

There are four friends â€“ Adam, Bella, Christopher and Drew. All of them are asked to choose any number in their mind. Now what is the probability that every one of them has the same number in mind? The chosen number must be in between 1 to 5.
Now the answer is 101/125.. which is way off what I find.. I would like to know where I go wrong with my reasoning?

I said let A, B,C,D represent the 4 people. Let Ai denote the event that A has number i in mind and so on. Then P(each have the same number)= P((A1,B1,C1,D1)U(A2,B2,C2,D2)U(A3,B3,C3,D3)U(A4,B4,C4,D4)U(A5,B5,C5,D5)).P((A1,B1,C1,D1) U (A2,B2,C2,D2) U (A3,B3,C3,D3) U (A4,B4,C4,D4) U (A5,B5,C5,D5)). Now each Ai,Bi,Ci,Di is disjoint from the event Aj,Bj,Cj,Dj, hence the above probability is just the sum of probabilities.. So we get P(Each have the same number)=∑5i=1P(Ai,Bi,Ci,Di)=∑5i=1P(Ai|Bi,Ci,Di)P(Bi|Ci,Di)P(Ci|Di)P(Di)=∑5i=1P(Ai)P(Bi)P(Ci)P(Di)=5∗(1/5)4=1/125.= \sum_{i=1}^{5}P(Ai,Bi,Ci,Di)= \sum_{i=1}^{5} P(Ai|Bi,Ci,Di)P(Bi|Ci,Di)P(Ci|Di)P(Di)= \sum_{i=1}^{5} P(Ai)P(Bi)P(Ci)P(Di)= 5*(1/5)^4= 1/125.

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101125\frac {101}{125} is obviously wrong. Way too high, even if there were only two people.
– lulu
Oct 21 at 0:31

For two people, A,BA,B: AA chooses whatever, then there is a 15\frac 15 probability that BB matches. Hence the answer for two people is 15\frac 15. Likewise, for nn people the answer is 15n−1\frac 1{5^{n-1}}.
– lulu
Oct 21 at 0:33

This is the solution given. The probability that Adam and Bella donâ€™t have the same number in their respective minds will be 4/5.Now let us include Christopher in the scene. The probability that Christopher doesnâ€™t have a digit same as either Adam or Bella will be 3/5.Now include Drew, the probability that she does not have any number which is same as either Adam or Bella or Christopher will be 2/5.So the combined probability that none of them has the same number in mind will be:4/5 * 3/5 * 2/5 = 24/125.Now the probability that they have the same number in mind will be:1 â€“ 24/125 = 101/125
– Socchi
Oct 21 at 0:36

Which is correct and why? Where do I go wrong?
– Socchi
Oct 21 at 0:36

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That analysis correctly answers a different question. That answers the question “what is the probability that at least two of them are thinking of the same number”. Very different.
– lulu
Oct 21 at 0:37

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You are not wrong.   The answer is not 101/125101/125.

The probability every one of the other three is mindful of the same number as Adam is 1/531/5^3

(Assuming independence and unbiased selections, of course.)

This is the solution given. The probability that Adam and Bella donâ€™t have the same number in their respective minds will be 4/5.Now let us include Christopher in the scene. The probability that Christopher doesnâ€™t have a digit same as either Adam or Bella will be 3/5.Now include Drew, the probability that she does not have any number which is same as either Adam or Bella or Christopher will be 2/5.So the combined probability that none of them has the same number in mind will be: 4/5 * 3/5 * 2/5 = 24/125.Now the probability that they have the same number in mind will be:1 â€“ 24/125 = 101/125

That is the solution to: “What is the probability that some of them are mindful of the same number?”   A different problem entirely.

“All of them do it” is not the complement of “None dome of them do it”.