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I'm trying to solve a birthday paradox problem, I'm really close to the answer but I guess I'm doing something wrong. The problem states that the probability of being born on a friday is $\frac{1}{3}$ and is equally probable to any other day of the week. Assuming 4 (or any number n people) where selected, what is the probability to have for example two born on friday and two born on other days of the week?

I know that the probability of two people sharing the same birthday is: $$\frac{(365*364*363*362)}{365^n}$$ Another note I was able to come to is that the chance of last two not sharing birthdays with the first two is: $$1-\frac{5}{7}*\frac{4}{7} \approx 0.591$$

Assuming that a person can be born on a Friday with probability 1/3 and with equal probability any other day of the week. What is the probability that among 4 randomly selected people, two were born on the same day and the other two in two other days?

The correct answer according to my exercise sheet is about $0.52$. Can anyone point me in the right direction?

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    $\begingroup$ The people's actual birthdays are irrelevant, so your first probability is moot to the problem. Just think of it in terms of days of the week; you need to have two people on Friday - and the probability of someone being born on Friday is 1/7 - and two people born on other days of the week (what's the probability of that?). You then need to consider all the different combinations of two people that could be born on Friday. $\endgroup$ – Steven Stadnicki Jun 29 '17 at 16:03
  • $\begingroup$ (That said, 0.52 is definitely not the probability that two were born on Friday and two on other days; can you be more specific/more precise about the statement of the problem?) $\endgroup$ – Steven Stadnicki Jun 29 '17 at 16:04
  • $\begingroup$ Oh wait... are the people distinct or not? My answer assumed that they were distinct people. $\endgroup$ – Franklin Pezzuti Dyer Jun 29 '17 at 16:04
  • $\begingroup$ @StevenStadnicki I added the original problem to the post $\endgroup$ – hallaksec Jun 29 '17 at 16:25
  • $\begingroup$ @Nilknarf I updated the OP $\endgroup$ – hallaksec Jun 29 '17 at 16:41
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It is of course unrealistic that the chance of a person born on a Friday is $\frac{1}{3}$, but so be it. We are also told that for all other days of the week the chance is the same, so that would be

$$\frac{1-\frac{1}{3}}{6}=\frac{1}{9}$$ (e.g. the chance of being born on a Wednesday is $\frac{1}{9}$)

Now, there are various ways to have 2 people being born on the same day, and the other two on 2 other days:

$A$. Two born on a Friday, and the other two on two different days other than Friday

$B$. Two born on the same day but other than a Friday, one born on Friday, and the last on a different day yet.

$C$. Two born on the same day but other than a Friday, and the other two on two different days other than Friday

$$P(A) = {4 \choose 2} \cdot \big( \frac{1}{3} \big)^2\cdot \frac{6}{9} \cdot \frac{5}{9} = \frac{180}{729}$$

(the $4 \choose 2$ is the number of ways to pick the two people born on a Friday)

$$P(B) = {4 \choose 2} \cdot 6 \cdot \big( \frac{1}{9} \big)^2\cdot 2 \cdot \frac{1}{3} \cdot \frac{5}{9} = \frac{120}{729}$$

(the extra $2$ term is to distinguish the two people: one born on Friday, the other not. The $6$ is to pick one of the 6 days that the first two are born on)

$$P(C) = {4 \choose 2} \cdot 6 \cdot \big( \frac{1}{9} \big)^2 \cdot \frac{5}{9} \cdot \frac{4}{9} = \frac{80}{729}$$

Hence, the chance that 2 people being born on the same day, and the other two on 2 other days is:

$$P(A)+P(B)+P(C)=\frac{380}{729} \approx 0.52$$

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  • $\begingroup$ Thank you for your input, I do agree the probability $1/3$ is unrealistic. $\endgroup$ – hallaksec Jun 29 '17 at 17:33
  • $\begingroup$ Thank you for your input, I do agree the probability $1/3$ is unrealistic but I guess the goal as to steer us $\endgroup$ – hallaksec Jun 29 '17 at 17:36
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First of all, you should not involve number of days in a year, because there is no assumption about uniform distribution on each days of a year.

In fact, since the assumption is evenly distributed across a week, it is slightly not evenly distributed across days in a year

$$\frac{\binom{4}{2} \times 1 \times 1 \times 6 \times 6}{7^4}$$

EDIT according to OP's edit

Notice that Friday has a different probability than other days.

$$\binom{4}{2}\times \frac{1}{3}\times \frac{1}{3} \times \frac{6}{9} \times \frac{5}{9} $$ $$+ \binom{4}{2} \times \binom{7-1}{1} \times \frac{1}{9} \times \frac{1}{9} \times \binom{2}{1} \times \frac{1}{3} \times \frac{5}{9}$$ $$+\binom{4}{2} \times \binom{7-1}{1} \times \frac{1}{9} \times \frac{1}{9} \times \frac{5}{9} \times \frac{4}{9}$$

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  • $\begingroup$ Thank you for your input I think I didn't explain the problem well, would you mind please having another look at the OP? $\endgroup$ – hallaksec Jun 29 '17 at 16:25
  • $\begingroup$ Sorry just get chance to check your edit - I saw there is an answer already, but I'll update mine anyway. $\endgroup$ – Yujie Zha Jun 29 '17 at 17:50

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