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There are 28 individuals in a room. Each person will celebrate their birthday in 2016 on the Sunday of the week concluding their birth. Nobody is born on February 29. What is the probability that 2 parties happen in 2016 on the same day.

Note: I asked my lecture to clarify and he said anybody in the last week of December will have their birthday in 2017. This makes the problem so confusing to me.

I thought it might be best to approach it from an inclusion-exclusion perspective? Or could I simply subtract from 1 the probability that nobody shares a birthday? How do I count this? There are 52 Sunday's in 2016, the Sunday after feb 29th only has 5 days preceding it, so will I have to count this one differently as well?

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    $\begingroup$ Unless your lecturer really wants you to get lost in fiddly details, I suspect the calculation they are after is just to treat the 52 birth weeks as equally likely for all 28 individuals, and to figure out how likely it is that two or more people have the same birth week. And yes, the way to do this is to figure out how likely it is that everyone's birthweek is different, and subtract that value from 1. $\endgroup$ – Paul Sinclair Mar 25 '16 at 3:02
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"Or could I simply subtract from $1$ the probability that nobody shares a birthday? How do I count this? There are $52$ Sunday's in $2016$."

First i calculate $P(k)$ the probability that nobody out of $k$ individuals share a birthday week and then its compliment $P(\bar{k})$.

For the sample space $\mathbf{n}(S)$ the number of possible ways $k$ individuals can have birthday weeks is

$$\mathbf{n}(S) = 52^k$$

The number of ways all have different birthday weeks, $\mathbf{n}(B)$, is the number of ways of selecting $k$ different weeks, the number of sequences without repetition.

$$\mathbf{n}(B)={_{52}P_k}=\frac{52!}{(52 - k)!} $$

$$P(k) = \frac{\mathbf{n}(B)}{\mathbf{n}(S)} = \frac{52!}{52^k (52 - k)!}$$

$$\therefore \:\: P(\bar{k}) = 1 - P(k) = 1 - \frac{52!}{52^k (52 - k)!}$$

$$P(\bar{28}) = 1 - \frac{52!}{52^{28} (52 - 28)!} = 1 - \frac{52!}{52^{28}\: 24!}$$

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