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E[X] = n − 1 ∑ i = 0xipi. And here x must mean the number of collisions involving i + 1 people, which is (n i). All n people born on different days means no collisions, i = 0; two people born on the same day means n collisions, i = 1; all n people born on the same day means n collisions, i = n − 1. Since the probabilities of three or more ...
With respect to the question in the title, by doing the second line, you are making your calculator attempt to compute a number greater than $100^{200}$. It won't. By doing the first line, you are making a multiplication of about $20$ numbers close to $1$.
The usual form of the Birthday Problem is: How many do you need in a room to have an evens or higher chance that 2 or more share a birthday. The solution is 1 − P(everybody has a different birthday) 1 − P (everybody has a different birthday). Calculating that is straight forward conditional probability but it is a mess. We have our first ...
On Wikipedia and on most probability books the approach to solve this is to first calculate P', the probability that no two people will share the same birthday, and then do 1 - P'. So you basically do (364/365) (364/365)... (343/365). This multiplication will give you 0.4927, 1-P' is 0.5073.
I began by breaking the problem down into 4 cases: Let E = the event that at least 2 people share the same birthday in a room of 4. Our sample size: 3654 365 4. Case 1: 4 people share the same birthday: 365 ways. Case 2: 3 people share the same birthday, 1 distinct birthday: 365 ⋅ 364 ⋅ C(4, 3) 365 ⋅ 364 ⋅ C (4, 3) Case 3: 2 people ...
The probability of getting at least one success is obtained from the Poisson distribution: P( at least one triple birthday with 30 people) ≈ 1 − exp(−(30 3)/3652) =.0300. P ( at least one triple birthday with 30 people) ≈ 1 − exp (− (30 3) / 365 2) =.0300. You can modify this formula for other values, changing either 30 or 3.
Deeper calculation gives rounded probabilities of at least three people sharing a birthday of 84 − 0.464549768 85 − 0.476188293, 86 − 0.487826289, 87 − 0.499454851, 88 − 0.511065111, 89 − 0.522648262 so the median of the first time this happens is 88 though 87 is close, while the mode is 85 and the mean is about 88.73891765. – Henry.
The second expression says that the expected number of birthday pairs is $\frac{3 \times 2}{2\times 2} =\frac32 = 1.5$; this is also $1 \times \frac34+3 \times \frac14$. So in this small example, you can see that both expressions are correct, but the first is less than double the second because of what happens when all three people share the ...
Combinations certainly give the number of possible birthday sets, which seems a reasonable way to solve the problem. However, the birthday problem is for a real group of people, and such groups allow for repetition of birthdays. Once repetition is allowed, the number of ways the group can have birthdays is 365^n, for an n-person group.
The birthday paradox and it's easy enough to calculate for small numbers. 23 people and 365 birthdays as used in the 50% examples. But how do you approach (or approximate) the birthday paradox for values like 52!? I understand 52! is a large (~226 bit) number but I would like to get a feel of the order of magnitude of the claim. Is it 1% or 0. ...