Was this question clarified? @gregmat

I think this may be the answer,

12/12 * 11/12 * 10/12 * 9/12 * 8/12

I checked that Sal from Khan Academy also does the same way

Was this question clarified? @gregmat

I think this may be the answer,

12/12 * 11/12 * 10/12 * 9/12 * 8/12

I checked that Sal from Khan Academy also does the same way

2 Likes

Are you sure?

Because Greg’s method seems proper to me

Yeah I’m pretty sure Greg’s method is wrong. You can actually write a short python program to count the number of cases where the five people’s birth months are different:

```
cnt = 0
for a in range(12):
for b in range(12):
for c in range(12):
for d in range(12):
for e in range(12):
x = len(set([a,b,c,d,e])) # remove duplicates
if x == 5:
cnt += 1
print(cnt)
```

This prints 95040 cases. There are `12^5`

total cases, so the answer is `95040/12^5`

. This matches `12/12 * 11/12 * 10/12 * 9/12 * 8/12`

Hmm. @gregmat can you look into this?

Hey I’m still doing research. I have found websites that do both methods. Wikipedia does the other one, while other websites use the method I use. Python seems to say that my method is wrong, but I can’t figure out as to why. I need to talk to a proper statistician.

Aah. Alright, let us know in case of any updates!

This website uses the approach I used:

https://betterexplained.com/articles/understanding-the-birthday-paradox/

I can confirm that the answer is

1 \times \frac{11}{12} \times \frac{10}{12} \times \frac{9}{12} \times \frac{8}{12}

The way to imagine it would be to consider 12 boxes, and you have to put balls into empty boxes. The first time, the probability is 1, because as no box is empty, you can put it in any box and you won’t encounter a case where you’re putting into a box that’s already filled. The second time, there is one box that is filled, and hence there’s a \frac{11}{12} chance that you\re going to put into a box that’s not filled. The same can be applied for the other cases.

@sisi a closed form expression exists - there is no need to use Taylor approximation - though it may be useful for estimation.

1 Like