Birthday Problem question

Fernando1

I'm reading a fairly elementary book on probability called 'Chance' by Amir D. Aczel. There's a chapter in it on the Birthday Problem. It gives the formula for calculating the number of people needed to have a 50% probability of two people having the same birthday;
1.2 x sq. root of #categories - so 1.2 x sq.root365.
And also the number of people needed for a 95% probability - 1.6 x sq. root365.
I'm wondering if the same would apply in trying to open a combination lock.
Say it's a 4-digit combination lock, so 10,000 possibilities. To have a 95% chance of opening the lock I would only need to choose 1.6 x sq.rt10,000 numbers, or 160 4-digit numbers?
This seems illogical to me but is it correct?

red_fox

Short answer is that you're right, it doesn't work for the lock.

Essentialy the lock is the same as asking how many people do you need to give a probability > 50% that one of them has the same birthday as you.

In the birthday problem you don't care which pair matches, so you so for n people you have n(n-1)/2 pairs. For the lock one part is fixed - the right combination, so with n numbers you have n pairs, which is far less in general - so it follows that the lock should be "harder" to open.

(tiny point: at n=1,2 you're more likely to open a 365 combination lock than have a pair of matching birthdays)

Fernando1

Thanks red fox. Nicely explained.