Double mathematics PhD challenge?

Question

I am afraid only those who got double PhD in mathematics can answer my mathematic questions correctly. Should you be the one?

This incident did happened recently and reported in the News. In a high way, 4 cars crashed. Among these drivers, their birthdays were August 8th, September 9th, October 10th and November 11th. They claimed that this will only happen in x/1. How do you calculate the amount of x?

In the real incident, the cars did not crashed in sequence. However, is it possible to calculate, and what will be the amount of x when these were crashed in sequence, for instance, the car owner whom birthday is August 8th was being hit by the owner whom birthday is September 9th....? Etc.

Those who can answer me all two questions will be chosen the best answer by me.

Answer 1

= 365!/(4! x 361!)   [365C4]

=1/727441715  chance in any order

the chance that it happens in order is [365P4]

=1/17458601160

this is a probability questions that does NOT require a PhD in maths, in fact just a low qualification in statistics

Answer 2

Double PhD ... not!

High school math ... yes.

If the question concerns these four specific dates,

then the probability of those (or any other specific dates) is:

1/365^4 = 1/17748900625 for the specific order 8/8, 9/9, 10/10, 11/11.

The four drivers could have those dates in any of 24 different

orders, so then it's 24/17748900625, but that's not a substantially larger number.

A more general case would any four consecutive dates of the

form n/n (Jan 1, Feb 2, Mar 3, etc).

There are 9 such sequences in the year, (assuming we disallow

wraparound cases such as 11/11, 12/12, 1/1, 2/2).

Then the probability is 9/17748900625 or 9*24/17748900625,

depending if specific order or any order is required.

All of these numbers are essentially 0,

the largest of them being 1.217 × 10^-8.

UPDATE:

Each guy can have any birthday.

Chances that first guy is 8/8 is 1/365.

Chances that second guy is 9/9 is 1/365.

Same for #3 and #4, both are 1/365.

Since there are 24 different assignments

for which guy is which, it's either 1/365^4

(specific order) or 24/365^4 (any order).

365C4 and 365P4 would be right if we had

the additional information that the drivers have

different birthdays, which we don't.

.