Probability of time and date of birthday being the same

One out of 365 (days in a year) times 1 out of 1440 (minutes in a day). I think, if my math is correct, that'd be one out of 525, 600
chances.

Be aware that it'd differ greatly from someone being born at exactly the same time as you on the same year day and minute. Is that what you're seeking?

Thank you for responding so quickly. To answer your followup question, this particular instance pertains a single individual.
Upon reflection I have two more questions for you.
First, does the fact that times of birth beyond the 31st minute of every hour have no date correlation, ie. there's no 32nd day in any month, pose a limiting factor?
Secondly, does the fact that there are two oppurtunities within a 24 hour period to achieve the same timestamp, i.e. AM and PM, pose another limiting factor?
Since Ive made myself dizzy, I'll eagerly await your reply.

I've no idea what you mean here but as near as I can grasp what I think you're asking...it is not a true condition. Meaning.. as soon as it's one second past midnight ..it then becomes the next calendar day - and, of course they have a date correlation..as does the 31st minute of every hour.

You've lost me there.
Furthermore...

Once again, you've lost me.

The odds are what I computed...as long as my math is correct and the scenario is what you've explained in your first post. If you want to make it more split-hair accurate, substitute 365.24 days instead of 365.

Also, the scenario you first asked about utilized a birth time that was in minutes, not in seconds. when you include seconds, you just change the odds to a finer resolution answer as well as a greater odds ...e,g...were the 2 individuals born on the same second vas within the saem minute time-frame?.

Are you asking what happens if the time involves seconds as well as hour and minutes, too?

Whatever you're asking about times with AM or PM...then you need to
re-phrase the question more clearly as I'm not clear on that either.

Your math isn't correct Ragman. There is no reason to multiply these two... that would be the right idea if you were asking the odds for a specific day... for example the odds that you were born on July 1 (and only July 1) at 7:01.

There is a 100% chance that you will be born on some day during the year. For each of these days... there are 1440 minutes, exactly two of which will meet the requirement (either AM or PM).

So the answer is simply 2 in 1440. There is a 1 in 670 chance that you will be born the same time as the date you are born.

Ok, if you say so. Those odds seem a bit too light....but at this time of the morning I can't tease out why I think so.

Perhaps it's a case of that I don't understand the conditions of the problem. Or worse than that- that I'm 100 years old and my math skills have waned.

{Edit : no I think you're right}

I am right.

Think of it this way. Roll two six sided dice. What are the odds the number on the first die will match the number on the second? The answer is not 1 in 36.

(I play poker. Making these calculations on probability is rather important to me ).

Gents,
Thanks for bearing with me... Let's revisit Maxdancona's earlier post..."There is no reason to multiply these two... that would be the right idea if you were asking the odds for a specific day... for example the odds that you were born on July 1 (and only July 1) at 7:01."

To clarify, my question is not related to odds of being born on a specific date and then being born at the same time as that date; my question is whatvare the odds of being born at the same time of day as the date that one happens to be born, i.e., being born at 12:25 on December 25th, being born at 1:01 on January 1st, being born on 7:01 on July 1st, etc.

Hopefully, that clarifies the question and if so, I am grateful to know know know the odds are a 1 in 670 chance of occurrence.

Happy New Year,
Kyle

I was away the day division was handed out.

Oops... apparently so was I (how embarrassing).

1440/2 is in fact 720. There is a 1 in 720 chance

I hope I can be as gracious as your good humoured acknowledgement of oops and error correction, if I should ever make a mistake.

Kyle B should note that he needs to make assumptions about the distribution of birth times eg. independent identically distributed random variables with a uniform distribution.

The key to calculating probability correctly is always the same, though the exact method will differ:

(1) Figure out the total number of ways an event can occur

(2) Figure out the total number of ways the desired event can occur

(3) Divide (1) into (2) to get the probability.

There are 60 x 24 = 1440 minutes in a standard day. There are 365 calendar days in a standard year. So the answer to (1) is 1440 x 365 = 525,600.

There are two times in each day (a.m.and p.m.) for the desired event to occur, so the answer to (2) is 365 x 2 = 730.

If we invert the division for the sake of clarity, then the answer to (3) is 525,600 / 730 = 720 to 1 odds.

This ignores Daylight Savings Time, leap years, leap seconds, and other irregularities. It also assumes, without reference to birth data, that all times of day and all days of the year are equally statistically likely for births.