Tuesday, February 18, 2014

"Math Explains Likely Long Shots, Miracles and Winning the Lottery [Excerpt]"

I'm only including the birthday problem here. For the seemingly long shot, like the lotteries, click here.
The birthday problem poses the following question: How many people must be in a room to make it more likely than not that two of them share the same birthday?

The answer is just 23. If there are 23 or more people in the room, then it's more likely than not that two will have the same birthday.

Now, if you haven't encountered the birthday problem before, this might strike you as surprising. Twenty-three might sound far too small a number. Perhaps you reasoned as follows: There's only a one-in-365 chance that any particular other person will have the same birthday as me. So there's a 364/365 chance that any particular person will have a different birthday from me. If there are n people in the room, with each of the other n − 1 having a probability of 364/365 of having a different birthday from me, then the probability that all n − 1 have a different birthday from me is 364/365 × 364/365 × 364/365 × 364/365 … × 364/365, with 364/365 multiplied together n − 1 times. If n is 23, this is 0.94.

Because that's the probability that none of them share my birthday, the probability that at least one of them has the same birthday as me is just 1 − 0.94. (This follows by reasoning that either someone has the same birthday as me or that no one has the same birthday as me, so the probabilities of these two events must add up to 1.) Now, 1 − 0.94 = 0.06. That's very small.

Yet this is the wrong calculation to consider because that probability—the probability that someone has the same birthday as you—is not what the question asked. It asked about the probability that any two people in the same room have the same birthday as each other. This includes the probability that one of the others has the same birthday as you, which is what I calculated above, but it also includes the probability that two or more of the other people share the same birthday, different from yours.

This is where the combinations kick in. Whereas there are only n − 1 people who might share the same birthday as you, there are a total of n × (n − 1)/2 pairs of people in the room. This number of pairs grows rapidly as n gets larger. When n equals 23, it's 253, which is more than 10 times as large as n − 1 = 22. That is, if there are 23 people in the room, there are 253 possible pairs of people but only 22 pairs that include you.

So let's look at the probability that none of the 23 people in the room share the same birthday. For two people, the probability that the second person doesn't have the same birthday as the first is 364/365. Then the probability that those two are different and that a third doesn't share the same birthday as either of them is 364/365 × 363/365. Likewise, the probability that those three have different birthdays and that the fourth does not share the same birthday as any of those first three is 364/365 × 363/365 × 362/365. Continuing like this, the probability that none of the 23 people share the same birthday is 364/365 × 363/365 × 362/365 × 361/365 … × 343/365.

This equals 0.49. Because the probability that none of the 23 people share the same birthday is 0.49, the probability that some of them share the same birthday is just 1 − 0.49, or 0.51, which is greater than half.
Scientific American via Instapundit 

17 comments:

Michael Haz said...

There must be at least 23 commenters here. Let's test the maths.

January 20th.

Lem the artificially intelligent said...

October 15

john said...

Sep 8

Leland said...

Dec 11

john said...

But - if 23 people enter their birthdays here, there is a 51% likelihood that two are the same. Not 100%

If I read that right.

XRay said...

Oct 22...

Plus that's why I hire out math, makes my head hurt.

john said...

Do the winners get a prize?

edutcher said...

August 8.

And I did go to undergrad school who was born on the same day.

The discipline is called discrete math.

DADvocate said...

I always thought it was a result of the fact that more people are born in the spring than any other time of year, i.e. the odds of being born on a particular day aren't even. But, the pairs solution makes sense too. I went to high school with a guy born on the same day and year. 4/25/51

sakredkow said...

Wow, I'm just skimming but ALL you guys were born on February 18?? What are the odds of THAT?!

DADvocate said...

BTW - discrete math is much preferred over indiscrete math.

AllenS said...
This comment has been removed by the author.
AllenS said...

I commented too soon. Nov 10.

Dust Bunny Queen said...

January 6

That math/birthday example makes the assumption that the gathering of people is completely random. It wouldn't be random as there are many other factors that would contribute to the occurrence of being in the same room that would remove the random factor.

As the the lottery. Your chances of not winning are 100% if you don't buy a ticket :-P

Lem the artificially intelligent said...

The odds are really against us, meaning, we have a good chance. We just need more bloggers asking the same question.

Leland said...

Statistically, most birthdays in the US occur in August, yet only 1 here. edutchers parents had great holiday cheer!

I'm about to head to lunch with a guy that is exactly 10 years my junior to the day. We both went to the same college, married nurses, and worked with the same first employer after college before leaving that industry and both ended up in a new industry with a different company on on own merits.

Lem the artificially intelligent said...

Coincidence is a pigment of the imaginaion.