Current Events > Can someone ELI5: The Birthday Paradox

(+) Yeah! (+)

I never understood this.

Paraphrased: 23 people. In a room of just 23 people theres a 50-50 chance of two people having the same birthday. In a room of 75 theres a 99.9% chance of two people matching.

I'm assuming years don't count (Example: I was born 6/15/1990, only the 6/15 part counts for this) not my real birthday

So if me and 22 other people are in a room there is a 50/50 shot someone was also born on 6/15.

How the hell is this possible? There's 365 (sometimes 366) days in a year so that is 365/6 possible birthdays.

I get that maybe some of it is chance. If you picked 100 people on the street you wouldn't get 100 different answers.
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WrkHrdPlayHrdr posted...

So if me and 22 other people are in a room there is a 50/50 shot someone was also born on 6/15.

It's a 50/50 chance any two of them have the same birthday. When you pick one out it's a much lower chance that one in particular will share a bithday.

This explains it well:

https://www.reddit.com/r/explainlikeimfive/comments/26778e/eli5_the_birthday_paradox/

This one is pretty good:
https://www.reddit.com/r/explainlikeimfive/comments/26778e/eli5_the_birthday_paradox/chofsgt

JE19426 posted...

WrkHrdPlayHrdr posted...

So if me and 22 other people are in a room there is a 50/50 shot someone was also born on 6/15.

It's a 50/50 chance any two of them have the same birthday. When you pick one out it's a much lower chance that one in particular will share a bithday.

So i'm not looking for MY birthday partner, just any two people may be share it?
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WrkHrdPlayHrdr posted...

So i'm not looking for MY birthday partner, just any two people may be share it?

Yep.

JE19426 posted...

This explains it well:

https://www.reddit.com/r/explainlikeimfive/comments/26778e/eli5_the_birthday_paradox/

This one is pretty good:
https://www.reddit.com/r/explainlikeimfive/comments/26778e/eli5_the_birthday_paradox/chofsgt

Why is it called a paradox when it's clearly not?
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rh000001 posted...

Why is it called a paradox when it's clearly not?

I think because dumb people like me look at the statement and go that can't be fucking right.

Might not be an actual "paradox" but I don't know the definition when used scientifically.
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rh000001 posted...

Why is it called a paradox when it's clearly not?

I think it's just because it looks paradoxical at first.

rh000001 posted...

Why is it called a paradox when it's clearly not?

Same reason why most people refuse to believe the Monty Hall problem, wherein if you have a choice of three options, one of the options is removed, and statistically you are absolutely better off changing your choice.

https://en.wikipedia.org/wiki/Monty_Hall_problem

It just doesn't sound right.
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rh000001 posted...

Why is it called a paradox when it's clearly not?

I've never met someone who was born with probabilistic intuition, personally; that has to be learned. To the uninitiated, certain solutions to problems in probability seem counter-intuitive. Also, keep in mind that a "paradox" doesn't necessarily have to be strictly contradictory, it just has to have the appearance of being illogical or absurd.
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K181 posted...

rh000001 posted...

Why is it called a paradox when it's clearly not?

Same reason why most people refuse to believe the Monty Hall problem, wherein if you have a choice of three options, one of the options is removed, and statistically you are absolutely better off changing your choice.

https://en.wikipedia.org/wiki/Monty_Hall_problem

It just doesn't sound right.

This just makes me rage.

Its 50-50. The odds of you being correct dont just magically lock in when you make your choice. It really doesnt fucking matter that it was 1/3 when you first picked. As soon as one choice is revealed to be a losing one, there is a 50-50 chance befween the remaining two choices.

The grouping of the revealed and the unchosen as if the revealed choice could still somehow magically turn out to be correct is absolutely senseless.
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Jerry_Hellyeah posted...

Its 50-50. The odds of you being correct dont just magically lock in when you make your choice. It really doesnt fucking matter that it was 1/3 when you first picked. As soon as one choice is revealed to be a losing one, there is a 50-50 chance befween the remaining two choices.

The host has more information than you and knows that the door he removes is a loser.

If the host didn't have any information, it would stay as 33/33/33 (i.e. he could have removed the winning door)

But since he knows he is picking one of two loser doors, the odds become 33/66.

Please remove your emotion from the problem and evaluate it with a rational mind.

If you draw 23 students at random from a Prob/Stats 101 class, chances are 100% that at least one of them will fight to the death defending his 50/50 answer to Monty Hall.
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clearaflagrantj posted...

Jerry_Hellyeah posted...

Its 50-50. The odds of you being correct dont just magically lock in when you make your choice. It really doesnt fucking matter that it was 1/3 when you first picked. As soon as one choice is revealed to be a losing one, there is a 50-50 chance befween the remaining two choices.

The host has more information than you and knows that the door he removes is a loser.

If the host didn't have any information, it would stay as 33/33/33 (i.e. he could have removed the winning door)

But since he knows he is picking one of two loser doors, the odds become 33/66.

Please remove your emotion from the problem and evaluate it with a rational mind.

Is it not quite irrational to think your 1 in 3 chance stays the same even after a losing selection is removed?
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Jerry_Hellyeah posted...

clearaflagrantj posted...

Jerry_Hellyeah posted...

Its 50-50. The odds of you being correct dont just magically lock in when you make your choice. It really doesnt fucking matter that it was 1/3 when you first picked. As soon as one choice is revealed to be a losing one, there is a 50-50 chance befween the remaining two choices.

The host has more information than you and knows that the door he removes is a loser.

If the host didn't have any information, it would stay as 33/33/33 (i.e. he could have removed the winning door)

But since he knows he is picking one of two loser doors, the odds become 33/66.

Please remove your emotion from the problem and evaluate it with a rational mind.

Is it not quite irrational to think your 1 in 3 chance stays the same even after a losing selection is removed?

Except the host will never remove the winning door. Because of this is why your odds are higher switching.
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Jerry_Hellyeah posted...

clearaflagrantj posted...

Jerry_Hellyeah posted...

Its 50-50. The odds of you being correct dont just magically lock in when you make your choice. It really doesnt fucking matter that it was 1/3 when you first picked. As soon as one choice is revealed to be a losing one, there is a 50-50 chance befween the remaining two choices.

The host has more information than you and knows that the door he removes is a loser.

If the host didn't have any information, it would stay as 33/33/33 (i.e. he could have removed the winning door)

But since he knows he is picking one of two loser doors, the odds become 33/66.

Please remove your emotion from the problem and evaluate it with a rational mind.

Is it not quite irrational to think your 1 in 3 chance stays the same even after a losing selection is removed?

I tell you I'm thinking of a number between 1 and 100

You guess it's "57"

I say it could be 57, or it could be 14, but it's not any of the other numbers

Your guess was 1/100, it still is 1/100, I eliminated every other number because I know those are all losers, all that's left is most likely my number.

Play the scenario out in your head, you can pick 57, 49, 3, 98, any number, I will then eliminate every number other than 14. The only scenario I wouldn't is where you originally selected 14, which is a 1 in 100 chance.

Jerry_Hellyeah posted...

Is it not quite irrational to think your 1 in 3 chance stays the same even after a losing selection is removed?

Let's try a different game... Pick any card you want from a 52-card deck, face down. Ace of spades is the winner, every other card is a loser. After you pick, I remove 50 of the remaining, losing cards from the deck and then I give you the option to switch your card for the sole remaining card. Do you switch?
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Jerry_Hellyeah posted...

K181 posted...

rh000001 posted...

Why is it called a paradox when it's clearly not?

Same reason why most people refuse to believe the Monty Hall problem, wherein if you have a choice of three options, one of the options is removed, and statistically you are absolutely better off changing your choice.

https://en.wikipedia.org/wiki/Monty_Hall_problem

It just doesn't sound right.

This just makes me rage.

Its 50-50. The odds of you being correct dont just magically lock in when you make your choice. It really doesnt fucking matter that it was 1/3 when you first picked. As soon as one choice is revealed to be a losing one, there is a 50-50 chance befween the remaining two choices.

The grouping of the revealed and the unchosen as if the revealed choice could still somehow magically turn out to be correct is absolutely senseless.

That's a nice opinion, but it's a wrong opinion. Statistically speaking, you are better off changing your pick. Simulations have repeatedly shown this.

Or, to put it different. Imagine you have a hundred doors to choose from, and you choose one. Then, 98 are removed and you're given a choice to change your pick. Do you think you really got the 1% shot right on your first try, or would you feel more comfortable with the other door being the correct one?

Same principal.

Here's a image from the link provided showing this:

https://upload.wikimedia.org/wikipedia/commons/0/0c/Monty_problem_monte_carlo.svg

I'm not denying that it seems counterintuitive, but the math is there.
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