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Thread: The birthday paradox

post 1626949943 12-29-2020, 05:54 PM
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Originally Posted By numberguy12⏩
Lol wut
What’s the math on the original post?
post 1626949953 12-29-2020, 05:54 PM
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Originally Posted By numberguy12⏩
Lol wut
Noobs ammirite?
Damn
post 1626950453 12-29-2020, 05:59 PM
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Originally Posted By WeBoutDat⏩
What’s the math on the original post?
Prob = 1-(365!/342!)/365^23
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post 1626950713 12-29-2020, 06:02 PM
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Originally Posted By numberguy12⏩
Prob = 1-(365!/342!)/365^23
Hmmm...sorry I asked.
post 1626950803 12-29-2020, 06:03 PM
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A guest lecturer was going to use this example to teach probability to a classroom that just happened to have 23 people in it. Unfortunately his calculation of around 50% likelihood that at least two people share the same birthday was totally off. What gives?
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post 1626950993 12-29-2020, 06:05 PM
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Originally Posted By numberguy12⏩
A guest lecturer was going to use this example to teach probability to a classroom that just happened to have 23 people in it. Unfortunately his calculation of around 50% likelihood that at least two people share the same birthday was totally off. What gives?
For month and day only, not including the year.
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post 1626951633 12-29-2020, 06:11 PM
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Originally Posted By WeBoutDat⏩
Hmmm...sorry I asked.
Calculate the chances everyone has a different birthday:

-First person can have whatever
- chances 2nd person has different: 364/365
-chances 3rd person has different than both first and second: 363/365
-chances 4th person has different than first 3: 362/365
....etc...
-chances 23rd person has different than first 22: 343/365.

So this probably is 364x363x362x...x343/(365x365x365x...365) otherwise denoted (364!/342!)/365^22. I just multiplied this by 365/365, or 1 to get the expression in the above.

Probability that at least two share the same is just the complement of the above, so 1-probability of everyone having different birthday found above.
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post 1626951973 12-29-2020, 06:14 PM
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Originally Posted By NealIRC⏩
For month and day only, not including the year.
Nope he was was thinking just in terms of month and day. And yet his estimation of 50% for the room of 23 people was still way off.
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post 1626952013 12-29-2020, 06:14 PM
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Originally Posted By numberguy12⏩
Prob = 1-(365!/342!)/365^23
post 1626953833 12-29-2020, 06:33 PM
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Originally Posted By numberguy12⏩
Calculate the chances everyone has a different birthday:

-First person can have whatever
- chances 2nd person has different: 364/365
-chances 3rd person has different than both first and second: 363/365
-chances 4th person has different than first 3: 362/365
....etc...
-chances 23rd person has different than first 22: 343/365.

So this probably is 364x363x362x...x343/(365x365x365x...365) otherwise denoted (364!/342!)/365^22. I just multiplied this by 365/365, or 1 to get the expression in the above.

Probability that at least two share the same is just the complement of the above, so 1-probability of everyone having different birthday found above.
Why wouldn’t the same equation apply to roulette wheel with 365 numbers on it? If it hit a number and I picked that it would hit it again in the next 23 rolls I would get 365 to 1 odds on my bet and would only have to place 23 bets for it to be a 50 percent chance of pay off.
post 1626954483 12-29-2020, 06:39 PM
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Originally Posted By WeBoutDat⏩
Why wouldn’t the same equation apply to roulette wheel with 365 numbers on it? I would get 365 to 1 odds on my bet and would only have to place 70 bets for it to be a 99.7% pay off.
364/365 chance of failing on any trial.
The birthday one shrinks with each trial because another number become excluded, while the roulette wheel fails with all 364 every time.
post 1626954793 12-29-2020, 06:43 PM
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Originally Posted By WeBoutDat⏩
Why wouldn’t the same equation apply to roulette wheel with 365 numbers on it? I would get 365 to 1 odds on my bet and would only have to place 23 bets for it to be a 50 percent chance of pay off.
Your example above with the 365 sided die is a correct analogy, and there’s nothing wrong with it. (assuming certain things about birthday randomness and independence). The guy that corrected you was not correct.
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post 1626954803 12-29-2020, 06:43 PM
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I changed it to odds of hitting same number again in 23 tries. Going to Vegas to get rich boyos!
post 1626955143 12-29-2020, 06:45 PM
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Originally Posted By numberguy12⏩
A guest lecturer was going to use this example to teach probability to a classroom that just happened to have 23 people in it. Unfortunately his calculation of around 50% likelihood that at least two people share the same birthday was totally off. What gives?
Oh. He did it in a 4-year cycle of counting Feb. 29 once.
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post 1626955663 12-29-2020, 06:51 PM
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How is that a paradox?
post 1626955793 12-29-2020, 06:53 PM
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Originally Posted By numberguy12⏩
Nope he was was thinking just in terms of month and day. And yet his estimation of 50% for the room of 23 people was still way off.
Nonrandom sample jumps out right off the bat.
post 1626955923 12-29-2020, 06:54 PM
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Originally Posted By WeBoutDat⏩
I changed it to odds of hitting same number again in 23 tries. Going to Vegas to get rich boyos!
Lol. No your roulette example is different and breaks down. You are now talking about a specific number you have to hit.

The chances of hitting that number at least once in 23 spins is 1-(364/365)^23, which is like 6%. Contrast to your die example where you said the probability of two of the same number coming up in 23 rolls (no one is specifying which number that is). That probability is around 50% like this birthday problem.
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post 1626956013 12-29-2020, 06:55 PM
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Originally Posted By Smegmalion⏩
A bat and a ball cost $1.10. The bat costs $1 more than the ball. How much is the ball?
depends...am I part of a BLM protest?
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post 1626956093 12-29-2020, 06:56 PM
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Originally Posted By Smegmalion⏩
A bat and a ball cost $1.10. The bat costs $1 more than the ball. How much is the ball?
5 c
post 1626956223 12-29-2020, 06:58 PM
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Originally Posted By numberguy12⏩
Calculate the chances everyone has a different birthday:

-First person can have whatever
- chances 2nd person has different: 364/365
-chances 3rd person has different than both first and second: 363/365
-chances 4th person has different than first 3: 362/365
....etc...
-chances 23rd person has different than first 22: 343/365.

So this probably is 364x363x362x...x343/(365x365x365x...365) otherwise denoted (364!/342!)/365^22. I just multiplied this by 365/365, or 1 to get the expression in the above.

Probability that at least two share the same is just the complement of the above, so 1-probability of everyone having different birthday found above.
Imagine being this wrong. YIKES!

Here is a quick lesson and you should really write this down bro, it works with any number of ppl in the room with you. From 1 to 178,963,529 to infinity.

The chances of someone having the same birthday as you is 50%

Either they do or they don't.

Chances of winning the lottery? 50%. Either you win or you don't.

Done.
post 1626956313 12-29-2020, 06:59 PM
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Originally Posted By LtGoose⏩
Imagine being this wrong. YIKES!

Here is a quick lesson and you should really write this down bro, it works with any number of ppl in the room with you. From 1 to 178,963,529 to infinity.

The chances of someone having the same birthday as you is 50%

Either they do or they don't. Done.
Trump supporters, ladies and gentlemen. So smart.
post 1626956423 12-29-2020, 07:00 PM
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Originally Posted By LtGoose⏩
Imagine being this wrong. YIKES!

Here is a quick lesson and you should really write this down bro, it works with any number of ppl in the room with you. From 1 to 178,963,529 to infinity.

The chances of someone having the same birthday as you is 50%

Either they do or they don't.

Chances of winning the lottery? 50%. Either you win or you don't.

Done.
Lol....if only
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post 1626956513 12-29-2020, 07:01 PM
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Originally Posted By numberguy12⏩
Lol....if only
post 1626956663 12-29-2020, 07:03 PM
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Originally Posted By Omnivium⏩
You call me retard and factorials scare you. SAD!
post 1626956873 12-29-2020, 07:06 PM
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last weekend, there were 5 of us in a room and 2 had the same birthday in the same year

chit was wild
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post 1626956933 12-29-2020, 07:07 PM
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Originally Posted By numberguy12⏩
A guest lecturer was going to use this example to teach probability to a classroom that just happened to have 23 people in it. Unfortunately his calculation of around 50% likelihood that at least two people share the same birthday was totally off. What gives?
Spoiler!

Because when he took a closer look at the back row...he noticed a pair of identical twins.


perhaps a little corny..
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post 1626957963 12-29-2020, 07:21 PM
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Originally Posted By NealIRC⏩
Nope.

Nope, that'd be (1/365)^23.
Originally Posted By mesobuild⏩
It's actually 0.5^22 because each time either hits or misses, 50/50.
lol dude any number less than 1 when raised to an exponent greater than 1 will become SMALLER in value.

i might leave this forum
post 1626958893 12-29-2020, 07:30 PM
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Originally Posted By Merovingian11⏩
I've known like 3-4 dudes that have the same DOB as me. Including my best friend from Junior High and High School.

Life is weird life is strange.

365/10, other than leap year.
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post 1626959093 12-29-2020, 07:32 PM
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Originally Posted By numberguy12⏩
Spoiler!

Because when he took a closer look at the back row...he noticed a pair of identical twins.


perhaps a little corny..
My statement that it is a nonrandom sample applies here kind of too. A classroom has people mostly of a certain age and conditions during that period may have favored some months over others or something.
post 1626959113 12-29-2020, 07:32 PM
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Originally Posted By optimisticAir⏩
lol dude any number less than 1 when raised to an exponent greater than 1 will become SMALLER in value.
You're right. It's 50% because any number of people ends up with either a match or no match. 2 choices. 50% chance.

Quod erat demonstrandum.
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