Reposting as i get an error when pulling it up

Cute math problem for the next club ride:
How many cyclists must be on the club ride before the odds of two riders having the same birthday hits 50% -Mike G

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Originally Posted by fastupslowdown

Cute math problem for the next club ride:
How many cyclists must be on the club ride before the odds of two riders having the same birthday hits 50% -Mike G

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23

I am more interested at what in your original post create that error on the forum engine ha ha.

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

People literally don't believe me when I tell them the answer to this, sometimes even after explaining the math.
Even more odd, it's practically 100% (99.999%) probability of a shared birthday for a group of 100.
Veridical paradoxes, man.

I guess the percent sign not followed by two characters that represent hex for another character?

Also, notice on the chart that a group of 57 has a probability of 0.99.

https://statisticsbyjim.com/fun/birthday-problem/

Back in the day, the University of Michigan had a ride-share board for intercity travel. I'd regularly go visit my girlfriend at Miami University in my Impala sedan.

So one trip, there's five of us - all strangers - in the car and I overhear a conversation in the back seat. A girl was telling a story and said, "...my birthday's August 23." Another girl immediately exclaims, "So's mine!"

So is mine.

That's not exactly probability at work but it was a very cool moment. 60% of us in that car shared a birthday.

Monty Hall:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Quote:

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Originally Posted by fastupslowdown

Monty Hall:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

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Yes! Always an advantage. Took me a while to get comfortable with this one but it makes sense.

Quote:

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Originally Posted by fastupslowdown

Monty Hall:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

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I'd do whatever Marilyn says.

And if you’d rather have the goat?

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Originally Posted by j44ke

And if you’d rather have the goat?

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Relevant XKCD: https://xkcd.com/1282/