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
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
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?
Tom Ambros
Also, notice on the chart that a group of 57 has a probability of 0.99.
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.
GO!
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?
And if you’d rather have the goat?
Relevant XKCD: https://xkcd.com/1282/
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