First of all: Thank you very much for the reply. I have a task to do, where I have to do the calculation that I described and I didnt find anything on the web, it does seem extremely hard.
Yes, you could place the 50% on the other side of the equal sign and then solve for n. But I think it would be easier to plug in different values for n and then zero in on the correct value. That is by the way a perfectly legitimate way of doing it. Ask your teacher if that will be acceptable.
i think the introduction is misleading. this is not the probability that someone in the room has the same birthday as you, rather this is the probability of 2 random people in the room having the same birth date.
maighe spot on. In truth birthday problems and their analogs are endless. One I like hard to find in print is the following: given the 23 people randomly selected what is the P of finding EXACTLY 2 with the same birthday? Before cerebrating take a guess(thinking fast as per Daniel Kahneman). Cryptic hint: I recall that I read that 23 has 1255 partitions; so exactly 2 and all the rest different is just one of those 1255 partitions, more or less.(could be two fives or two elevens or two twenty ones).
i just love this playlist
great video!
Glad you enjoyed it
How do you change this equation when you want to find out how many people you need for a 50% chance?
It would be easier to change n and use trial and error.
First of all: Thank you very much for the reply. I have a task to do, where I have to do the calculation that I described and I didnt find anything on the web, it does seem extremely hard.
Yes, you could place the 50% on the other side of the equal sign and then solve for n. But I think it would be easier to plug in different values for n and then zero in on the correct value. That is by the way a perfectly legitimate way of doing it. Ask your teacher if that will be acceptable.
It is funny that it only needs 57 to get 99% chance, but 366 to get 100% chance!
+Mark Allen lol..not this year! Need 367 for certainty.
@@enderyu there are only 366 days in a leap year, so you'd need 367 to guarantee two people having the same birthday
i think the introduction is misleading. this is not the probability that someone in the room has the same birthday as you, rather this is the probability of 2 random people in the room having the same birth date.
That is correct.
maighe spot on. In truth birthday problems and their analogs are endless. One I like hard to find in print is the following: given the 23 people randomly selected what is the P of finding EXACTLY 2 with the same birthday? Before cerebrating take a guess(thinking fast as per Daniel Kahneman). Cryptic hint: I recall that I read that 23 has 1255 partitions; so exactly 2 and all the rest different is just one of those 1255 partitions, more or less.(could be two fives or two elevens or two twenty ones).
Oops not two 21’s.
@@alanrapoport2090 who asked
Whoever.
cool
First :)