Is it true that if you have a random group of at least 25 people your odds of finding someone with the same birthday (Month,Day) are about 85% to 90%? Does this hold true for pocket change also?
IIRC the odds are even money after 23 people. Would doubt that your odds are 90% after 25 people. If someone else in the TM doesn’t provide the probability calculations then I’ll see if I can find my uni stats notes later.
Well, it’s higher than you would expect, but not quite that high. For 25 people, the probability at least two have the same birthday is about 56.8% (assuming birthdays are uniformly distributed randomly, ignoring leap day). One way of thinking why it’s so high is, notice that there are 300 different possible pairs of people in a group of 25–a lot of pairs of birthdays to compare.
About the pocket change, are you asking what’s the probability of two people out of a group of 25 having the same amount of pocket change? Similar principal, but you would have to make some assumptions on the distributions of different amounts of pocket change across the population. For example, if you were to assume everyone had $3.64 or less in pocket change at all times, and any amount was as likely as any other amount, then it would be exactly like the birthday problem.
That’s pretty straightforward, once you see the trick: work out the probability that you don’t find a match.
Take the first person’s birthday. It’s certain not to make a pair, so the probability is 1. Then the chance of the next person not matching is 364/365 (ignoring leap years - that would be messy), since there are 365 possible dates and we want one that hasn’t been used. The next person has a chance of 363/365 to miss the two we used already, the next 362/365, and so on. Since we need all those conditions to be true, we multiply the probabilities:
1 * 364/365 * 363/365 * 362/365 * …
until we reach the number of people we want. Since we calculated the probability of not getting a pair, the chance that we do get one is 1 minus the result above.
I didn’t do the arithmetic, but IIRC the even-odds point is between 23 and 24, as wooly says.
Long time lurker, but one of my favorite math “tricks” beings me out…
The previous post nails the key to understanding how to calculate such things. What are the odds of A happening? 100%, minus the odds of A not happening.
For the birthday thing, what are the odds that two or more people out of X people will have the same birthday…
Uhhhhh, X being between 1 and 365.
(forgive my attempts at notation…)
1 - Product(N=1 to X) [{366-N}/365]
There’s probably a simpler way of expressing it, and it certainly looks better on paper, but it works.
Threw that into a TI-92(?) for an algebra class I taught once. They were entirely unimpressed, I’m afraid.