HERC777 wrote: > and 1,000,000 other people all flip 100 coins themselves. > > on average, will someone flip the same 100 long sequence I did? > > how long a sequence will they match up to on average? > > Herc
There are 1M people and 100 coins. There are 2**100 possible sequences of 100 coins. The chance that 1 out of the 1M will flip the same sequence as you is (with ** indicating exponentiation)
1M * (2**-100) (if I remember correctly you add the probabilities of independent events to get the probability that they will all occcur).
I don't know exactly what this is but its pretty small (certainly less than 1 in 2).
For the second question:
The probability for matching the first 100 is above. For matching the first 99 is 1M * (2**-99) For matching the first 98 is 1M * (2**-98) (way less than 1) . . . For matching the first 1 is 1M * (2**-1) (way over 1)
So somewhere in between there:
For matching the first 20 is 1M * (2 **-20), or around 1M / 1024000 (latter approximate) - slightly less than 1. So around there.
I'm not sure of my logic here. I'm not sure about 1M * (2**-1). I thought probabilities only went up to 1?