
Re: say I flip a coin 100 times...
Posted:
Apr 13, 2005 3:55 PM


Ralph Hartley wrote: > Robert Low wrote: >> Ralph Hartley wrote: >>> ken quirici wrote: >>>> 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). >>> >>> >>> You *multiply* the probabilities of independent events to get the >>> probability that they will all occur. (What you said was wrong, but >>> what you *did* was correct). >> Surely not. > Because 10^6 is much smaller than 2^100 we can ignore the possibility of > multiple matches.
Oh, OK, I see what you're doing. I hadn't realised you were making such heavy use of the actual numbers involved. Fair enough. (Though it does rather require you to have a good feel for the situation to justify the approximation. My approach can also be used by those who, like me, fail to notice such things :))
Nevertheless, what the OP did was wrong in principle (as is shown by some of the computations he did) but a good approximation in the case of interest.
> That gives us 10^6 mutually exclusive events each with > probability 2^{100}. Adding the probabilities gives 10^6 * 2^{100}. > This is a much better approximation than "2^{100} is about 8*10^{31}", > but it counts trials in which there are 2 matches twice (similarly for > three matches etc.)
Well, I could give 2^{100} to more significant figures, but it didn't seem worthwhile in the context.
And as you point out, I could have used the binomial theorem to get just as good an answer.
Nice when all the different approaches work out the same in the end, eh?
 Rob

