But the probability that a true random generator will output 0000000000 should be the same as the probability that it will output 0010111101, because all sequences of equal length are equally likely.
The numbers produced are independent of each other and for our purposes we don’t care about the order. The relevant thing is how likely it is is to produce a given total number of zeroes or ones.
p(just one 1) = 10/2^10; A whole heap more likely!
So the chance that the generator is fair is rather slim. You can calculate just how slim by simply applying bayes rule (and doing some integration).
On a related note if you role two six sided dice you are just as likely to get two sixes as you are to get a three and a five. But if you are playing Settlers of Catan and put all your settlements next to the twelve instead of the eight then you are probably going to lose.
if you role two six sided dice you are just as likely to get two sixes as you are to get a three and a five.
Nitpick: this is true if by “a three and a five” you mean (that the dice are labeled and) “die A comes up 3, and die B comes up 5″, but it’s false as written (and in games like Settlers, the identities of simultaneously thrown dice are not tracked).
(and in games like Settlers, the identities of simultaneously thrown dice are not tracked).
They are tracked actually, at least in the latest version—the only one worth playing. The red one determines whether or not you qualify for a bonus card. So yes, obviously I mean “a three and then a 5”. If also considering a five and then a three then the point becomes even stronger.
With a fair random generator:
p(0000000000) = 1/2^10
p(1000000000) = 1/2^10
p(0100000000) = 1/2^10
p(0010000000) = 1/2^10
The numbers produced are independent of each other and for our purposes we don’t care about the order. The relevant thing is how likely it is is to produce a given total number of zeroes or ones.
p(just one 1) = 10/2^10; A whole heap more likely!
So the chance that the generator is fair is rather slim. You can calculate just how slim by simply applying bayes rule (and doing some integration).
On a related note if you role two six sided dice you are just as likely to get two sixes as you are to get a three and a five. But if you are playing Settlers of Catan and put all your settlements next to the twelve instead of the eight then you are probably going to lose.
Actually p(just one 1) = 10/(2^10).
Nitpick: this is true if by “a three and a five” you mean (that the dice are labeled and) “die A comes up 3, and die B comes up 5″, but it’s false as written (and in games like Settlers, the identities of simultaneously thrown dice are not tracked).
They are tracked actually, at least in the latest version—the only one worth playing. The red one determines whether or not you qualify for a bonus card. So yes, obviously I mean “a three and then a 5”. If also considering a five and then a three then the point becomes even stronger.
Hm. So the only relevant measure is the prevalence of zeros, because the generators are stateless (n+1st digit does not depend on the nth digit)?
But what if the generator B was not necessarily stateless?