Angela rolls a fair die nine times and each time she rolls a three. What is the probability that on her next roll, she will roll another three?

[Anonymous]

[abc3643]

In principle, it's 1/6th. That's because the purpose of the educational exercise is to tell you that rolling a die is an example of an independent event that doesn't depend on past rolls. In other words, if the die is fair, then each face will have equal probability of occurring on any roll and that probability does not change based upon previous values of rolls of the die.

So, you put down 1/6 as the answer.

HOWEVER! In reality, this is not a fair die. The probability of rolling 9 straight 3s is 1/6^9 = 9.9229 x 10^-8 or about 1 in 10,000,000.

the James maximum entropy principle

[abc3643]

Jesus fucking Christ I hate this phone!

The Jaynes Maximum Entropy Principle would determine that the most likely role you're going to have is going to be a 3. You see, that probability of 1/6 is actually an assumption about the fairness of the die but the evidence is that the die is not fair.

So if you want to be a wise-ass, site the Jaynes Maximum Entropy Principle should be used for determining the probability because it will determine an unbiased probability distribution for that die.

[abc3643]

en.m.wikipedia.org/.../Principle_of_maximum_entropy

[abc3643]

Jaynes, E. T. (1957). "Information theory and statistical mechanics". Physical Review. 106 (4): 620–630.
http://bayes.wustl.edu/etj/articles/theory.1.pdf

Jaynes, E. T. (1957). "Information theory and statistical mechanics II" (PDF). Physical Review. 108 (2): 171–190.
http://bayes.wustl.edu/etj/articles/theory.2.pdf

[iBluestar]

Well so far its a 100% chance since all she's rolled is a 3 but if we were to assume it's out of ten she would have a 90% chance

[jarbaugh]

I can't tell if this is real...

[Anonymous]

1/6 the past doesn't effect the odds

[SuccessfulHornDog]

1/6 .

[QueenofCups]

One in six.

[Anonymous]

Just ask Angela

[Asker]

Haha very funny