You have 999 dimes in a jar which are normal, and one dime which is 2 sided (both heads).
you now pick a coin at random from the jar and flip it 10 times.
amazingly you got 10 heads in a row.
what is the probability you are holding the 2 headed coin?

N.

Not sure on this, but I'll take a stab:

P( I pick a normal dime and it comes up heads 10 times) = (999/1000) x (1/2) ^10

So probability I picked the two headed coin = 1 - [(999/1000) x (1/2) ^ 10] = .99902.


Hmmm, seems too high to me. Where is the mistake in my thinking?


Jack Shalom

Jack there is indeed a mistake.
think bayes formula...

Agreed. Relying on Bayesian probability is a mistake.

Could it be around half?

1 / (1 + 999 * 0.5^10)

/Tomas

Tomas got it! (again)

from a belly "feeling" it should be 50% since 1/1000 is almost as 1/(2^10)
do we need the exact solution?

If you can pick the coin out of the jar, and you're able to flip it, wouldn't you also be able to look at the coin itself? Why not just look on both sides? If it has no tail, you defintely have the coin! Otherwise, you don't.

scott com'on,
do you also reason with other logical puzzles the way you did here?
it would be no problem to devise a set of conditions that will limit the person from looking at both sides...
this is a mathematical puzzle and should be solved that way.

Nir

I didn't use Bayes' though, but the definition of conditional probability: P(A|B) = P(A[union]B) / P(B)

/Tomas