

Nir Dahan Inner circle Munich, Germany 1390 Posts 
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. 
landmark Inner circle within a triangle 4755 Posts 
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
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Nir Dahan Inner circle Munich, Germany 1390 Posts 
Jack there is indeed a mistake.
think bayes formula... 
Scott Cram Inner circle 2677 Posts 
Agreed. Relying on Bayesian probability is a mistake.

TomasB Inner circle Sweden 1143 Posts 
Could it be around half?
1 / (1 + 999 * 0.5^10) /Tomas 
Nir Dahan Inner circle Munich, Germany 1390 Posts 
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? 
Scott Cram Inner circle 2677 Posts 
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.

Nir Dahan Inner circle Munich, Germany 1390 Posts 
Quote:
On 20070131 12:44, Scott Cram wrote: 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 
TomasB Inner circle Sweden 1143 Posts 
I didn't use Bayes' though, but the definition of conditional probability: P(AB) = P(A[union]B) / P(B)
/Tomas 
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