Scenario: There’s a bag with 4 regular coins (tail-head) and 1 irregular coin (head-head). At random, you take a coin out of the bag and proceed to flip it five times and it comes up heads all five times.
Question: What is the probability that the coin you’re flipping is the irregular coin?
1 to 5, it doesn’t matter how often you flip the coin, the probability of getting the irregular one is always 1-5.
This is a textbook case of where you should apply Bayes’ Theorem. And I’ll leave it at that for now, in case this is a homework problem.
Well, let’s see…
IF you do pick the double-head, then the probability of getting 5 heads in a row is 100%
IF you pick one of the normal coins, then the probability of getting 5 heads in a row is .5 to the fifth power, or 3.125%
So, once you get 5 heads in a row, the probabilities remaining are:
20% (double-headed coin,)
80% * 3.125% = 2.5% (ordinary coin that happened to get 5 heads in a row.)
Everything else has been eliminated. So, the chances that the coin you hold in your hand is 20/22.5, or 88.89% - exactly eight out of nine.
Capi1978, no, that’s not true. The original odds of 1 in 5 can definitely be altered based on the results of flipping the coin. Simple and obvious answer, if you pick a coin at random, flip it, and get tails, the odds that you picked the double-headed coin have dropped from 1 in 5 to zero.
Chronos, I’ll take the chance of having done somebody’s homework - I just hope I did it right. 
I was intrigued by that top picture. Bayes’ theorem spelt out in blue neon at the offices of Autonomy in Cambridge.
This is all correct.
it was a hw problem, but not mine. my sister asked me because her friends were debating it (with some outlandish conclusions).
the way i approached it was similar to chrisk’s. i figured that out of all the different ways 5 heads could come up, 32 permutations were from the trick coin and 4 from the normal coins. 32/36 = 8/9.
i’m trying to do it with bayes’ theorum but i’m not getting it.
P(Q|W) = (P(W|Q)*P(W))/(P(Q))
so, P(fake given 5 heads in a row) = P(5 heads in a row given fake) * P(5 heads in a row) / P(fake)
= (1) * (1/51/32+1/51/32+1/51/32+1/51/321/532/32) / (1/5)
= (1/5)*(36/32)/(1/5)
= 36/32 = 9/8
?
I was eager to see solutions to this, because it does sound clever.
But if the question you’re asking is the probability that the the coin is the irregular one, then the fact that you flipped it at all is irrelevant, right? The chances of picking it out of the bag are still 1:5.
Edit: Sorry, read and thought a little harder. Disregard me!
Of course there are several more complicated answers above, but this is what I’m thinking. 
The equation for the Bayesian update is P(Q|W) = P(W|Q)P(Q)/P(W).
::embarrassed smiley::
