I have two coins, one normal and one having both heads. Randomly picked one and tossed it 3 times. All three were HEADs.
What is the probability of seeing a tail on next toss?
Going with independent event approach,
normal: 1/2 x 1/2 = 1/4
biased: 1/2 x 0 = 0
total: 1/4 + 0 = 0.25
Is this correct? Or is there a different approach?
Best Answer
If the coin is fair then the number of heads in a row before you flip has no affect on the probability of another head so the the answer for a fair coin is $\frac{1}{2}$.
If this is a coin with both heads then you clearly has no chances to see tail at all.
Using Bayes' theorem you can say that $$P(\mbox{your coin is double-headed}) = \frac{1\cdot \frac{1}{2}}{1\cdot \frac{1}{2} + \frac{1}{8}\cdot \frac{1}{2}} = \frac{8}{9}$$
Just to add more generality, using the same approach you can show that the probability that the coin is double-headed after $N$ heads in a row and no tails is $$P = \frac{p}{p+(1-p)\cdot2^{-N}},$$ where $p$ is the probability you think the coin is double-headed before you receive any information via flips. After some rather big $N$ it will be very close to $1$ for any reasonable $p$.