TITLE: 3 coins - 1 biased
QUESTION [7 upvotes]: We have 3 coins, of which 2 are fair and the 3rd is biased (gives heads with probability 4/7).
Two friends throw, in turn, all coins together, one time each, and the winner is the one who gets more heads. What is the probability of a draw?
My first attempt:
In order to have a draw, both flips must give the same output: Either
TTT or HTT or THT or TTH or THH or HTH or HHT or HHH, where the first two letters correspond to the fair coins and the 3rd letter to the biased.
So we have 8 possible outputs for EACH draw and we calculate the individual probability for each of them:
1st: 1/2*1/2*3/7=3/28
2nd, 3rd, 7th: same
All the rest: 4/28 = 1/7. 
All of them must add to 28/28, which indeed is OK.
But then, in order to have a draw in both flips, we must have:
TTT with TTT (zero heads)
or 
HTT or THT or TTH with either of 
HTT or THT or TTH
or 
THH or HTH or HHT with either of
THH or HTH or HHT
or HHH with HHH.
Probability for first is 3/28 * 3/28 
Same also for the last one, 4/28 * 4/28
I am a bit confused as regards the other two cases and also how to calculate the total probability for a draw.
(Also, I am not sure for what I have done so far!)
I am not at all familiar with probabilities and combinatorics :(

REPLY [2 votes]: For an individual person the probabilities are:

zero heads $\frac3{28}$ 
one head $\frac3{28}+\frac3{28}+\frac4{28}=\frac{10}{28}$ 
two heads $\frac4{28}+\frac4{28}+\frac3{28}=\frac{11}{28}$ 
three heads $\frac4{28}$ 

making the probability of a draw 

$\left(\frac{3}{28}\right)^2+\left(\frac{10}{28}\right)^2+\left(\frac{11}{28}\right)^2+\left(\frac{4}{28}\right)^2 = \frac{246}{784} \approx 0.3138$

If all three coins had been fair, the answer would have been $0.3125$ so not much different