<s>[INST] Ben and Jordan have three coins between them. Two of them are fair but one of them has a 4/7 chance of showing heads. Now, Ben and Jordan both flip each coin once and write down the outcomes.
What is the probability they both get the same number of heads?

My approach:
4 possible outcomes: 3H, 3T, 2H&1T, 2T&1H.
$P(3H) = (\frac{1}{2})^2\frac{4}{7} *\binom{3}{1} = \frac {36}{84} $
$P(3T) = (\frac{1}{2})^2\frac{3}{7} *\binom{3}{1} = \frac{27}{84}$
$P(2H,1T) = \frac{2}{3}(\frac{1}{2})^2\frac{4}{7} + \frac{1}{3}(\frac{1}{2})^2\frac{3}{7} = \frac{11}{84}$
$P(2T,1H) = \frac{2}{3}(\frac{1}{2})^2\frac{3}{7} + \frac{1}{3}(\frac{1}{2})^2\frac{4}{7} = \frac{10}{84}$
P(same number of heads) = $(\frac {36}{84})^2 + (\frac{27}{84})^2 + (\frac{11}{84})^2 + (\frac{10}{84})^2 = \frac{1123}{3528} \approx 32\% $
What do you think?
 [/INST] You don't need to multiply by binomial coefficient or in any other way consider different orders of coins. We can simply say that first and second coins are fair, and the last is weighted.
For example, $3H$ means simply that both fair coins and weighted came heads, thus $P(3H) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{4}{7} = \frac{1}{7}$.
Similarly, $2H, 1T$ means that either both fair coins came heads and weighted came tail, or first fair came tail and the rest came heads, or second fair came tail and the rest came tail, so $$P(2H, 1T) = \frac{1}{2}\cdot \frac{1}{2} \cdot \frac{3}{7} + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{4}{7} + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{4}{7} = \frac{11}{28}$$
After finding correct probabilities for all possible numbers of heads, you need to sum their squarees, as you did.
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