Posterior distributions and loss functions. I am unsure why I got this question wrong:

So if my posterior probability of a head-based coin is currently 60%, then that means I'll choose heads and so my EV of a loss is 4 (40% * 10). However, if I wait and flip another coin, I immediately incur a loss of 1 + whatever my new EV is. What's my new EV seems to be the heart of the question.
How do I know what my new posterior is after I flip another coin?
 A: As indicated in the hint, you can find the posterior using Bayes rule. For example, if $P\left( B_H \right)=0.6$ is your current posterior probability that the coin is head-biased, and $P\left( {H|B_H} \right)=0.75$ the likelihood of obtaining head in the next flip (under the hypothesis of a head-biased coin), then the posterior probability of a head-biased coin after having observed head in the next flip is  
$$
P\left( {{B_H}|H} \right) = \frac{{P\left( {H|{B_H}} \right)P\left( {{B_H}} \right)}}{{P\left( {H|{B_H}} \right)P\left( {{B_H}} \right) + P\left( {H|{B_T}} \right)P\left( {{B_T}} \right)}} \approx0.82
$$
Hence the expected loss (after having observed head and decided that the coin was biased toward heads) would be 
$$L(B_H,H)=1 + \left[1-P\left( {{B_H}|H} \right) \right] \cdot 10 \approx 2.8$$
You can do the same calculation and compute the expected loss after having observed $T$ (tail). I find that $P\left( {{B_T}|T} \right) \approx 0.67$, and that the expected loss (after having observed tail and decided that the coin was biased toward tails) is $L(B_T,T)\approx 4.3$.
However I am not entirely sure of what they mean with minimum posterior expected loss in the question, in particular when they write that the "minimum expected loss of making the decision now is 4": I think this is confusing, and should be just the "expected loss of making the decision now". The minimum expected loss is instead associated with the other possible course of action, that is making the decision after another flip. You can compute the expected loss of flipping another time the coin by taking the average of the expected loss of each of the outcomes of the flip (which we computed above), weighted by their probability. The probability $p(H)$, for example, is given by the denominator of the equation above
$$ p(H) = {P\left( {H|{B_H}} \right)P\left( {{B_H}} \right) + P\left( {H|{B_T}} \right)P\left( {{B_T}} \right)} = 0.55$$
The expected loss of flipping the coin another time is then
$$
p(H)\cdot L(B_H,H) + p(T)\cdot L(B_T,T) = 3.5
$$
So the correct answer should be the third one.
