Suppose I have 500 silver coins and 500 gold coins, and all are fair. If a heads on a silver coin gives me 1 dollar, and a heads on a gold coin gives me 3 dollars. Then we know the expected number of dollars received after flipping all 1000 coins is $1\cdot250 + 3\cdot250=1000$.
The question: Suppose I flipped all the coins and got 1100 dollars. What is the most likely configurations of the number of silvers and number of golds?
Thoughts: I know there are many possible ways this could happen. The extra \$100 outside of the expected 1000 could come from, say, 100 extra silver heads, or 33 golds and 1 silver. Intuitively, both of these configurations are unlikely because there is only way of scoring 100 extra silvers, e.g. So something in between all silvers and (almost) all golds should be more likely.
I could brute force this problem by using the binomial distribution to calculate the probability of any number of heads: $P(k \text{ heads}) = (500 \text{ choose } k)\left(\frac12\right)^{500} $. So we could find the probability of any configuration ($j$ silver heads, $k$ gold heads). But I would have to plug in every possible combination that adds up to 100.
Note: this was an interview question, and was meant to be solved intuitively rather than with equations.
Let $N_S$, $N_G$ denote independent binomial random variables with parameters $(500,\frac 12)$. You want to maximize the conditional probability $$P(N_S=i, N_G=j|N_S+3N_G=1100)$$ over $0\leq i,j\leq 500.$
This quantity is proportional to $$\binom{500}{i}\binom{500}{j} 1_{i+3j= 1100},$$ hence it suffices to maximize $$a_j:=\binom{500}{1100-3j}\binom{500}{j} ,$$ where $200\leq j\leq 366$.
By analyzing the ratio $a_{j+1}/a_j$, we see that $a_j$ is maximal for $j=280$, hence $i=260$.
Thus the most probable configuration is $260$ silver coins and $280$ gold coins.