# maximum likelihood or else

Suppose you throw 1000 darts where each dart has 0.5 probability of scoring. For the first 500 darts each is worth 1 point, for the second 500 darts each is worth 3 points. If you score 1500 points, how many 3 point darts have you scored most likely?

If I use the likelihood method, I need to maximize the following function: $$\binom{500}{k} \binom{500}{1500-3k}$$ where $k$ ranges from $334$ to $500$. The maximum happens at $k = 398$.

However, if I try to find the expectation I get a different value. From the question we can model the first dart as a binomial distribution with $B_1(500,0.5)$ and the second dart as a binomial distribution $B_2$ with same parameters. Then we need to find $$E [B_2 | B_1 + 3B_2 = 1500]$$ which turns out to be $375$.

My question is which approach should I choose given that I obtain different values for each?

• "Most likely" has to be the mode, since it has the highest probability mass. – TenaliRaman May 21 '14 at 13:26
• +1 to @TenaliRaman's comment. This is an example where the mode of the probability mass function is different from its integral, which is the expectation. Which one you need really depends on what you will do with the result. – Stephan Kolassa May 21 '14 at 13:36