We have two fair coins, A and B. We flip each coin 50 times, and for every toss of A in which we get a head, we receive \$5, and for B, we receive \$1. Suppose we played this game, and we received \$250. What is the most likely number of heads for A and B that we received for the 50 tosses of each?
I think we can approximate the binomial distributions as normal, and we have a bivariate normal for the likelihood. However, I'm not sure whether this is right/where to go next.
I am not sure what approach you have been taught, I am just attacking it by commonsense
We can ignore the probabilities because they will be $(1/2)^a(1/2)^{(50-a)}(1/2)^b)(1/2)^{(50-b)} = (1/16)^{100}$
We just need to maximise the combos that yield $250$\$ for which we want $\binom{50}{A}\binom{50}{B}$ to have the maximum value
The ideal situation would be for both $A$ and $B$ at $25$ each to maximise the combinations, but this is not possible because of the $250$\$ constraint. Also, it is easier to push $B$ to give a higher combo because for every step of $A$, $B$ needs to move $5$ steps to maintain $250$\$ constraint
So we start from $A$ and $B$ as close to each as possible at $A=42,B=40$ and reduce $A$ by steps of $1.$ I get the maximum at $A=44, B = 30$