I am doing a course on probability–
Here is an example from the course which I am quiet confused with calculations for favorable outcomes —
Q is: To determine number of dice ('n') to roll to get optimum expected value. Problem statement is as follows —
"Suppose we modify a dice rolling game in your favor. You pay 3k(dollars) to play, and now you win and get a 10k(dollars) payoff if you get either a 5 or a 6 , but no 1s; otherwise, you lose. how many dice should you roll to get optimum expected value"
(for brevity I am only showing favorable outcome calculation as that is the crux of the solution in my opinion)
Solution says favorable outcomes are calculated as $(5^n – 3^n)/6^n$
I am trying to understand why favorable outcomes are calculated as $(5^n – 3^n)/6^n$ instead of $(1-4^n)/6^n$.
Isnt it that Except 5 & 6 other numbers are not favorable.
Or to put it other way. If we get any of the other 4 (1,2,3,4) we end up not getting any reward.
So I am saying if say we were asked
"Suppose we modify a dice rolling game in your favor. You pay 3k(dollars) to play, and now you win and get a 10k(dollars) payoff if you get either a 5or6 a , but no 1s & 2s; otherwise, you lose."
Does the favorable outcomes now become $(4^n – 2^n)/6^n$
Regards
Best Answer
We want to count the rolls where at least one $5$ or one $6$ comes up, but no $1$s. First of all, there are $5^n$ rolls that have no $1$s. From those, we have to exclude the rolls where no $5$ or $6$ shows up, that is those rolls where the only numbers that come up are $2,3,4$. There are $3^n$ of these. This explains the first calculation.
For the second calculation, the same sort of reasoning shows that excluding both $1$s and $2$s does indeed lead to $4^n-2^n$ favorable outcomes.