My first language is not English, sorry if there are grammatical errors.
In a probability reference class, my professor gave the typical example of what is the most repeated number when rolling a certain number of dice.
Of course it's pretty simple to figure this out by rolling dice amounts of 6, it's just 3.5×N=MRQ (Most repeated quantity).
But what if the dice weren't necessarily 6s, this is still easy with for example a given amount of 10s, it's as simple as 5.5×N=MRQ. N=Number of dice. Being necessary to calculate the 5.5, simply MA/2+0.5.
Creating the formula (MA/2+0.5)×N=MRQ.
But, in a case where the dice have different MA (maximum amounts), how could this be applied?
With this question in mind, myself and a colleague of mine set to work on some way that would allow us to calculate the MRQ, on dice with different MA, regardless of the number of dice (N).
Along the way we came across the Range(R) problem, since in different cases there were more than one possibility with the same probability of being reached, for this reason we focused on finding the TMRQ (Theoretical Most Repeated Quantity), a concept that would represent the MRQ situated among all other MRQs.
Finally, after successfully finding a pattern mainly by brute force, we arrived at what we believed to be the definitive formula, which we tested in different ways and it worked perfectly in all of them.
TQMR=TMA×((AMA+1/)(2×AMA)) TMA=Total maximum amount, being the highest possible number when rolling the dice. AMA=Average maximum amount, being the result of TMA/N
We were satisfied with this formula, and we thought that everything would end there, but in the next class, while we were showing our formula to the teacher, at the end of our explanation a classmate asked us if it would apply to a dice that rolled decimal, not whole numbers.
When verifying that our formula did not apply, we began to realize that our formula did not apply with Negative Numbers, Not integers, and dice with jumps different from 1.
Without being discouraged by this discovery, we once again got involved in discovering a formula, this time definitive. We finally found this.
TQMR=TMA×((AMA+AS)/2×AMA) AS=Average start, calculable as TS/N S=Smaller number of each dice TS=Sum of all S
Finally, remembering a suggestion from another classmate, it occurred to me to try in cases in which there was a non-linear jump, or even chaotic as prime numbers. For example review 2 dice that will work with scales of multiplying by 2 up to 16 2,4,8,16 When calculating it by hand, it gives as a result that 6,10,14,18,20,24 are equally probable, seeing this one might think that 15=TQMR.
Using the formula returns 18, which actually represents the true center on that scale of number functioning. Since if 1=2 2=4 3=8 4=16 5=32 On this numerical scale the true midpoint between 6 and 24 would be 18.
It occurred to me to post this here, to see if someone could explain why it works, what the 2 represents in the formula. It would also be useful to know of any cases in which this formula is wrong and needs correction, or if someone has a precedent for a formula that seeks to calculate the same thing.
I cannot see how $18$ can be considered "halfway" between $6$ and $24$, especially if you measure it on a scale such as your suggestion, \begin{array}{rr} 1 & 2 \\ 2 & 4 \\ 3 & 8 \\ 4 & 16 \\ 5 & 32 \\ \end{array} According to this, $4$ is the midpoint between $2$ and $8$, because $2\times 2=4$ and $4\times 2=8$. Similarly, $6\times 2=12$ and $12\times 2=24,$ so the midpoint of $6$ and $24$ would be $12$ on that scale.
So that part of the question is hard to interpret. But it may not matter by the time we finish looking at this.
Interpreted according to the standard precedence of operations,
$$TMA \times ((AMA+AS)/2\times AMA) = TMA \times \left(\frac{AMA+AS}{2}\times AMA\right). $$
But since you say the formula gives the result $18$ for your two dice with faces numbered $2,4,8,16$, I infer that your intended formula was
$$TMA \times ((AMA+AS)/(2\times AMA)) = TMA \times \frac{AMA+AS}{2\times AMA}. $$
We can simplify this by observing that $TMA = N\times AMA$ and $TS = N \times AS$:
\begin{align} TMA \times \frac{AMA+AS}{2\times AMA} &= N\times AMA \times \frac{AMA+AS}{2\times AMA} \\ &= \frac{N\times (AMA+AS)}{2} \\ &= \frac{TMA+TS}{2} \\ \end{align}
So your formula can be written $TQMR=\dfrac{TMA+TS}{2}$.
The obvious "meaning" of the $2$ in the formula is that you are taking the arithmetic mean of the two most extreme sums (the least sum and the greatest sum). This will give good results as long as the numbers on each die are an arithmetic series, or at least are arranged symmetrically around a "central" value, because the arithmetic mean of all the most frequent results will be equal to the mean sum of all rolls, which also is given by the same formula.
For your two dice with faces $2,4,8,16$, the formula says $TQMR=(TMA + TS)/2 = (16 + 16 + 2 + 2)/2 = 18,$ as you wrote.
Now let's try three dice with faces $2,4,8,16.$ The most frequently occurring sums are $14, 20, 22, 26, 28$, each occurring $6$ times out of $64$ rolls on average. The formula says $TQMR=(3\times16 + 3\times2)/2 = 27$; is that the true midpoint of that list of numbers?
Now let's try four dice with faces $2,4,8,16.$ Now the formula says $TQMR=(4\times16 + 4\times2)/2 = 36$. But the most frequently occurring sum is $30,$ occurring $24$ times out of $256$ rolls on average. No other sum occurs as frequently; the second most frequent sum is $28$, which occurs $20$ times out of $256$ rolls on average.
We can easily find more extreme examples. Consider two dice that each have faces labeled $1,2,3,999$. The formula says $TQMR=(999 + 999 + 1 + 1)/2 = 1000$. But the most frequent sum is $4$, occurring $3$ times out of every $16$ rolls on average.
Now try two dice with faces labeled $1,997,998,999$. The formula still says $TQMR=1000,$ but the most frequent sum is $1996$.
And we haven't yet even considered dice with different numbers of faces, or faces with non-integer values.
If you want to consider any number of dice, not necessarily the same, with arbitrary numbers on their faces, the problem is quite complicated. You will have to consider every number on every die, not just the least and greatest numbers.