@tomleslie thanks for your effort.

I don't have the GOT package either, Certainly I would use the linear transformation, but it states its only valid for Omega Ratios of over 1. The link which explains the linearisation technique is no longer valid...

I have a suplementary question. I'm not sure about the monthly returns data. I'd assumed they just subtract 1 to get the percent monthly return, Cell R1C1 is 1.283 (28.3% -seems high).next month R1C2 is 0.429 (- 57.1%). But R1C8 is -0.311....??

I want to put my own data though this for 3 stocks, annual return for each (4 years worth):

Matrix(4, 3, [[0.049214060475074796, -0.039495815964258746, 0.11449725779665731], [0.10937360103809103, -0.07403192172846383, 0.05145193083198468], [-0.02002071584993901, 0.032795876292319816, -0.137325940737269], [0.061433054336773145, -0.07926813859959723, 0.5713767521086268]])

@Carl Love 

This is a picture of the excel worksheet (Evolver). Note the low variance... It is calculated independant of the budget. It obtains (slightly) differerent covariant matrix to Maple..

Your method of unitary budget solves the problem of maximizing growth while keeping variance low.

Using your second method: 0 variance produces only moderate growth (3.2%). Have I missed somethng?

@Carl Love You're right of course. I mistook a returns matrix for the covariance matrix. The clue was the 3x3 matrix for Q, whereas a returns schedule of only 3 years would be considered inadequate (would require 7 minimum)...

What i'm after is a procedure to maximize the return g (rather than the stated 10%), and at the same time minimize the variance X^t*Q*X.

@acer thanks 

@Carl Love Whatever size you define in Kitonums code, the resulting gif is exported as a 400x400

@tomleslie I wanted option 3 in Carl Loves post, works well!

@Carl Love edited. It seems to me there are two probabilities. The probability of the particular configuration occurring, which is 1/1001. Then testing whether that configuration was a result of proper randomisation. You applied Fishers test and got probability 6/1001. Because that is a lot less than 0.05 it is significant. i.e there is evidence the pick wasn’t done randomly. I didn’t really think about a test for significance when I originally posted the question.

I’m not sure if you are tearing your hair out at my interpretation but I’m going to bed now.

@Carl Love I don't have Maple 2019. But thanks anyway. I put it through your 2018 code and it works....

So my old maths prof was correct for the probability...1 / 1001 = 0.000999

@mmcdara 

Thankyou for your effort. Just to clarify, each contestant was given an opaque bag containing a buff.

Once they were dispensed, they were told to reveal their buffs (simultaneously). So that would be Scenario 2.

@acer On yours it worked, but I tried this in my worksheet, but label 9 still displays USD....

Feel free to peruse the problem i'm trying to solve, its a work in progress..

Solar_Battery_Payoff.mw

Yuri, how can I use you code to solve a 2 bucket problem with unlimited water.

https://mindyourdecisions.com/blog/2013/02/04/the-water-jug-riddle/

Happy Xmas!

@Joe Riel Sorry for being thick, but whats the instructors number prediction? Would it be an envelope containing one (or more?) of the 8 most frequent sums 1755,1638,1656,...?

is our friend ?

Markiyan Hirnyk 

http://www.mapleprimes.com/posts/200573-Putnam-Done-With-Maple

@Rouben Rostamian  for your effort!

As the error says, you have a symbolic upper limit z. You need to change it to a number