I don’t understand the outcome

[J2neuby]

Hi all.
There is a starting total value of 10,000 and a series of random calculations that affect this running total.

A fixed percentage of the total value is used for each series of calculations with two possible outcomes, each with an equal opportunity of occurring (50/50):

One outcome is to multiply the total value by the percentage and subtract that amount from the total value to grant a new total value for the next calculation.

The other outcome is to multiply the total value by the percentage and then by 1.5 and then add that to the total value to grant a new total value for the next calculation.

The series of calculations are truly played out randomly, but if we were to argue that after 100 runs, there would be exactly 50 of both types of outcomes, then based on the percentage used, a final outcome can be obtained and compared to running the series over again with a different percentage.

My question is: why isn’t there a linear outcome (ie greater percentage used grants a greater total outcome)? For me, there seem to be greater total outcomes as you increase the percentage from 0.001, and then it peaks at a certain percentage, and then begins to diminish the total value when using even greater percentages than that peak percentage. The question is, why is this so, instead of linear?

 

[lex]

These 2 operations can be simplified. In one case you are multiplying the previous number by (1-p) and in the other by (1.5p+1)
(p being your 'percentage').
So if you do one operation 10 times and the other 10 times, your final answer will be [MATH](1-p)^{10}(1.5p+1)^{10}[/MATH] times the starting value. (Order doesn't matter obviously).
This is clearly not a linear function of p. It is a polynomial of degree 20.
In fact the graph is:


The max occurs when [MATH]p=\frac{1}{6}[/MATH]
and then the value is[MATH] \left(\frac{25}{24}\right)^{10}[/MATH] times the original.

If you do each operation n times. The max will occur when [MATH]p=\frac{1}{6}[/MATH]and then the value is[MATH] \left(\frac{25}{24}\right)^{n}[/MATH] times the original.

 

[J2neuby]

Thank you so much!

 

[J2neuby]

What could be done to increase the 1/6 to 1/5 or better?

 

[lex]

What could be done to increase the 1/6 to 1/5 or better?

You could do several things, but increase the 1.5 to 2 or 3 etc... and that will increase the value from 1/6
(In what sense is that 'better', or was that just a turn of phrase)?

If instead of multiplying by 1.5, you multiply by a number (labelled) 'a',
then the max will occur at [MATH]p=\frac{a-1}{2a}[/MATH]
and, if you do each of the two operations 'n' times:
the max value will be [MATH]\left[ \raise 7pt {a} \left(\frac {a+1}{2a}\right)^2 \right]^n \hspace1ex[/MATH] times the original value.

 

[J2neuby]

You could do several things, but increase the 1.5 to 2 or 3 etc... and that will increase the value from 1/6
(In what sense is that 'better', or was that just a turn of phrase)?

If instead of multiplying by 1.5, you multiply by a number (labelled) 'a',
then the max will occur at [MATH]p=\frac{a-1}{2a}[/MATH]
and, if you do each of the two operations 'n' times:
the max value will be [MATH]\left[ \raise 7pt {a} \left(\frac {a+1}{2a}\right)^2 \right]^n \hspace1ex[/MATH] times the original value.

Thank you kindly. Just better if looking for the highest ending total outcome

 

[J2neuby]

These 2 operations can be simplified. In one case you are multiplying the previous number by (1-p) and in the other by (1.5p+1)
(p being your 'percentage').
So if you do one operation 10 times and the other 10 times, your final answer will be [MATH](1-p)^{10}(1.5p+1)^{10}[/MATH] times the starting value. (Order doesn't matter obviously).
This is clearly not a linear function of p. It is a polynomial of degree 20.
In fact the graph is:
View attachment 27355

The max occurs when [MATH]p=\frac{1}{6}[/MATH]
and then the value is[MATH] \left(\frac{25}{24}\right)^{10}[/MATH] times the original.

If you do each operation n times. The max will occur when [MATH]p=\frac{1}{6}[/MATH]and then the value is[MATH] \left(\frac{25}{24}\right)^{n}[/MATH] times the original.

How did you solve for the p value that would be at the peak of the curve? I.e. how did you know that p = 1/6 would produce the maximum output?

likewise, where did 25/24 come from? How was the polynomial reduced to that fraction? You answers were quite correct, but I lack understanding of how you reduced the polynomial with p’s into its maximum output value.
Thanks in advance!

 

[lex]

Assuming p>0 then your final value after n of each operation is:

[MATH](1-p)^n(\tfrac{3}{2}p+1)^n\\ =[(1-p)(\tfrac{3}{2}p+1)]^n[/MATH]which will be max when [MATH] (1-p)(\tfrac{3}{2}p+1)[/MATH] is max.

[MATH]p(x)=(1-p)(\tfrac{3}{2}p+1)\\ =-\tfrac{3}{2}p^2+\tfrac{1}{2}p+1[/MATH]
(1) [MATH]p(x)[/MATH] is max when [MATH] p'(x)=0[/MATH][MATH]-3p+\tfrac{1}{2}=0\\ p=\tfrac{1}{6}[/MATH](then substitute back into [MATH]p(x)[/MATH]...etc...=[MATH]\left(\tfrac{25}{24}\right)^n[/MATH])

or
(2) [MATH]p(x)=-\tfrac{3}{2}p^2+\tfrac{1}{2}p+1\\ =-\tfrac{3}{2}(p^2-\tfrac{1}{3}p)+1\\ =-\tfrac{3}{2}((p-\tfrac{1}{6})^2-\tfrac{1}{36})+1\\ =-\tfrac{3}{2}(p-\tfrac{1}{6})^2+\tfrac{25}{24}\\ =\tfrac{25}{24}-\tfrac{3}{2}(p-\tfrac{1}{6})^2[/MATH]Since a square is ≥0
the max of this expression is when [MATH]p=\tfrac{1}{6}[/MATH], and then [MATH]p(x)=\tfrac{25}{24}[/MATH]
Therefore [MATH][(1-p)(\tfrac{3}{2}p+1)]^n[/MATH] is max when [MATH]p=\tfrac{1}{6}[/MATH],
and has value [MATH]\left(\tfrac{25}{24}\right)^n[/MATH]

 

[J2neuby]

Assuming p>0 then your final value after n of each operation is:

[MATH](1-p)^n(\tfrac{3}{2}p+1)^n\\ =[(1-p)(\tfrac{3}{2}p+1)]^n[/MATH]which will be max when [MATH] (1-p)(\tfrac{3}{2}p+1)[/MATH] is max.

[MATH]p(x)=(1-p)(\tfrac{3}{2}p+1)\\ =-\tfrac{3}{2}p^2+\tfrac{1}{2}p+1[/MATH]
(1) [MATH]p(x)[/MATH] is max when [MATH] p'(x)=0[/MATH][MATH]-3p+\tfrac{1}{2}=0\\ p=\tfrac{1}{6}[/MATH](then substitute back into [MATH]p(x)[/MATH]...etc...=[MATH]\left(\tfrac{25}{24}\right)^n[/MATH])

or
(2) [MATH]p(x)=-\tfrac{3}{2}p^2+\tfrac{1}{2}p+1\\ =-\tfrac{3}{2}(p^2-\tfrac{1}{3}p)+1\\ =-\tfrac{3}{2}((p-\tfrac{1}{6})^2-\tfrac{1}{36})+1\\ =-\tfrac{3}{2}(p-\tfrac{1}{6})^2+\tfrac{25}{24}\\ =\tfrac{25}{24}-\tfrac{3}{2}(p-\tfrac{1}{6})^2[/MATH]Since a square is ≥0
the max of this expression is when [MATH]p=\tfrac{1}{6}[/MATH], and then [MATH]p(x)=\tfrac{25}{24}[/MATH]
Therefore [MATH][(1-p)(\tfrac{3}{2}p+1)]^n[/MATH] is max when [MATH]p=\tfrac{1}{6}[/MATH],
and has value [MATH]\left(\tfrac{25}{24}\right)^n[/MATH]

Thank you very much for the detailed response. It was quite helpful.