latent vine
There's a 41% chance for 0
52% chance for 1
6% chance for 5
1% chance for 10
Assuming I roll these odds 12 times, how would I calculate the average return and it's standard deviation
(If you're curious about the video game it's The Binding of Isaac's Wooden Nickel item)
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There's a 41% chance for 0
52% chance for 1
6% chance for 5
1% chance for 10
As...
+close
HES FAST
rancid python
There's a 41% chance for 0
52% chance for 1
6% chance for 5
1% chance for 10
As...
Oh, BoI! NIce.
Well, as each roll is independent, we can use some properties of expectation and variance.
Since X1, ..., Xn are i.i.d., we have:
E(X1 + X2 + ... + Xn) = nE(X)
σ(X1 + X2 + ... + Xn) = √(n)σ(X)
So, you just need to find E(X) and σ(X), where X is the distribution of pennies from one roll of WN. Shouldn't be a problem.
Also, since this is a simple discrete distribution, you can also easily model the distribution of pennies from any number of throws by convolving the distribution with itself a bunch of times. Should be easy if you know coding.
If you don't, I can try coding it up. Shouldn't be hard.
latent vine
hmph im not much of a coder but i am an excel gremlin
E(x) is expected return and σ is sigma?
rancid python
E(x) is expected return and σ is sigma?
E is expectation, σ (sigma) is standard deviation.
rancid python
hmph im not much of a coder but i am an excel gremlin
Hm... Might be possible in Excel, but probably not the best way.
I'll try coding it.
rancid python
hmph im not much of a coder but i am an excel gremlin
Ok, done. Here's the Scilab code, values and PMF. Sorry for taking so long, took a while to figure out how to rotate x-axis labels.
For Y = X1 + ... + X12:
E(Y) = 11.04
σ(Y) ≈ 5.107
n = input('Number of rolls of Wooden Nickel: ');
seq = [1];
x = 0:(length(roll) - 1);
EX = sum(x.*roll);
sigmaX = sqrt(sum(((x - EX)^2).*roll));
disp('Expected value', n*EX);
disp('Standard deviation', sqrt(n)*sigmaX);
for i=1:n
seq = convol(seq, roll)
end
x = 0:(length(seq) - 1);
bar(x, seq, 1);
a = gca();
a.tight_limits = "on";
a.x_ticks.labels = "$\rotatebox{90}{"+string(x)'+"}$";
a.font_size = 3;
xtitle('P = f(N)', 'Pennies', 'Probability')```
As seen from the PMF, the mode is 7.
rancid python
You're welcome!
Oh, wait. I think I didn't paste the whole code.
Ok, it's good now.
Almost left out the most crucial part 😅
rancid python
how come there are so many 0s in the [roll]
Those are the probabilities of getting 2-4 and 6-9 pennies.
Which are 0, since there aren't any coins with those values.