1 out of 1 members found this post helpful.
Did you find this post helpful?
Yes |
No
Originally Posted by
21forme
Don, what is the significance of ln2 in solving this? How was that derived?
Let's start with something simple: A coin-toss game.
We want to know how many tosses we need to get an edge. We know that to have an edge, we need an EV of > 0; therefore our losses should be less than our wins, or our wins should be the majority of our outcomes.
For simplicity, let's assume we want to know our break-even value. That would be 0.5-0.5 = 0. So, we start off with the following formulae:
1-(1/2)^x = 0.5
We then solve for x:
-(1/2)^x = 0.5-1 = -0.5
(1/2)^x = 0.5
Here, we then solve for x using this law of Logarithms : Log(n^k)= k*Log(n); where Log(n) is the natural Log of n
Log(0.5^x) = Log(0.5)
x*Log(0.5) = -0.69314718
x = -0.69314718/Log(0.5)
x = 1
Now we know we need 1 roll per game to average an EV of zero.
Now, where does the Ln2 come from?
Look back at our Law of Logs formulae : Log(n^k)= k*Log(n); where Log(n) is the natural Log of n
Substitute n for 0.5
Log(0.5^k)
If you remember algebra, 2^-1 = 0.5; so sub that into the equation:
Log(2^-1)
and solve for the rest:
Log(2^-1) = -1*Log(2) = -1*(-0.69314718) = ~0.69
Therefore Log(2) ~ 0.69
Bookmarks