444 avg WR - 41.29 Erik Akkersdijk

cmhardw said:
Yes really. I don't get your question? Are you asking for the demonstration or the proof of this?
Click to expand...
Yes, sorry, I realise that wasn't clear now. It's not apparent to me at all how e would be relevant at all to the calculation of that probability.
 
hawkmp4 said:
cmhardw said:
Yes really. I don't get your question? Are you asking for the demonstration or the proof of this?
Click to expand...
Yes, sorry, I realise that wasn't clear now. It's not apparent to me at all how e would be relevant at all to the calculation of that probability.
Click to expand...

Let \( N \) be a natural number. Now assume that the probability of some event occurring is \( \frac{1}{N} \)

We will run a trial of N attempts at this event, and we are interested in the probability that the event occurs at least once. This is done by finding 1 minus the probability that the event never occurs.

\( 1 - \left(1-\frac{1}{N}\right)^N \)

Now let's think about the number \( e \) for a moment. There are numerous definitions of \( e \), but one of which is the following:

Invest a sum of money for one year at an annual interest rate of 100%, compounded continuously. We are interested in the accumulation factor of that sum of money.

\( a(n) = \left(1+\frac{i}{n}\right)^n \) where n is the number of times you compound in 1 year, and i is the interest rate expressed as a decimal.

Then:
\( \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n \) will express the accumulation factor for investing a sum of money for one year at an annual percentage of 100%, compounded continuously. This limit definition is also equal to \( e \). So \( e \) or approximately 2.71828 is the accumulation factor for this sum of money.

Now back to our probability question. We are interested in the probability:
\( 1 - \left(1-\frac{1}{N}\right)^N \)

Not only that, but let's examine what happens when we let N go to infinity. So we are really interested in:

\( \lim_{N \to \infty} 1 - \left(1-\frac{1}{N}\right)^N \)
\( \lim_{N \to \infty} 1 - \left(\frac{N-1}{N}\right)^N \)
\( \lim_{N \to \infty} 1 - \left[\left(\frac{N}{N-1}\right)^{-1}\right]^N \)


Now let's introduce a variable substitution. Let \( N=K+1 \)

Since we are examining the limit as N approaches infinity, this will be similar to investigating when K approaches infinity, as K is defined simply as the number before N.

\( \lim_{K \to \infty} 1 - \left[\left(\frac{K+1}{(K+1)-1}\right)^{-1}\right]^{K+1} \)

\( \lim_{K \to \infty} 1 - \left(\frac{K+1}{K}\right)^{-K-1} \)
\( \lim_{K \to \infty} 1 - \left[\left(1+\frac{1}{K}\right)^{K+1}\right]^{-1} \)

\( 1 - \left[\lim_{K \to \infty}\left(1+\frac{1}{K}\right)^{K+1}\right]^{-1} \)
\( 1 - \left[\lim_{K \to \infty}\left(\left(1+\frac{1}{K}\right)^{K}\times\left(1+\frac{1}{K}\right)\right)\right]^{-1} \)
\( 1 - \left[\left(\lim_{K \to \infty}\left(1+\frac{1}{K}\right)^{K}\right) \times \left(\lim_{K \to \infty} \left(1+\frac{1}{K}\right)\right)\right]^{-1} \)
\( 1 - \left(e \times 1 \right)^{-1} \)

\( 1 - e^{-1} \)
\( 1 - \frac{1}{e} \approx 0.632121 \)

Chris
 
Last edited:
Ah! Okay. Thank you.
I think I misunderstood Lucas- I thought that he was saying that the fact that 64 trials gave a number related to e was coincidence. I understand now. Bad reading on my part.
Thank you again, for yet another wonderful explanation.
 
Edward said:
*Sees another world record*
Wat8.jpg
*Sees that it's not Faz this time*
Both+Ears.jpg

Congrats Erik!
Click to expand...
LOL :)
 
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