Talk:Wolf Sledge/@comment-71.250.212.253-20200218235405/@comment-25254386-20200219201753

K, the problem with that is that it is only the expected for 3.95%, it's completely independent of the other 3.95% or two 13.55%s. Yes, in 26 runs you'd expect to see $26*3.95%$ successes for that one 3.95%, but that does not mean you'll see $26*3.95%$ and $26*13.55%$ successes for the other 3.95% or two 13.55%s respectively.

Think about this example for a sec: The chance to land on heads or tails of a perfect, two sided coin is 50% for each side. And the expected number of tosses to land on heads is $1/0.5$. In other words, in 2 tosses you could expect to land on heads at least one time. This also applies to tails.

So, in two tosses, can you expect to see both heads and tails at least once? Turns out no, that expected value is not two, but rather three tosses. That can be proven easily with logic by breaking it down into steps: And thus there will be a total of three flips on average.
 * 1) With the first toss you are guaranteed to land on either side. Let's say you land on heads. You've now seen one head and zero tails after one toss.
 * 2) The goal now is to see at least one tails, so we must toss the coin until it turns up just one time regardless of how many heads we see. So now the question becomes how many times should we expect to toss the coin until at least one tails shows up? We've already answered this though, tails has a 50% to happen, so the expected value is simply $1/0.5$.

The logic of simply calling the expected value of multiple independent success the expected of the success with the smallest probability is not correct. It doesn't actually account for the other probabilities, even though you'd think it should if the expected value encompasses the individual values.