aka. Things That Everybody Should Know About Probabilities, part 2
Doubling the number of attempts doubles my likelihood to succeed
Ok. This is not that common, but annoying enough. And I wanted to make sure this is a series, and one post is not so series-ish, huh?
Lemme tell you a story. Imagine there was a lottery every weekend, and every time one (1) ball out of ten, numbered from 1 to 10, would be picked. Now, there are exactly 10 total outcomes each time (duh). Think about the probability of winning if you participate only once, using a single ticket. Yes, of course it is 1 out of 10, or 10%. Or 0.1 as mathematicians would say, using the normalized form.
Now try to guess what is the probability of winning — at least once — if you participate exactly twice, using a single ticket both times. Is it 2 out of 10, that is, 0.2?
No. Definitely not.
If you disagree, then, according to that logic, participating ten times should equal 10 out of 10, equal to 100%, meaning absolute certainty of winning at least once! But we all know that you still wouldn’t necessarily win, even if you participated in our contrived lottery ten times. Why is that?
The answer is simple: by picking your choice another time you are not “covering twice the ground” compared to the last time — each lottery outcome and ticket is completely unrelated to next lottery, and you won’t win if the ball you picked last time would be picked now. The case is completely different if you buy ten tickets to single lottery, enumerating all the possibilities (that is, every ball from 1 to 10), but rest assured the winning coefficient has been setup so that you would lose by filling an exhaustive pile of tickets; in real-life situation you would be payed at most 8:1, or maybe as little as 6:1.
But now I digress. What is the actual probability for winning if you’d participate ten times, filling a single ticket. If it is better than one out of ten (which even our intuition tells it has to be) but worse than absolute certainty (again, our intuition should now nod approvingly)… what are the odds?
The solution is simple once you understand the concept of complement probability, which is a very simple one: if any event x has a probability p, the probability “not x” must be q = 1 – p — that is, the sum of probabilities “either x or not x happens”
must be one. So the answer for the question “what is the probability of winning at least once when participating with single ticket ten times” is “what is the complement of the probability of not winning at all when participating with single ticket ten times”. Now, because each time the probability to not win, ie. lose which is the technical term the mathematically inclined mathematicians have a habit of using, is 9 out of 10 = 0.9, according to the principle of product we get
(10 times)
which says simply “probability of NOT happening lose at first attempt AND lose at second attempt AND lose …”, which is roughly 0.65, meaning that it is actually more likely to win at once than lose every time. Yup.
Also, make sure that the complementary case for “not winning” is not just “winning” but more precisely “winning at least once”, which becomes more obvious if you elaborate the first option more: by “not winning” you probably mean “not winning even once”, because winning say, three times or every time is likely a welcome situation. Now, the complement of “not winning even once” comes more clear, which is “winning at least once”.
Now, I probably got confused myself. Flame on.