When you have over 750,000 miles on just one of the big carriers, you are always aware that with each flight the odds get better of a problem occurring. Air travel is very safe, but the actuarial tables don’t lie. Fly often enough long enough and something will certainly happen and after awhile you wonder “ is this the one that does me in”?
This is what's known as the "Gambler's Fallacy" — the notion that if you've flipped tails five times in a row, the next flip is more likely to go heads.
In fact, each coin flip is independent: the coin doesn't know what has happened in the past.
Now, other than making fun of a common fallacy — certainly a worthy effort in itself — there's another reason to post this. As I was thinking about it, a way to point out the problem with this reasoning occurred to me that I had never thought of before. (Original? I doubt it. But new to me.)
Think about the whole population on the plane. In fact, assume that Frank, with his 750,000 miles, is sharing the plane with some five year old who has never been on a plane before in his life. If this reasoning were correct, the odds of the plane crashing have to be much higher for Frank ... but much lower for the five year old.
Since the plane can't both crash for Frank and not crash for the five year old, there has to be something wrong with the original reasoning.
3 comments:
The nice thing about probabilistic misunderstandings is that, unlike political arguments, you can dramatically demonstrate the reality to the people involved. If they think their odds of flipping a head increase after flipping five tails in a row, place a little bet with them. The prospect of losing a few dollars will tend to greatly clarify their minds.
Well, that one was simple enough so that even I could understand it. (My general background in determing probabilities comes from assiduously studying outcomes generated by randomly casting paired cubes.)
I'm trying to come up with a formula to determine the probability of a change of control of the House. The current breakdown is 232/R-201/D-1/I with the I voting as a D. 218 is the magic number for control so the R's have to lose 15 seats in order for the D's to take over. There are really only 30 seats that will be contested heavily and the probability for each seat is slightly different. What does the determinative formula look like?
Rick, what you're describing is called a "discrete markov process". Since it's more or less the middle of the night here, I wouldn't take this toooooo seriously, but my first look at it says you're going to have 2**30 possible outcomes (ack, we can't have superscripts in our HTML) and it'll be a differential equation in 30 terms. Nothing that can't be solved with a computer, but with different probabilities you're not going to get a simple formula.
If it were exactly 50/50 it's much simpler.
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