For the same reason that the probability of having a 100-year storm in any given 100-year period is not 100%.
On average, any 100-year period will contain a single 100-year storm - but some of those periods will have two or three such storms, and some of those periods will have zero. If you go through the same sort of math I did upthread, you find out that the odds of any 100-year period containing one more more such storms is only 37%.
The same sort of thought process can be applied to smaller blocks of time as well.
It seems to me that the odds of it happening in any given year are 100 to 1. It doesn’t matter if it happened two years in a row or hasn’t happened for 150 years.
Twenty is 1/5 of 100. So, the chances of it happening within twenty years should be five times greater than 1/100, which would be 1/5 or “5 to 1 odds”.
No, the average number that you would expect would be 20 times greater than the average number you’d expect in a single year. But the catch is that it’s also possible to have more than one in 20 years.
Not quite. Let’s talk about dice for a moment, just to abstract away some of the issues with storms.
The probability of a die coming up “6” is 1/6. What’s the probability of getting at least one six if you throw three dice? You’d think it was 3/6 or 0.5. But it’s not. It’s 1-125/216 = 91/216 - or about 42%. However, the average number of 6s is 0.5, just as you expected. This is because the average number of sixes includes the cases where more than 1 six comes up (there’s a 1/216 chance of 3 sixes and a 16/216 chance of 2 sixes); since the average number of sixes is 0.5, and this average includes cases with more than 1 six, the probability of exactly one six has to be less than 50%
Similarly, since more than one big storm can occur in the 20 year period, and the average number of storms per 20 years should be 0.2, the probability of exactly one storm is less than 0.2.
Okay, a probability question that’s unrelated to the one in the OP.
You have five regular dice (six-sided cubes, with the sides numbered one to six). You roll the five dice and then you’re free to group the rolled results as you wish. One die is put aside and the remaining four are grouped into two pairs.
In what percentage of random rolls will you be able to group the dice so the two pairs of dice will have the same total?
This is not true assuming the events are independent and identically distributed.* The next year is the one in which it is most likely to happen next, but it’s just as likely to happen in year on as in year n. That’s what identically distributed means. The next year is the one mostly for the first occurrence because for the first occurrence in year two, the event must occur in year two and not occur in year one.
*Not assuming iid could result in virtually any asnwer
It’s partly specific to floods (yes, due to climate cycles), but there’s part of it that’s generally applicable, and just has to do with model error.
The 100-year floodplain is based on a model. A pretty complicated one that incorporates watersheds and climate and ground permeability and so on. But reality is even more complicated, and may diverge from the model. One reason you should expect more unlikely events after experiencing one is that your model for how likely the event is is wrong, and those events are actually more likely than you thought.
If you flip a coin once and it comes up heads, what’s the probability that it will come up heads on the next flip? The model says 50%, but the actual probability is greater than 50%, because it incorporates the possibility of systematic error like the flipping process not being uniform, the coin not being a fair coin, etc.
If you flipped a coin 100 times and got 100 heads, it’s certainly possible that it’s a fair coin and process and you hit an extreme outlier, but I would sure not put an even money bet on flip 101 coming up tails!
I got the same answer when I brute-forced it (systematically checking all the possibilities). Is that how you did it, or is there a way to figure it out?
“One of these days in your travels, a guy is going to show you a brand-new deck of cards on which the seal is not yet broken. Then this guy is going to offer to bet you that he can make the jack of spades jump out of this brand-new deck of cards and squirt cider in your ear. But, son, do not accept this bet, because as sure as you stand there, you’re going to wind up with an ear full of cider.”
There may be a nice way to figure it out, but honestly the first thing I did was just count them, same as you, since in this particular instance there are not that many possibilities. As a side note, note that in general partition problems are NP-hard, but that does not necessarily have any bearing on this puzzle, which is asking for a probability.
And in fact, this is a very common carnival game, usually called “Chuck-a-Luck”. You can put a dollar down on any number, and if your number comes up on any of the dice, you take back your dollar plus an additional dollar.
To see why it’s not actually a fair game (as if the fact that it’s offered by carnies isn’t evidence enough), imagine that there are six people playing, one on each number (in fact, a common situation, at a busy booth). If all three dice are different, then three people lose, and give their dollars to the three people who win, and the house takes nothing. But if two of the dice happen to match, which is fairly often, then three people lose a dollar, but only two people gain a dollar, and the house takes a dollar profit. And in the rare-but-not-all-THAT-rare case that all three match, then the house takes two dollars.
(as an aside, some carnivals now play this game, instead of with dice, with a big wheel that spins and has three “random” dice shown on every space of the wheel. But if you count it up, the wheel spaces have a significantly higher number of rolls with doubles and triples than you would expect from random dice. So they’re taking a game with already very good odds for the house, and making it even more skewed.)
I started that way but I realized there were 7776 possible combinations of just the five dice being rolled. And then I was going to have to figure out all of the possible combinations of grouping them 1-2-2 dice (which I calculated at 116,640 possibilities). And then adding up the possible totals of the two pairs and comparing them.
So I figured there must be a simpler algorithm and I hoped somebody more probability-savvy than myself would know it.
You’re right that there are 7776 possible dice rolls, but it’s not tricky to ask the computer to loop through all of them. Each roll has 5 different ways of picking the single die to set aside, and then each of those has 3 ways of pairing up the four other dice.
That’s a bit glass-half-empty, especially if you are allowed to use a computer There are only 126 ways to roll 4 dice if you don’t care about the order, and 252 outcomes for all 5 dice. And once you know, e.g., 1+4 = 2+3, that covers the rolls 1,1,2,3,4 ; 1,2,2,3,4 ; 1,2,3,3,4 ; 1,2,3,4,4 ; 1,2,3,4,5 and 1,2,3,4,6. That does not mean this is the Right Way to solve the puzzle, but it shows how, with this low number of dice, you could even tally all the outcomes manually if you wanted.
Say we have one flood every 100 years and we have one this year. What are the chances that the next in 100 year flood happens next year. One in a hundred.
What are the chances that the next one happens the following year. Well, 99 chances in 100 that it doesn’t happen next year X one in a hundred. So .99 in 100.
The next year, 99/100 X 99/100 X 1/100 = .98 in 100. And so on.
It’s just an amusing fact that people have trouble getting their heads around. So, when you get unlikely things happen people think, “It will be a long time before that happens again.” But the most likely year for it to happen next is always next year.
And once you understand it, you come up with interesting questions when people talk about how unlikely, say the COVID-19 pandemic was. In which year after this one are we most likely to have another pandemic?
There are only 126 ways to roll 4 dice if you don’t care about the order, and 252 outcomes for all 5 dice. And once you know, e.g., 1+4 = 2+3, that covers the rolls 1,1,2,3,4 ; 1,2,2,3,4 ; 1,2,3,3,4 ; 1,2,3,4,4 ; 1,2,3,4,5 and 1,2,3,4,6. That does not mean this is the Right Way to solve the puzzle, but it shows how, with this low number of dice, you could even tally all the outcomes manually if you wanted.