If you've already done it for 12 months without it happening though, the next 3 months are no more dangerous for you than for someone starting from scratch.
That's true, but usually when we are deciding which actions to take, we're not comparing "I take actionA" versus "I take actionB," rather than comparing "I take actionA" versus "some random other person takes actionA."
OK, the next 3 months are no more dangerous for you than if you hadn't spent the last 12 months doing it. What you did in the past has no bearing on the chances going forward. I'm not sure if it's more clear to say it like that or not. Clearly, humans have a lot of trouble speaking and thinking clearly about statistics.
The next three months are no riskier than your first three months were. They don't become more risky because they will add up to 15 months total -- once you've already finished the first 12 without incident.
For the dice roll example that is true. But other examples it isn’t. For example the MTBF of a device that has run for x hours approaching the MTBF is probably more likely to fall in the next x hours. Or if there is some cyclic behavior. Like waiting outside for a hot day.
That's not how this works as a rational investment choice.
It's true that you can never win a lottery you don't enter, but the expected value of that ticket is vastly lower than what you paid for it. That means, as an investment, your $10 will be expected to do better in literally anything with a positive return.
If you are buying > $10 worth of dreaming (for you), fine - but that's consumption.
You forget: once you roll three 6s in a row, you're dead, and you don't roll any more. Your expected calculation assumes that people keep rolling after they get 666.
Though I'm not sure where they got their figure from, because there isn't an “expected time to live”; there's a 90% probability to live time, a 5% probability to live time…
There’s a difference between expected value of number of days you’ll survive and the number of days a given fraction of the subjects will survive, but I don’t see either supporting the claim “If you do it every day, you have about 15 months to live”.
(215/216)^450 ≈ 0.124
, so about one in eight will survive for 15 months or more. The “5% probability to live” time is around day 645 (about 1¾ years):
(215/216)^645 ≈ 0.0501
the “half will survive at least for” point is around 5 months:
RandomSwede's comment is accurate, but maybe the below can help add some 'flesh' to their response.
Basically, the problem is that you can't just multiply it all together.
(1/6) ^ 3 is correct, and the probability of rolling 3 sixes is indeed 1/216 today, but if you repeat independent events, you don't just add up the probability.
Imagine instead of dice it's coins, and it's only two. Your odds of getting HH today are 1/4, but the odds of getting HH by day four are not now 4/4. We know that it's possible, although unlikely, you could flip coins for the rest of your life and NEVER get two heads. So we know that you can't ever have odds of 4/4 (or 1), only odds that approach 1. So that means that we can't say 216 days from now will be 216/216.
Instead, you need to work out the probability of the event NOT happening, and then repeatedly NOT happening independently (so we can multiply together to get the probability.
For our four coins, the probability of NOT getting HH is 3/4. On Day 2, the probability of NOT getting HH on both occasions will be (3/4)×(3/4), (9/16, 56.25%). By day 3, it will be (3/4) × (3/4) × (3/4), or 27/64. On day 4, it'll be 81/256, or 31.6%. Now we can subtract from 1, to work out that by day 4, the odds of us having hit HH are almost 70%.
As RandomSwede explains, there's a 50% chance that you will have rolled three sixes by day 149. By day 496, you're down to 10%.
runs <- 10000
x <- vector(mode = "numeric", length = runs)
for (i in 1:runs){
while (sum(sample(1:6, size = 3, replace = TRUE)) != 18){
x[i] <- x[i] + 1
}
}
summary(x)
quantile(x, c(0.5, 0.8, 0.9))
> summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0 62.0 149.0 216.2 300.0 1902.0
> quantile(x, c(0.5, 0.8, 0.9))
50% 80% 90%
149 350 495
A simple simulation. Run 10K times. Count the number of times it takes for three dice to add up 18.
The numbers very much agree with you. The median is 149. The 90th is 495 in the simulation, which is close enough to 496. There is very much a long tail in the data. So, the median and the average will not be the same. Is it a coincidence that mean is a 216?
No, I don't think this is a coincidence, but I'm not completely confident in saying that.
Thinking about it doesn't make me feel like I'm solving a maths problem. I start stacking ideas and concepts in a way which makes me feel like I'm overlaying them in a way which is incorrect.
It makes me feel like I'm solving a riddle, which hints to me that maybe it's actually a question of semantics and definitions rather than a maths problem.
Dice (typically) do not have a memory, so whatever happened yesterday will not influence what happens today. If you roll it daily, your chance of surviving at least N days is (215/216)^N, for the specific case of "rolling three 6 on three 6-sided dice" that puts you at ~50% at 149 days and at ~10% at 496 days.
At sufficient scale, even incredibly unlikely things become quite probable.
runs <- 10000
x <- vector(mode = "numeric", length = runs)
for (i in 1:runs){
while (sum(sample(1:6, size = 3, replace = TRUE)) != 18){
x[i] <- x[i] + 1
}
}
summary(x)
quantile(x, c(0.5, 0.8, 0.9))
> summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0 62.0 149.0 216.2 300.0 1902.0
> quantile(x, c(0.5, 0.8, 0.9))
50% 80% 90%
149 350 495
A simple simulation. Run 10K times. Count the number of times it takes for three dice to add up 18.
The numbers very much agree with you. The median is 149. The 90th is 495 in the simulation, which is close enough to 496. There is very much a long tail in the data. So, the median and the average will not be the same. Is it a coincidence that mean is a 216?
Off the top of my head, I don't know. It MAY be related to the fact that 6*3 is 216, but I don't have deep enough statistics knowledge to say for sure. You coudl try it again with 3 8-sided dice and rolling 24, that should give you ~50% at 344 iterations, and ~90% at 1177 iterations. If my supposition that the mean is related to the possible rolls, then the mean should end up being 512.
Iteration counts gathered with Python and a (manual) binary search (actually faster than writing code).