How to extrapolate from 1-year health risk?

[Melisande]

Would anyone here happen to know how to extrapolate health risk? For example, if I have a 5% chance of having a stroke this year (for the sake of argument — my true risk is thankfully lower than this), I don’t think this means that in 20 years I will have 100% chance of having a stroke. In other words, you don’t simply multiply the risk factor by the number of years, do you? But if not this, then what do you do?

Yes, I only thought of this question after I left the doctor’s office, unfortunately.

[AnswerIs42]

5% chance of having a stroke means 95% chance of not having a stroke in any one year.

Assuming each year is independent, then 0.95 ^ 20 = 0.358

So 35.8% chance of not having a stroke in any year in the next 20 years, which is 64.2% chance of having a stroke in at least one year.

[soccerluvof4]

5% chance of having a stroke means 95% chance of not having a stroke in any one year.

Assuming each year is independent, then 0.95 ^ 20 = 0.358

So 35.8% chance of not having a stroke in any year in the next 20 years, which is 64.2% chance of having a stroke in at least one year.

Just shows other than General math I was not very good at it! haha

[Melisande]

5% chance of having a stroke means 95% chance of not having a stroke in any one year.

Assuming each year is independent, then 0.95 ^ 20 = 0.358

So 35.8% chance of not having a stroke in any year in the next 20 years, which is 64.2% chance of having a stroke in at least one year.

What is the ^ operation? Is that a root or something? I'm just wondering how to replicate that with my real risk number -- 1.9% per year.

And how did you know that? Could you describe the reasoning behind the calculations?

[AnswerIs42]

What is the ^ operation? Is that a root or something? I'm just wondering how to replicate that with my real risk number -- 1.9% per year.

And how did you know that? Could you describe the reasoning behind the calculations?

^ is "to the power of". If you've got independent probabilities, and you want to calculate the probability of them all happening, then you need to multiply the probabilities by each other. So it's 95% chance it doesn't happen in the first year, then 95% chance it doesn't happen in the second, etc - so you calculate 0.95 * 0.95 * 0.95 ...etc until you have 20 of them.

So for the real figure of 1.9%, which means 98.1% chance of it not happening in a year:

0.981 ^ 20 = 0.681

So 68.1% chance of it not happening in 20 years, therefore 31.9% chance of it happening at least once (assuming every year is independent of the last).

[Bracken_Joy]

Another way of looking at it involves fractions. Here's a page on calculating the probability of independent events: https://www.mathgoodies.com/lessons/vol6/independent_events

"Summary: The probability of two or more independent events occurring in sequence can be found by computing the probability of each event separately, and then multiplying the results together." (the sequence here being 20 years, so you're multiplying the probability together 20 times- which is what the power/exponent is doing there).

Hopefully that helps and doesn't just muddy the waters more =)

[Melisande]

What is the ^ operation? Is that a root or something? I'm just wondering how to replicate that with my real risk number -- 1.9% per year.

And how did you know that? Could you describe the reasoning behind the calculations?

^ is "to the power of". If you've got independent probabilities, and you want to calculate the probability of them all happening, then you need to multiply the probabilities by each other. So it's 95% chance it doesn't happen in the first year, then 95% chance it doesn't happen in the second, etc - so you calculate 0.95 * 0.95 * 0.95 ...etc until you have 20 of them.

So for the real figure of 1.9%, which means 98.1% chance of it not happening in a year:

0.981 ^ 20 = 0.681

So 68.1% chance of it not happening in 20 years, therefore 31.9% chance of it happening at least once (assuming every year is independent of the last).

OK, great. Thanks so much!

[Dr Kidstache]

Just a quick addition to the excellent math above - your risk of stroke is unlikely to be static over time. It could go down or it could go up - your personal physician can better explain what the modifiers are that will change your risk over time. I'm sorry that you've received news that you're at an elevated risk of stroke.

[Abe]

The formulas above assume a constant risk over time with each year an independent risk period. That isn’t exactly how it works since age is a risk factor for stroke. The increase in risk with age is low enough per year that it’s probably close enough for now, but you’ll need to be reassessed regularly. I’m not an expert in stroke but the most recent population data indicates the risk increases 10% per year with age. This increase should be taken into account with any age related health risk. Then there’s also conditional lifetime risk, wherein the fact you haven’t had a stroke is taken into account along with your age.

[Freedom2016]

Are physicians able to give health risks by year?

I had breast cancer so I'm familiar with breast cancer risk calculations, and it's always stated as a "lifetime risk." Which changes the calculations considerably!

[Abe]

We can to some extent but it’s less accurate than lifetime risk. Also it gives a false impression of linear risk with time, which as you know isn’t accurate for cancer, or even most diseases. I prefer lifetime (for my patients with family history of cancer) or conditional lifetime risk (risk of recurrence x years since their treatment for a patient that has had cancer) for those reasons.

[Melisande]

Just a quick addition to the excellent math above - your risk of stroke is unlikely to be static over time. It could go down or it could go up - your personal physician can better explain what the modifiers are that will change your risk over time. I'm sorry that you've received news that you're at an elevated risk of stroke.

Thanks, I was already aware of this, though. Just wanted to simplify the question.