Adjusting a probability based on another related probability

[Cloud8]

Hi all,

I was hoping someone would be able to help me out with something. I am trying to figure out how to adjust a probability using another related probability. Here is the scenario I’m working with:

Say someone is considering pursuing flying small airplanes and would plan to fly a total of 1000 hours. Assume the odds of someone being involved in a fatal accident over the course of these 1000 hours of flying is 1 in 140. But, assume 320 of these 1000 flying hours would involve necessary travel that would be traveled by car (42,000 total miles) if not by flying. Assume the odds of someone dying in a car crash for this amount of mileage is 1 in 1179.

It seems like one could then argue “Indeed the odds of a small airplane fatal accident for 1000 hours are 1 in 140. However, if this person chose flying, there would no longer be a chance of them being involved in a fatal car accident during this required travel. Thus, one should account for this which would produce an ‘equivalent’ odds of less than 1 in 140”. I understand the car and airplane probabilities are independent but if one is trying to figure out additional added risk of this person getting into flying then it seems this removed risk of driving for required travel should be factored in somehow.

The best method I can come up with is to just take the overall flying fatality probability (1 in 140) and subtract out the driving fatality probability (1 in 1179). This would be 1/140 – 1/1179 = 1 in 159 equivalent odds. Based on simple intuitive reflection something around 1 in 160 feels like it could be right (it shouldn’t be something like 1 in 300 for example). But I’m just not convinced it is this simple or that you can just subtract out the probability.

Am I on the right track or if not, how should this analysis be done? Or maybe is this just an apples and oranges type of comparison where an "equivalent" odds/probability isn't really possible?

Thanks

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Below are more details if it is helpful…

• I am assuming the fatal car accident rate is 1.26 deaths per 100 million miles driven
• I am assuming the fatal small airplane accident rate is 0.72 per 100,000 hours flown
• I am using the Poisson Distribution where survivability = e ^ (-x * rate), where x is either miles (car) or hours (small airplane).
• The fatal accident probability for 320 hours flying is 1 in 435. Thus it seems(?) that a potentially helpful generalization can be made that flying is about 3 times more fatal than equivalent driving (1 in 1179).

 

[HallsofIvy]

You have a choice of flying or driving? What is the probability you will fly and what is the probability you will drive?

(I would say "three times as dangerous", not "three times as fatal". There is no such thing as "more fatal"!) (And notice the "as" rather than "more". "Three times as much as x" means "3x". "Three times more than x" means "x+ 3x= 4x".

 

[Dr.Peterson]

You repeatedly use the word "odds" where I suspect you mean "probability", because you say "1 in 140" and "1/140" rather than "1 to 140". If the odds in favor of an accident were 1:140, then the probability would be 1/141. This is a common mistake in nontechnical usage. (Similarly, "3 times more" is used so often, even in math texts, to mean "3 times as much" that I just consider it an idiom, but check the data if possible to see if the latter is what was meant.)

As for the question itself, I don't think I'd use the term "equivalent probability". I'd just say that flying increases the probability of an accident from 1/1179 = 0.085% to 1/140 = 0.71%, that is, 8.4 times as much.

On the other hand, that's comparing flying both for necessity and pleasure with driving only for necessity, and doing nothing for pleasure. What is done when not driving, in the absence of the plane, might have its own risks. Apples and oranges do come to mind ...

 

[Cloud8]

"Three times as much as x" means "3x".

I will try to say three times as fatal now, thanks.

You have a choice of flying or driving? What is the probability you will fly and what is the probability you will drive?

There is no "probability of flying or driving". My scenario is set such that I know when someone would fly vs drive, there wouldn't be any sort of deliberation between the two.

Perhaps I could rephrase my scenario as such if it is helpful to clarify what I'm working with: Assume someone is already planning to drive 42,000 miles in a car for required travel. This person is considering getting into flying and if they did they would substitute this driving for flying (320 hours flying vs 42,000 miles driven). If this person did pursue flying they would not only substitute this driving for flying, but would also do an additional 680 hours of flying for other reasons (fun, etc.). Thus, they would fly a total of 1,000 hours. What I am looking for is what it seems you are mentioning, is some kind of increase in probability of a fatal accident if they pursued flying. But it would be an increase in probability that takes into account there being less driving as a result.

You repeatedly use the word "odds" where I suspect you mean "probability", because you say "1 in 140" and "1/140" rather than "1 to 140".

Good point. I will keep this in mind as well.

As for the question itself, I don't think I'd use the term "equivalent probability". I'd just say that flying increases the probability of an accident from 1/1179 = 0.085% to 1/140 = 0.71%, that is, 8.4 times as much.

On the other hand, that's comparing flying both for necessity and pleasure with driving only for necessity, and doing nothing for pleasure. What is done when not driving, in the absence of the plane, might have its own risks. Apples and oranges do come to mind ...

I think I see where you are getting your 8.4 times as much conclusion. But for the reasons you describe I probably wouldn't want to phrase it that way because it feels like it may be a little too "apples and oranges" like you say.

What do you think about the following instead?

Before I get to the conclusion below in bold I want to preface it with this. If we look at the chance of it being better that someone would have flown vs driving you would want to do the following:

1. You want to look at the required travel for flying vs. driving. The situation where it would have been better to fly vs drive is only when you avoid the situation where someone would have been in a fatal car crash had they driven AND surviving the same amount of equivalent flying hours.
2. Find the probability of a fatal car crash happening for these 42,000 miles = 1/1179
3. Find the probability of surviving a plane crash for these 320 hours = 99.77%
4. The probability of it being "better" is 1/1179 * 99.77% = 0.08462% or a probability of 1/1182.

Thus perhaps a good way to communicate the overall conclusion to the entire scenario is as such: If this person were to pursue flying there would be a 1 in 140 chance of a fatal plane crash and there would only be a 1/1182 probability of flying being more safe due to less driving that would occur because of flying.

Thanks to both of you for your responses so far!