Accident Probabilities: Bicycles vs. Cars vs. Trucks

[Agent Smith]

TL;DR Summary

Bicycle vs. Light Vehicles (your ordinary car) and Heavy Vehicles (trucks)

I read a news article shouting for more regulations on trucks as a study revealed that compared to accidents involving cars, the death rates for accidents involving trucks was higher (among cyclists).

I imagined this scenario (it squares with the news article mentioned):
P(accident involving cars) = 0.8
P(dying in a car accident) = 0.3

P(accident involving trucks) = 0.2
P(dying in a truck accident) = 0.7

P(accident involving a car and dying in that accident) = P(C) = $0.8 \times 0.3 = 0.24$
P(accident involving a truck and dying that accident) = P(T) = $0.2 \times 0.7 = 0.14$

I can't parse this well, despite having computed the probabilities. Does P(C) > P(T) weaken/nullify the justification for more regulations on trucks?
Anything else I should be asking?
Gracias.

 

[pbuk]

I read a news article shouting for more regulations

Firstly, shouty news articles are not known for the soundness of their statistical basis.

Secondly you need to be clear about what is meant by the word "truck". In the US (and possibly elsewhere) this is a truck - in fact it is advertised as a "full size truck":

In much of the rest of the English speaking world, this is a truck.

(Source for both Wikimedia)

Vehicles like those in the second picture have caused a number of fatalities due to turning right (in countries that drive on the right-hand side of the road) and crushing a cyclist who has decided to creep up the inside of the vehicle at a junction.

Does P(C) > P(T) weaken/nullify the justification for more regulations on trucks?

No, the fallacy of this can be seen by considering whether the fact that (in most countries) more people die from falling down stairs than from automatic weapon fire weakens the argument for regulations on automatic weapons.

 

[Agent Smith]

@pbuk, thanks for the vocab clarification.

But, per the calculations, you're more likely to do die in a light vehicular accident than in a heavy vehicular accident. I tried something else:

P(accident involving a car and not dying) = P(NC) = $0.8 \times 0.7 = 0.56$
P(accident involving a truck and not dying) = P(NT) = $0.2 \times 0.3 = 0.06$

P(NC) > P(NT)

Cars are far safer given the above BUT ...

P(accident involving a car and dying in that accident) = P(C) = $0.8 \times 0.3=0.24$
P(accident involving a truck and dying that accident) = P(T) = $0.2 \times 0.7 = 0.14$

P(C) > P(T)

 

[pbuk]

... misread

 

[pbuk]

@pbuk, thanks for the vocab clarification.

It's not just vocab: the risks associated with a 40 foot 32 tonne vehicle are clearly different from those associated with a 20 foot 2.5 tonne vehicle. Which are you talking about?

But, per the calculations, you're more likely to do die in a light vehicular accident than in a heavy vehicular accident. I tried something else:

Different calculations but leading to the same conclusions. Similarly, stairs are far safer than automatic weapons but (among most populations) P(incident involving stairs $\land$ dying) (in the UK I think its about 1.5x10^-5 per annum) $\gg$ P(incident involving an automatic weapon $\land$ dying) (about 5x10^-7 per annum for deaths from all firearms).

 

[FactChecker]

I can't parse this well, despite having computed the probabilities. Does P(C) > P(T) weaken/nullify the justification for more regulations on trucks?
Anything else I should be asking?

IMO, you should consider how much effort and cost the regulations require. I would prefer that the objective be to maximize lives saved per effort/cost.

 

[Dale]

I imagined this scenario (it squares with the news article mentioned):
P(accident involving cars) = 0.8
P(dying in a car accident) = 0.3

P(accident involving trucks) = 0.2
P(dying in a truck accident) = 0.7

P(accident involving a car and dying in that accident) = P(C) = $0.8 \times 0.3 = 0.24$
P(accident involving a truck and dying that accident) = P(T) = $0.2 \times 0.7 = 0.14$

Let's write things more clearly. Let $C$ be the event that there was an accident with a car, let $T$ be the event that there was an accident with a truck, and let $D$ be the event that the cyclist died. So we have $$P(C)=0.8$$$$P(D|C)=0.3$$$$P(T)=0.2$$$$P(D|T)=0.7$$ So by the definition of conditional probability $$P(D \cap C)=P(C) P(D|C)=0.24$$$$P(D\cap T)=P(T) P(D|T)=0.14$$

So, we can see that $P(D|T)>P(D|C)$ meaning that it is more likely that a cyclist dies given that they were in an accident with a truck than it is that a cyclist dies given that they were in a car accident. We can also see that $P(D\cap C)>P(D\cap T)$ meaning that it is more likely that a cyclist is in a car accident and dies than it is that a cyclist is in a truck accident and dies.

Does P(C) > P(T) weaken/nullify the justification for more regulations on trucks?

No. Probabilities don't tell you what you should or shouldn't do, and they don't justify or un-justify regulations. They simply tell you risks and uncertainties. What you decide to do or not to do can be perfectly justified based on other considerations. This usually depends on your goals, your tolerance for risk, and the cost-benefit of all the possible choices.

Here, it is unclear what the goal of the proposed regulations is. I would guess that the goal is to reduce $P(D)$, the probability of a cyclist dying. This is important to clarify because none of the above computations give you $P(D)$. It is possible that a regulation could reduce $P(D\cap T)$ without changing $P(D)$. It is also possible that a regulation could reduce $P(D)$ without changing either $P(D\cap T)$ or $P(D\cap C)$.

Here, it is also unclear what the tolerance for risk is. This is sometimes known as the value of a statistical life or the value of preventing a fatality. In typical liberal democracies that is typically in the range of 1 to 10 million dollars. In other words, 1000 people would typically be willing to each pay $1000 to $10000 to reduce their individual risk of death by p=0.001.

Here, it is also unclear what the cost-benefit of the proposed regulations are and who bears the cost. Maybe three different regulations exist, all of which lead to the same decrease in $P(D)$ and one would cost $100 per car driver and the other would cost $1000 per truck driver and the last would cost $1000 per bicyclist. The total cost to society would depend on the number of car drivers, truck drivers, and cyclists. And the willingness to pay may differ for the different groups. Car drivers may not be willing to pay $100 to reduce the risk for cyclists while cyclists may be willing to pay $1000 to reduce the risk to themselves.

Probabilities cannot make decisions for you. They can just quantify risk.

 

[Agent Smith]

It's not just vocab: the risks associated with a 40 foot 32 tonne vehicle are clearly different from those associated with a 20 foot 2.5 tonne vehicle. Which are you talking about?

This is true. missed that completely. Comes down to physics, of which I know little, I suppose. Would you agree that cars clock higher speeds than trucks?

As to your point on stairs vs. guns, it seems analogous to the scenario in my OP if the chance of falling down the stairs and dying > the chance of getting shot and dying. Contd. below

IMO, you should consider how much effort and cost the regulations require. I would prefer that the objective be to maximize lives saved per effort/cost.

That's correct. Wouldn't the probabilities affect that? Contd. below

@Dale thanks for clarifying the situation. Yes, given the probabilities, the chance of dying in a car accident < chance of dying in a truck accident (0.3 < 0.7), but the chance of getting in a car accident and subsequently dying > the chance of getting in a truck accident and dying in it (0.24 > 0.14).

I can say these perhaps:
1. If I have a truck accident, I'd probably die as P(dying) = 0.7
2. If I have a car accident, I'd probably survive as P(surviving) = 1 - P(dying) = 1 - 0.3 = 0.7
3. I'd probably die in a car accident (P(car accident & dying) = $0.8 \times 0.3 = 0.24$) than a truck accident (P(truck accident & dying) = $0.2 \times 0.7 = 0.14$) and $0.24 > 0.14$

I saw this at the bottom of this thread:

 

[Agent Smith]

@berkeman thank you for editing my post for clarity.

 

[FactChecker]

IMO, you should consider how much effort and cost the regulations require. I would prefer that the objective be to maximize lives saved per effort/cost.

That's correct. Wouldn't the probabilities affect that? Contd. below

Good. That is exactly right. There are a variety of steps that could be taken for either cars, trucks, or both. The probabilities you indicated are involved in the total cost/benefit of each.

 

[Agent Smith]

Good. That is exactly right. There are a variety of steps that could be taken for either cars, trucks, or both. The probabilities you indicated are involved in the total cost/benefit of each.

 

[Agent Smith]