Need pro statistician help to answer question

jpod · 4 Mar 2016 at 10:23

One individual has 4 teeth that are injured.

i There are 2 rare complications you can develop over time, each has a chance of 5% of developing in a lifetime

ii There is one other rare complication, the risk is 7.5% over a lifetime.

Q/ What is the percentage chance of this individual developing a complication?

Is it 7.5% x 4 = 28% chance over their life

Please explain the correct answer for this maths numpty so I can understand the field of statistics we are referring to. And thank you.

jpod · 4 Mar 2016 at 10:24

PS they have no letter box.

jpod · 4 Mar 2016 at 11:27

Is the risk per tooth, or per injury?

Is the second complication directly related to the first or independent of it?

Again, is it per tooth or related to the general injury?

There is only one injury, there are three potential complications which are independent of each other.

jpod · 4 Mar 2016 at 11:27

nst68 said:
Is there an open window?

No, and their mouth is closed

jpod · 4 Mar 2016 at 11:38

Glen's letter box is wide open fellas.

jpod · 4 Mar 2016 at 12:08

Hmmm...

There are 3 possible complications, all independent, their chance of developing is 5%, 5% and 7% over the patient's lifetime

There are four teeth damaged by one injury, with those risks above.

What is the % chance the patient will suffer complications in their lifetime?

jpod · 4 Mar 2016 at 18:56

The complications are pulp chamber obliteration (the root canal space closes up), root resorption (root dissolves off) and dental pulp necrosis (dental nerve dies).

The ladybird terms are for our OC mates if you are a dentist Hikari.

So there is 1 injury to 4 teeth
Each has a 5% chance of developing complication 1 and 2 and a 7.5% chance of developing complication 3.

The risks are unrelated.

So each tooth has a 16.5% chance of developing a complication over their life span.

In total the individual has 51.4% chance of developing any of the complications in 1 or 4 teeth.

Essentially therefore 'it is more likely than not' that the individual will suffer at least one complication.

Is that correct?

(My head hurts, I only count up to 32, and thanks for posting it has helped me understand this better).

jpod · 4 Mar 2016 at 22:38

The risks are published, a % change of those risks developing over time.

You are entirely right it is a crude estimation but the combined brain power of OCUK has determined the risk is much higher than I imagined as a maths rookie.

Believe it or not the fact that a chance of complications is 'more likely than not' is tremendously useful information for my purposes and it does not have to be perfect science - just a 'sketch'

The complications are unrelated to each other. They have different aetiologies/mechanisms.

Now stop doing maths and thinking on a Friday evening and go and frag some aliens or cruise chicks and suck face

It is good to be sceptical Dowie.

jpod · 6 Mar 2016 at 12:15

Tis interesting my professional colleague.

Cheers for posting and to all who illuminated the maths problem.

Pod

jpod · 22 Mar 2016 at 23:00

Hello
Just having a read, I cannot answer you question, perhaps Dowie can.

For my purposes I just needed to know 'it was more likely than not'.

Hopefully a maths Pro will be along sometime soon (I just do subtraction

).

jpod · 23 Mar 2016 at 17:47

Hi Dowie

When there were four teeth, each with its own % chance of developing a complication how does 'to the power of four' come into it. Is there an easy way to imagine that - explain like I am 10?

Ta pod

jpod · 23 Mar 2016 at 22:54

Thanks for that, got it, tremendously helpful and it may interest you to know I used the information for my work.

Where do your ubermaths skills come from?

jpod · 29 Mar 2016 at 17:42

Another two questions from the desk of pod

Problem 1
There is a complication risk of 5%, independant complication, there are two teeth:

Chance = 1 - (0.95x.0.95)/\2 = 0.81

The chance of complication in a single tooth is 19%

Q/ What is the chance of complications in both teeth?

Is it 1/2 of 19%? So 9.5%

Problem 2

There are three teeth, two complications, all independent

Complication A has a 5% risk and requires treatment X

Complication B has a 4% risk but does not require treatment, however there is thereafter a 1% risk of treatment Y

What is the % chance of requiring treatment in 1 tooth, 2 teeth and all the teeth? I definitely cannot figure that out.

Please explain like I am 10, thanks Pod

jpod · 29 Mar 2016 at 18:06

dowie said:
Problem one is just 0.05 * 0.05

Problem two - are you saying there is a 1% chance in general or a 1% chance given the event with a 4% chance occurs?

Hi Dowie.

Don't forget the explain like I am 10 bitty.

Problem 1
How come there is no square to the power of 2? Don't understand.

So the answer 0.05 x 0.05 = 0.0025

So the chances of 1 tooth developing a problem is 0.25%? and what is the chance of both teeth developing a problem?

Problem two - are you saying there is a 1% chance in general or a 1% chance given the event with a 4% chance occurs?

A 1% chance given the event with a 4% chance.

Thanks for your maths prowess, and where does it come from? I did ask a while back, I am curious.

jpod · 29 Mar 2016 at 18:48

Understood, thank you.

Ok problem 1 (two teeth)
Chance of both teeth developing a problem is 0.05*0.05 = 0.025 which is 2.5% chance

Chance of one tooth developing a problem is 2 * 0.05 * 0.95 = 0.005 which is a 9.5% chance

That explanation is brilliant I get it, ty.

Chance of no problem is 0.95 * 0.95 = 0.9025 which is a 90% chance

Ok Problem 2, going to have a think about that one.

jpod · 29 Mar 2016 at 18:54

The probabilities are per tooth by the way. There is one patient, one individual.

jpod · 29 Mar 2016 at 19:20

Howdy, thanks for posting, have a look at the new problem, post #38

They are all independent risks, problem one has the individual patient has 2 teeth, problem 3 the patient has 3 teeth.

jpod · 29 Mar 2016 at 19:37

Lol

I have had a go and I can't figure it out. I am too confused, sorry.

Problem 1
2 teeth one patient, independent risks of 5%
So is this correct

Chance of no problems 1 - 2*(0.95 * 0.95)

Chance of one tooth out of two with problems 1 - 2*(0.05 * 0.95)

Chance of two teeth with problems 1 - 2*(0.05 * 0.05)

(Can't understand why there is no to the power of 2)

jpod · 29 Mar 2016 at 20:03

estebanrey said:
Both Teeth = 0.05 x 0.05 = 0.25%
Neither Teeth = 0.95 x 0.95 = 90.25%
Either Tooth (remaining option) = 100 - 90.25 - 0.25 = 9.5%

Understand that. Thank you, appreciated.

So problem 2
3 teeth, same patient, independant risks
Risk 1 5% chance = 0.05
Risk 2 4% chance, 0.04, of those 1% require treatment, chance = 0.0004

I need to figure out the % chance of the complications with the need for treatment:
No tooth complication
1 tooth comp
2 teeth comp
3 teeth comp

Is no tooth complication 1 - (0.95*0.9996)/\3 = 0.14 or 14% chance?

jpod · 29 Mar 2016 at 22:23

(Shh...he is very strict, don't upset him or he will make it 4 decimal placemats).

Fellas that is exactly what I needed, thank you for taking the time to post such crystal clear proofs. I am very grateful indeed. It may interest you to know that once again your uber maths prowess has benefited my professional practice and patient care.

PS
I got the point about the irrelevant consideration, 5% v the very small %, made perfect sense.

Once I have had a look if I get stuck may I post again.