Need pro statistician help to answer question

[dowie]

If you're looking for the probability of at least one tooth developing at least one complication it's 70% I think (0.05 + 0.05 + 0.075)x4 = 0.7

That is what I worked it out to be? At least a 70% chance that at least one of his teeth will develop one complication.

that is obviously wrong

for example what if he had 6 dodgy teeth? According to that calculation you'd then have a probability greater than 1

see post #13

chance of at least one complication occurring = 1 - chance of none occurring

=1 - (.95 * .95 * .925)

approx 16.5%

if the probabilities are actually per tooth then for 4 teeth it is:

= 1 - (.95 * .95 * .925)^4

approx 51.4%

though this is all rather moot as the OP is probably making some dubious assumptions here

 

[balky12]

that is obviously wrong

for example what if he had 6 dodgy teeth? According to that calculation you'd then have a probability greater than 1

see post #13

chance of at least one complication occurring = 1 - chance of none occurring

=1 - (.95 * .95 * .925)

approx 16.5%

if the probabilities are actually per tooth then for 4 teeth it is:

= 1 - (.95 * .95 * .925)^4

approx 51.4%

though this is all rather moot as the OP is probably making some dubious assumptions here

Bingo... no wonder I keep losing poker

 

[D3lirious]

Note to self: stay out of maths threads

 

[Destination]

that is obviously wrong

for example what if he had 6 dodgy teeth? According to that calculation you'd then have a probability greater than 1

see post #13

chance of at least one complication occurring = 1 - chance of none occurring

=1 - (.95 * .95 * .925)

approx 16.5%

if the probabilities are actually per tooth then for 4 teeth it is:

= 1 - (.95 * .95 * .925)^4

approx 51.4%

though this is all rather moot as the OP is probably making some dubious assumptions here

This is accurate and if it is the situations I am thinking about, then a potential complication developing in any of the teeth would be possible over the course of a lifetime.
So In this case the 51.4% answer would apply, as in with four teeth injured, and 3 potential complications occurring to each over a lifetime, there is basically a 50% chance that nothing will happen, and other amounts for 1 to 1 tooth, 1 to 4 teeth, 3 to 1 tooth, 3 to all teeth in sliding steps, but the 51.4% answer is the one jpod is looking for.

Out of interest, what trauma complications are you warning regarding at a 5/5/7.5 rates?
Professional interest.

 

[joeyjojo]

1 - chance of none occuring

= 1 - (.95 * .95 * .925)

= approximately 16.5 %

Correct. The OP says

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

which is the same as "1 - the chance of no complications".

If you aren't sure why OP, consider all the possibilities:

0 complications occur
1 complication occurs
2 complications occur
all 3 complications occur

It's got to be one of these, so the chance of developing "1 or more" complications (which is what you asked) is 100% minus the chance of 0 complications.

Unless the chances are per tooth!

 

[jpod]

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).

 

[dowie]

The risks are unrelated.

To each other or between each of the teeth? I'd question the latter.

if this is a real world problem then you're making some dubious assumptions to begin with and the calculated probability is a bit dubious as a result

I'd doubt very much that these events are completely independent - I mean suppose all 4 teeth are next to each other and were injured at the same time/from the same event, given that one of them develops a complication are you certain that (given we now know of that prior event) that prior event doesn't now change the chance of the adjacent teeth (with the exact same injury, from the same time) developing that particular complication too. As opposed to say an injured tooth in a different part of the mouth, from a different event/injury that occurred at a different time.

There is also probably a time component you've not posted or are unaware of - I'm not buying a flat percentage chance over a 'lifetime'... i.e. a kid with a whole lifetime left to live sustaining an injury to their adult teeth vs say a 65 yr old with not long left both having the same chance of developing these complications. Or taken to the extremes a cancer patient with less than a year to live gets the same injury - doubt he/she has the same chance of developing any of these three complications in their 'lifetime' as some kid with a whole lifetime left in which to develop any of them.

So you've got this rather crude catch all 'lifetime' percentage and you're assuming things it isn't clear that it is reasonable to assume then we're performing a calculation to come up with this just over 50% figure... given the dodgy assumptions and the errors that would potentially be compounded here - no you can't really state 'it is more likely than not'. The real world doesn't work like that.

 

[jpod]

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.

 

[Destination]

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).

Aye, i'd say in theory there is a slightly greater chance that there will some lifelong complication, than not for this individual, given the fact four teeth are involved.
As you well know, however, the stock % are from population numbers, and involve all degrees of trauma, so depending on scale of injury there would be some mitigation of chance.

I had never considered the lifelong risk to actually work out as high as 50/50 for a 4 tooth trauma. Quite interesting really, but then you do see plenty of folks with that single yellow front tooth, no hint of a pulp chamber, and absolutely no recollection of injury, until quizzed in detail and they mention they fell of their bike when they were ten, thirty years ago.

 

[jpod]

Tis interesting my professional colleague.

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

Pod

 

[deusexlumina]

that is obviously wrong

for example what if he had 6 dodgy teeth? According to that calculation you'd then have a probability greater than 1

see post #13

chance of at least one complication occurring = 1 - chance of none occurring

=1 - (.95 * .95 * .925)

approx 16.5%

if the probabilities are actually per tooth then for 4 teeth it is:

= 1 - (.95 * .95 * .925)^4

approx 51.4%

though this is all rather moot as the OP is probably making some dubious assumptions here

This has puzzled me and made me feel a bit stupid, so would you mind explaining something to me? (sincerely!)

Why are you multiplying? surely that's the probability of each of the complications occurring together.

Indeed, if you have 6 teeth and calculate it by addition you'll get an answer greater than 1, but that's ok because its no longer a probability - by multiplying by 4 (or by 6 in your case) you've just calculated the average fraction of all teeth that will be affected, unless I'm wrong?

If each probability is independent, and each tooth is considered a "trail" with a yes/no outcome (with yes being there's a problem -17.5%) then isn't this best represented by the binomial distribution?

In which case, I find:

P(0 teeth infected) = 46.3%
P(1 tooth infected) = 39.3%
P(2 teeth infected) = 12.5%
P(3 teeth infected) = 1.7%
P(4 teeth infected) = 0.09%

These add up to 100%, so seems correct in my head. Need someone to talk some sense into me though if i got it all wrong..

edit: so going back to the OP, the chance of developing any complication is (100-46.3)%, giving 53.7%.

 

[jpod]

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 ).

 

[dowie]

Why are you multiplying? surely that's the probability of each of the complications occurring together.

nope 0.95 * 0.95 * 0.925 is the probability of them all not occurring

ergo the chance of one or more occurring is 1 - (0.95 * 0.95 * 0.925)

it is just the quicker way of calculating it...

you could work it out in a long winded fashion if you like;

lets say A = 5%, B = 5%, C = 7.5%

A notB notC = 0.05 * 0.95 * 0.925
notA B notC = 0.95 * 0.05 * 0.925
notA notB C =...
A B notC =
notA B C =
A notB C =
A B C =

add them all together and you'll get the same answer as simply 1 - (0.95 * 0.95 * 0.925)

If each probability is independent, and each tooth is considered a "trail" with a yes/no outcome (with yes being there's a problem -17.5%)then isn't this best represented by the binomial distribution?

In which case, I find:

P(0 teeth infected) = 46.3%
P(1 tooth infected) = 39.3%
P(2 teeth infected) = 12.5%
P(3 teeth infected) = 1.7%
P(4 teeth infected) = 0.09%

These add up to 100%, so seems correct in my head. Need someone to talk some sense into me though if i got it all wrong..

edit: so going back to the OP, the chance of developing any complication is (100-46.3)%, giving 53.7%.

nope, they are three independent events you can't just add the individual probabilities together like that... see above. That is where you're going wrong, you've stated the probability of there being at least one complication is 17.5% when it is approximately 16.5% as per the above.

(edited for clarity)

 

[jpod]

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

 

[dowie]

it was just a shorter/neater way of writing 1 - (0.95 * 0.95 * 0.925 * 0.95 * 0.95 * 0.925 * 0.95 * 0.95 * 0.925 * 0.95 * 0.95 * 0.925) i.e. all three independent events for each of the 4 teeth not happening

 

[jpod]

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?

 

[Trucamo]

Some nice use of the complement rule, much easier than going the other way. One thing I always struggled with statistics was deciding when things would work out easier with the complement rule.

I used to get myself into a right mess trying to work out P(A or B) rather than just 1 - (P(not A) and P(not B)). The latter looks more complicated but actually works out easier for multiple independent events such as this.

 

[jpod]

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

 

[dowie]

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?

 

[jpod]

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.