Vagueness and the Aftermath - A sporadic diary

[Mohican]

[Rod Paradise]

Proability of zero right= (31/32)^32 = 0.362
Probability of one right = (31/32)^31 * (1/32) * 32 = 0.374
Probability of two right= (31/32)^30 * (1/32)^2 * 32 * 31/2 = 0.187
Probability of three right = (31/32)^29 * (1/32)^3 * 32 * 31 * 30/(2*3) = 0.060
Probability of four right = (31/32)^28 * (1/32)^4 * 32 * 31 * 30 * 29 / (2*3*4) = 0.014

etc

1 is the most likely

QED

Save yourself 9 grand Longines

OK now I'm thinking about it without distractions - where does the red bit come from?
"Probability of one right = (31/32)^31 * (1/32) * 32 = 0.374"


Isn't it
Probability of 0 right = (31/32)^32 * (1/32)^0 = 0.362
Probability of 1 right = (31/32)^31 * (1/32)^1 = 0.011
Probability of 2 right = (31/32)^30 * (1/32)^2 = 0.0003
Probability of 3 right = (31/32)^29 * (1/32)^3 = v small
Probability of 4 right = (31/32)^28 * (1/32)^4 = even smaller
.
.
.
.
Probability of 32 right = (31/32)^0 * (1/32)^32 = tiny

So the single most likely outcome is 0.

But the expected number of screws in the right hole would be the average value of all of these -  (with the help of excel) this turns out to be 0.011337

So that's the answer - 0.011337 screws end up in the same hole they started as

Starting to head the right way, and I'm not just picking holes - but would the total of the probabilities not equal 1?

[RED-DOG]

I think David Attenborough's jib is safe.

[RED-DOG]

Have you ever had a car with a fuel gauge that shows full for ages and then drops to less than half a tank all of a sudden? Well that's what my energy levels are like lately. I'm full of beans when I get out of bed and I run about flinging myself at everything until early afternoon and then phut, I'm in limp mode.

[Doobs]

Proability of zero right= (31/32)^32 = 0.362
Probability of one right = (31/32)^31 * (1/32) * 32 = 0.374
Probability of two right= (31/32)^30 * (1/32)^2 * 32 * 31/2 = 0.187
Probability of three right = (31/32)^29 * (1/32)^3 * 32 * 31 * 30/(2*3) = 0.060
Probability of four right = (31/32)^28 * (1/32)^4 * 32 * 31 * 30 * 29 / (2*3*4) = 0.014

etc

1 is the most likely

QED

Save yourself 9 grand Longines

OK now I'm thinking about it without distractions - where does the red bit come from?
"Probability of one right = (31/32)^31 * (1/32) * 32 = 0.374"


Isn't it
Probability of 0 right = (31/32)^32 * (1/32)^0 = 0.362
Probability of 1 right = (31/32)^31 * (1/32)^1 = 0.011
Probability of 2 right = (31/32)^30 * (1/32)^2 = 0.0003
Probability of 3 right = (31/32)^29 * (1/32)^3 = v small
Probability of 4 right = (31/32)^28 * (1/32)^4 = even smaller
.
.
.
.
Probability of 32 right = (31/32)^0 * (1/32)^32 = tiny

So the single most likely outcome is 0.

But the expected number of screws in the right hole would be the average value of all of these -  (with the help of excel) this turns out to be 0.011337

So that's the answer - 0.011337 screws end up in the same hole they started as

Not sure about the merit of removing something you are not sure of!

In the probability 1 is right, then the formuma before the removed 32 is the probability that the first hole is right and the other 31 are all wrong.  As there are 32 different holes you need to multiply by 32 to get the full answer. 

Likewise with 2, the formula before 32 x 31/2 is the probability that the first two holes are right but all the others are wrong.  We need to multiply by 32 x 31/2 as this is the total number of different combinations of 2 holes (or in Tal speak 32!/(30! X 2!))

The total probabilities should add to 1.  The mean is 1.  You should be able to get this by probability (answer =0) x 0 + probability (answer = 1) x 1 etc. up until 32.

[david3103]

Proability of zero right= (31/32)^32 = 0.362
Probability of one right = (31/32)^31 * (1/32) * 32 = 0.374
Probability of two right= (31/32)^30 * (1/32)^2 * 32 * 31/2 = 0.187
Probability of three right = (31/32)^29 * (1/32)^3 * 32 * 31 * 30/(2*3) = 0.060
Probability of four right = (31/32)^28 * (1/32)^4 * 32 * 31 * 30 * 29 / (2*3*4) = 0.014

etc

1 is the most likely

QED

Save yourself 9 grand Longines

OK now I'm thinking about it without distractions - where does the red bit come from?
"Probability of one right = (31/32)^31 * (1/32) * 32 = 0.374"


Isn't it
Probability of 0 right = (31/32)^32 * (1/32)^0 = 0.362
Probability of 1 right = (31/32)^31 * (1/32)^1 = 0.011
Probability of 2 right = (31/32)^30 * (1/32)^2 = 0.0003
Probability of 3 right = (31/32)^29 * (1/32)^3 = v small
Probability of 4 right = (31/32)^28 * (1/32)^4 = even smaller
.
.
.
.
Probability of 32 right = (31/32)^0 * (1/32)^32 = tiny

So the single most likely outcome is 0.

But the expected number of screws in the right hole would be the average value of all of these -  (with the help of excel) this turns out to be 0.011337

So that's the answer - 0.011337 screws end up in the same hole they started as

Not sure about the merit of removing something you are not sure of!

In the probability 1 is right, then the formuma before the removed 32 is the probability that the first hole is right and the other 31 are all wrong.  As there are 32 different holes you need to multiply by 32 to get the full answer. 

Likewise with 2, the formula before 32 x 31/2 is the probability that the first two holes are right but all the others are wrong.  We need to multiply by 32 x 31/2 as this is the total number of different combinations of 2 holes (or in Tal speak 32!/(30! X 2!))

The total probabilities should add to 1.  The mean is 1.  You should be able to get this by probability (answer =0) x 0 + probability (answer = 1) x 1 etc. up until 32.

Doobs answers looks right on an intuitive basis, the sum of the probabilities has the feel of a series that will total 1 which is reassuring in itself.

[EvilPie]

Doobs answers looks right on an intuitive basis, the sum of the probabilities has the feel of a series that will total 1 which is reassuring in itself.

Got to agree with this.

JonMW's answer surely can't be right. If zero = 0.362 then every non-zero must add up to the rest and his series is never going to get there.

[KarmaDope]

My eldest has just finished his first year of a Maths degree - going to ask him this one and see if I've got my nine grands worth.....

To be honest I think it's more of an A Level problem - it's just that everyone on here who once had that level of competence has long since forgotten it

My fiancee got her A level in maths much more recently - doesn't help though, her answer is, "What's wrong with you? Why does anyone care?"

Very thin Cowboy, very thin.

 

[RED-DOG]

A brief demonstration of my re vamped tool box in action.

Ratchet shaped hole alerts me that my 1/2" drive ratchet is missing. (I didn't lose it, I broke it.

Click to see full-size image.

I buy a new 1/2" ratchet from the bloke with the tool stall on the market.

It's 1/8" too long for the space in my box so I do a quick adjustment and were good to go.

Click to see full-size image.

Note to self, toolbox needs hoovering out.

[RED-DOG]

So today, I have mostly been a plummer.

Click to see full-size image.

This is my plumbers headlight. (also useful for casual surgery).

Click to see full-size image.

And this is my glamorous assistant.

Click to see full-size image.

There are more pictures but Mrs Red says they are too graphic to post.

Suffice to say, everything came out OK.

[Jon MW]

Doobs answers looks right on an intuitive basis, the sum of the probabilities has the feel of a series that will total 1 which is reassuring in itself.

Got to agree with this.

JonMW's answer surely can't be right. If zero = 0.362 then every non-zero must add up to the rest and his series is never going to get there.

It doesn't have to total 1; Doobs answer was looking at the most probable outcome (I get it now - the other stuff was all the permutation; I never liked permutations - or probability for that matter); but the question was:

...How many screws are likely to have ended up in the original hole?

So we're looking for a single value - which will be the average value of all the probabilities.

It did trouble me though that we are doing 32 cases of a 1/32 shot - plus 1 being the most likely outcome when permutations were involved.

This made me think I was looking at the wrong average - with all those indices the average of the 32 terms is probably not the mean, but the 32nd root.

If I recalculate the series I made before it adds up to 0.374121....; the 32nd root of that is 0.96974....

So I'll revise my previous answer to say the number of screws in the right place is 0.96974....
(the reason it's not 1 is just like with the roulette wheel - there aren't 32 outcomes, because you might get 0 in the same place).

If that's not it then the answer is going to be somewhere in here https://en.wikipedia.org/wiki/Expected_value

But I can't really look much more at it as we have a honeymoon to continue planning for and apparently that is much more actually important

[Karabiner]

Far too much screwing around on here lately so how about a protest song:

https://www.youtube.com/watch?v=qfZVu0alU0I

[Doobs]

Doobs answers looks right on an intuitive basis, the sum of the probabilities has the feel of a series that will total 1 which is reassuring in itself.

Got to agree with this.

JonMW's answer surely can't be right. If zero = 0.362 then every non-zero must add up to the rest and his series is never going to get there.

It doesn't have to total 1; Doobs answer was looking at the most probable outcome (I get it now - the other stuff was all the permutation; I never liked permutations - or probability for that matter); but the question was:

...How many screws are likely to have ended up in the original hole?

So we're looking for a single value - which will be the average value of all the probabilities.

It did trouble me though that we are doing 32 cases of a 1/32 shot - plus 1 being the most likely outcome when permutations were involved.

This made me think I was looking at the wrong average - with all those indices the average of the 32 terms is probably not the mean, but the 32nd root.

If I recalculate the series I made before it adds up to 0.374121....; the 32nd root of that is 0.96974....

So I'll revise my previous answer to say the number of screws in the right place is 0.96974....
(the reason it's not 1 is just like with the roulette wheel - there aren't 32 outcomes, because you might get 0 in the same place).

If that's not it then the answer is going to be somewhere in here https://en.wikipedia.org/wiki/Expected_value

But I can't really look much more at it as we have a honeymoon to continue planning for and apparently that is much more actually important

There is absolutely no reason why it should be the 32rd root. 

Just look at 1 hole.  We can see that the mean of one hole is 1/32:
Probability we get this hole correct on its own is 1/32.
The mean is then 1x 1/32 + 0 x 31/32 = 1/32 (probability of 0 correct = 1-1/32).  You can't get two screws in one hole, so there are no other possible outcomes. 

Therefore mean number of correct screws in hole 1 is 1/32. 
Going on to the next hole, the maths is exactly the same.
Mean number of correct screws in hole 2 = 1/32.   

We can carry this on for all 32 holes and each one has a mean of 1/32.

Before you get carried away thinking but this changes because we know the answer to hole 1.  The probability will still be the same, as once you do this (Probability 2 is correct given hole 1 is correct and Probability 2 is correct given hole 1 is wrong) the answer should be the same. 

Another way to think about this is our choice of what hole we call hole 1 is arbritary, so they must all have the same mean as we could happily have picked any hole first.  We can't change the mean just by choosing to label that particular hole as hole 5 instead of hole 12.       

Sum them all up and you get 32 x 1/32 = 1.

 

[Rod Paradise]

Doobs answers looks right on an intuitive basis, the sum of the probabilities has the feel of a series that will total 1 which is reassuring in itself.

Got to agree with this.

JonMW's answer surely can't be right. If zero = 0.362 then every non-zero must add up to the rest and his series is never going to get there.

It doesn't have to total 1; Doobs answer was looking at the most probable outcome (I get it now - the other stuff was all the permutation; I never liked permutations - or probability for that matter); but the question was:

...How many screws are likely to have ended up in the original hole?

So we're looking for a single value - which will be the average value of all the probabilities.

It did trouble me though that we are doing 32 cases of a 1/32 shot - plus 1 being the most likely outcome when permutations were involved.

This made me think I was looking at the wrong average - with all those indices the average of the 32 terms is probably not the mean, but the 32nd root.

If I recalculate the series I made before it adds up to 0.374121....; the 32nd root of that is 0.96974....

So I'll revise my previous answer to say the number of screws in the right place is 0.96974....
(the reason it's not 1 is just like with the roulette wheel - there aren't 32 outcomes, because you might get 0 in the same place).

If that's not it then the answer is going to be somewhere in here https://en.wikipedia.org/wiki/Expected_value

But I can't really look much more at it as we have a honeymoon to continue planning for and apparently that is much more actually important

There is absolutely no reason why it should be the 32rd root. 

Just look at 1 hole.  We can see that the mean of one hole is 1/32:
Probability we get this hole correct on its own is 1/32.
The mean is then 1x 1/32 + 0 x 31/32 = 1/32 (probability of 0 correct = 1-1/32).  You can't get two screws in one hole, so there are no other possible outcomes. 

Therefore mean number of correct screws in hole 1 is 1/32. 
Going on to the next hole, the maths is exactly the same.
Mean number of correct screws in hole 2 = 1/32.   

We can carry this on for all 32 holes and each one has a mean of 1/32.

Before you get carried away thinking but this changes because we know the answer to hole 1.  The probability will still be the same, as once you do this (Probability 2 is correct given hole 1 is correct and Probability 2 is correct given hole 1 is wrong) the answer should be the same. 

Another way to think about this is our choice of what hole we call hole 1 is arbritary, so they must all have the same mean as we could happily have picked any hole first.  We can't change the mean just by choosing to label that particular hole as hole 5 instead of hole 12.       

Sum them all up and you get 32 x 1/32 = 1.

 

If the screw in hole 1 is screw 2? Probability 2 is 0. If hole 1 is wrong, one of the other holes is guaranteed drawing dead - so how can you sum them all up?

[Doobs]

Doobs answers looks right on an intuitive basis, the sum of the probabilities has the feel of a series that will total 1 which is reassuring in itself.

Got to agree with this.

JonMW's answer surely can't be right. If zero = 0.362 then every non-zero must add up to the rest and his series is never going to get there.

It doesn't have to total 1; Doobs answer was looking at the most probable outcome (I get it now - the other stuff was all the permutation; I never liked permutations - or probability for that matter); but the question was:

...How many screws are likely to have ended up in the original hole?

So we're looking for a single value - which will be the average value of all the probabilities.

It did trouble me though that we are doing 32 cases of a 1/32 shot - plus 1 being the most likely outcome when permutations were involved.

This made me think I was looking at the wrong average - with all those indices the average of the 32 terms is probably not the mean, but the 32nd root.

If I recalculate the series I made before it adds up to 0.374121....; the 32nd root of that is 0.96974....

So I'll revise my previous answer to say the number of screws in the right place is 0.96974....
(the reason it's not 1 is just like with the roulette wheel - there aren't 32 outcomes, because you might get 0 in the same place).

If that's not it then the answer is going to be somewhere in here https://en.wikipedia.org/wiki/Expected_value

But I can't really look much more at it as we have a honeymoon to continue planning for and apparently that is much more actually important

There is absolutely no reason why it should be the 32rd root. 

Just look at 1 hole.  We can see that the mean of one hole is 1/32:
Probability we get this hole correct on its own is 1/32.
The mean is then 1x 1/32 + 0 x 31/32 = 1/32 (probability of 0 correct = 1-1/32).  You can't get two screws in one hole, so there are no other possible outcomes. 

Therefore mean number of correct screws in hole 1 is 1/32. 
Going on to the next hole, the maths is exactly the same.
Mean number of correct screws in hole 2 = 1/32.   

We can carry this on for all 32 holes and each one has a mean of 1/32.

Before you get carried away thinking but this changes because we know the answer to hole 1.  The probability will still be the same, as once you do this (Probability 2 is correct given hole 1 is correct and Probability 2 is correct given hole 1 is wrong) the answer should be the same. 

Another way to think about this is our choice of what hole we call hole 1 is arbritary, so they must all have the same mean as we could happily have picked any hole first.  We can't change the mean just by choosing to label that particular hole as hole 5 instead of hole 12.       

Sum them all up and you get 32 x 1/32 = 1.

 

If the screw in hole 1 is screw 2? Probability 2 is 0. If hole 1 is wrong, one of the other holes is guaranteed drawing dead - so how can you sum them all up?

Because it is a mean.  You aren't working out what happens each time, just what happens on average.