Vagueness and the Aftermath - A sporadic diary

[Tal]

If we allocate each screw to a card and shuffle a 32 card deck we have 32! possible outcomes 32*31*30.... *2*1 which is 2.6313084e+35
That's 263108400000000000000000000000000000 different sequences of screws
What's the chance of any of them being in the 'right' place?


 

Well there are 31! scenarios where Screw 7 goes back into Hole 7, which is logical because the chance of Screw 7 being in the right hole is 1 in 32 and that's what we get when we divide 32! By 31!

I'm desperate to know what the right answer is, though, with maths.

[Rod Paradise]

If we allocate each screw to a card and shuffle a 32 card deck we have 32! possible outcomes 32*31*30.... *2*1 which is 2.6313084e+35
That's 263108400000000000000000000000000000 different sequences of screws
What's the chance of any of them being in the 'right' place?


 

Well there are 31! scenarios where Screw 7 goes back into Hole 7, which is logical because the chance of Screw 7 being in the right hole is 1 in 32 and that's what we get when we divide 32! By 31!

I'm desperate to know what the right answer is, though, with maths.

Is it not 32! - 31! scenarios where it's right? Or am I misunderstanding completely? 31! seems way too many for correct answers.

[JohnCharver]

Popped in thinking reddogs diary would be different but no just another vegas diary all about screwing.

[RED-DOG]

Once you've sorted the screws thing I have another question.

My dogs can tell which bowl has water in without getting up to look. They can detect the scent of a rabbit on the wind from several hundred yards away. So why do they have to put their noses within 2" of a strange dog turd, even if it stinks to high Heaven?

[Rod Paradise]

Popped in thinking reddogs diary would be different but no just another vegas diary all about screwing.

 

[Rod Paradise]

Once you've sorted the screws thing I have another question.

My dogs can tell which bowl has water in without getting up to look. They can detect the scent of a rabbit on the wind from several hundred yards away. So why do they have to put their noses within 2" of a strange dog turd, even if it stinks to high Heaven?

Doggy Facebook innit?

[Jon MW]

off the top of my head 1-(n-1/n)^n seems right, but this margin is too small to prove it  

Oh no!

I didn't read the question properly - what I've put is obviously the chance of getting any in the right place for n screws and holes.

I'd go with Tal's logic as well; there are 32 chances to get 1/32 so the most likely outcome is 1 (assuming you don't check them one at a time as you go along)

This'll teach me not to try any maths while I'm trying to pretend to work

The small sample I ran showed that not to be the case though.

it's 32 of 1 in 32 if you throw the screw back into the bowl of screws after checking it. if you don't you've either got one right or proved 2 wrong after the first try throwing the chances of the next one off the 1/32 chance significantly, and so on down the line.

If you don't know whether you've got the first one right the chances of getting the second one right is still 1 in 32 (etc for the rest). Like if you pick a card from a deck the chances of an ace are 1 in 13, but if you split the deck and pick a card from either side the chances are still 1 in 13. With the screws it's more like you effectively decide which screw goes with which hole beforehand then you fit them all in. Even if you do them one by one its 'like' you are doing that.

Having said that I still think we're missing something just can't figure out what it is.
EDIT: The problem I have with it is that if each (independent) probability is 1/32 then the most likely answer is obviously that it won't be in the right hole; and so if you have 32 of them then wouldn't the total outcome be zero because they're all so unlikely to be in the right one (?)

I'm sure we're missing the extraction of the first drawn screw from the pool - which affects the probability of every subsequent draw. You can't ignore the previous result as it significantly affects the chances of correct draws.

We're drawing every card in the deck - the chance of getting any one card is 1 - we will get it, but once we've drawn it if it's wrong it's still used & cannot be right later on and also makes one of the other cards wrong, that has to have a negative effect, hence the most likely result of 0 correct.

I'm sure we're missing the extraction of the first drawn screw from the pool - which affects the probability of every subsequent draw.
As it was originally written every screw is just randomly put into a hole - you're not checking whether any are right as you do it. If you checked if they were right and put the screw back into the bucket if it was wrong then it would be dependent probabilities based on all the previous choices.

Because it isn't being checked as you go along it is 'like' all the screws are being chosen all at the same time, from a probability point of view there is no first one they all get put in at the same time.

Whether the first one is correct does, obviously, affect the probability of the subsequent one's being correct - but because you don't know whether it's correct or not - actually .... it doesn't.

This can be illustrated with just the aces from a pack of cards.
If you have the aces and pick one at random, the chance of picking the heart is 1/4

If you split them into two piles of 2, then pick one at random from one of the piles then obviously the probability of picking the heart depends on whether the heart is in that pile or the other one. It's either 1/2 or 0.

But the probability of picking the heart in that circumstance is the probability of picking the heart in that pile of 2 rather than the other one (1/2) multiplied by the probability of then picking the heart from that pile of 2 (1/2) - hence the probability of picking the heart when you've split the aces into 2 piles is still 1/4 (this can be done in the abstract sense to prove that if you don't the result of one part of the picking it doesn't affect the probability of doing it all together - but I think that's a pretty clear illustration, apart from all the 2's which are hard to get rid of unless you do an unnecessarily complicated example)

Annoyingly I think I've remembered/worked out how to do it - but it takes a bit of thought; might get back to it at lunch.

[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

[Jon MW]

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

That's sort of what I was thinking of - but not quite.

We were looking at the wrong probability space before (my fiancee additionally said she'd just write out the whole probability space, I don't know how she got an A - I'm guessing grade inflation; it did make me think of this though)

Anyway this is the right probability space and it's true 1 is the most likely answer - but "I think" the answer is you add up all those probabilities and divide by 33 - and that will be the expected result (and will be much lower than 1 as might be imagined intuitively)

[Rod Paradise]

I still don't see how 1 can be the most likely.

Ran 1000 tests, got 741 with 0.

I realise it's a small sample, but it's showing a trend.

Ignoring the change in pool size & the amount of draws which are drawing dead has to be wrong.

[RED-DOG]

I'm struggling to keep up but it's fascinating.

[Doobs]

I still don't see how 1 can be the most likely.

Ran 1000 tests, got 741 with 0.

I realise it's a small sample, but it's showing a trend.

Ignoring the change in pool size & the amount of draws which are drawing dead has to be wrong.

Your model is wrong or my answer is wrong.  The sample is big enough. 

[Rod Paradise]

I still don't see how 1 can be the most likely.

Ran 1000 tests, got 741 with 0.

I realise it's a small sample, but it's showing a trend.

Ignoring the change in pool size & the amount of draws which are drawing dead has to be wrong.

Your model is wrong or my answer is wrong.  The sample is big enough.  

I've double checked my workings, explanation of method below.

Model in excel - positions 1 to 32 allocated a 'random' no using =RAND()*. Rand No is between 0 & 1 to 10 decimal places. These are then ranked giving integers from 1 to 32. Where the integer matches the position we score one correct.

I then copied that array 50 times. Used SUMIF() to total 0,1,2 etc right.

Copy and paste value results into a table, every time you paste value the arrays recalculate, repeat 10 times for 500 tests.

Now done 1500 with the following results:

0 - 1111
1 - 267
2 - 94
3 - 21
4 - 5
5 - 1
6 - 1
7 - 0

* =RAND() is not a true random, but close enough for this kind of test).

PS I'm happy if you find a mistake - I want to know how this works.

[Rod Paradise]

I'm thinking it's the 'drawing dead' that makes a difference.

The examples of drawing a card above, while correct for repeated attempts at AH, ignore the fact that once you draw for AH, you can't draw for it again, you had a 1 in 32 chance if it was the first draw, and now you  can't get it right. Draw 2 is for KH.

How that is worked in mathematically I don't know but I am sure that if you'd to pick not only your lottery number, but the position in the draw it would come out your odds of winning go down.

[Jon MW]

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