Now, regular users will recognise that I know more about designer shopping and general good taste than maths, but the good people who employ me unforunately don't want this area of my expertise; if someone can help me with the following maths problem you'll save me the embarrassment of having to remove my shoes and socks to count my toes in the office...

I'm trying to work out the probability of two events coinciding. I don't work with fish, but will use this as an example otherwise I'm going to confuse you as much as me.

If there is a pool of water with 950,000 fish swimming in it, and I catch 1, the probability of any one fish being caught is 0.0001%. If I catch a hundred fish, this percentage increases to 0.01%

If the fish are then put back and someone else comes along and catches 100 fish, what is the probability that one of the fish will have been caught twice?

It really is doing my head in. I've asked loads of folk here for help, but keep getting different replies. Please help before I show myself to be the proper numptie that I am! :D

Ta,

K x

Kristian, you really need to get out more.

Where on earth do you work????????????

I'm not asking for fun, this is serious! Can anyone help, or shall I look up Stephen Hawkings on BT.com? :)

The chance of one fish being caught AND the same fish being caught again is the multiple of the probabilies of the fish being caught each time.

Therefore: 1 in 950000 multiplied by 1 in 950000

(1/950000) X (1/950000) = 1 in 902,500,000,000

Pretty slim.

Originally posted by valentine
Where on earth do you work????????????
Well, I don't think Kristian wanted everyone to know this but...

...he works at a very big salmon farm, let me explain...

...earlier today he caught one of the fish (a salmon obviously), anyway, it bit one of the very nice "mother of pearl" buttons of his favourite Hawaiian shirt, and then escaped back into the (very large) pond...

...we've been trying to catch the offending fish ever since, but it seems to be taking forever, and some of the guys are starting to ask "what are the chances of us actually catching the same fish?"

Right, does that answer your question, nosey.

:mad:









;) :D ;)

If you're serious, I'll ask my neighbour for you he's a statistician/mathematician by training who loves this sort of stuff and I'm sure he'll come up with something, hopefully the right answer, a suitable formula and how to do the calculation as well. :)

is carol vorderman in the phonebook

Originally posted by tim_rutter
The chance of one fish being caught AND the same fish being caught again is the multiple of the probabilies of the fish being caught each time.

Therefore: 1 in 950000 multiplied by 1 in 950000

(1/950000) X (1/950000) = 1 in 902,500,000,000

Pretty slim.


I have now read the question again and realise that I am being thick.

So the second guy comes along and the 100 fish have been caught already and put back. Therefore the odds of catching a caught fish is 100/950000 = 1/9500. However, he has 100 goes and therefore you can multiply these odds by 100 i.e. 100/9500 which is:

1/95!!!

If this makes it any clearer, it's important that any one fish is not caught twice; this is the probability I need.

Where's JoeP when you need him? ;)

And for the record, this really is not a wind-up thread.

Cheers,

K x

he said you've got your calculatins wrong from the start and will ahve a look at it later.

Doesn't the probability depend on how stupid the fish are and whether or not they swim out of the way before the 2nd person comes along

Sorry I'm not taking this serious am I

Ok I'm gonna make the numbers easier.

You have 100 fish in a pond. You catch ten

Dave comes along and catches a fish. The odds of him NOT catching a caught fish is 90/100 i.e 9in10.

He then catches another fish, the odds of him NOT catching a caught fish is then 89/99 - one fish removed from pond already.

This continues till he has caught 10 fish. THe odds at the end of him having NO CAUGHT fish is

90/100 x 89/99 x 88/98 x 87/97..............x 81/91

And this works out as: 33%


i.e there is a 33% chance of no fish being caught twice

Does this help....this is the solution to calc prob of throwing dice...I am sure the same formula will still apply!!

The probability of one dice being a particular number is 1/6. Generally speaking, if you choose a condition and another condition (such as the first dice being 2 and the second dice being 4), then you multiply the two probabilities together - 1/6 x 1/6 = 1/36. But life is harder than that! People rarely want the first dice being one number and the second another. They just want the two numbers on the two dice - they don't care which is which. So the condition becomes the first dice has the first number and the second dice has the second number, or the first dice has the second number and the second dice has the first number. The condition or means that you add the probabilities, so the answer is 1/6 x 1/6 + 1/6 x 1/6 = 2/36

The or conditions can produce even more problems, since you get these throws which can obey more than one condition, so they turn up in the calculation more than once, and the surplus ones have to removed. For example, if you want a 2 or a 5, then there is a 1/6 probability of the first dice being 2, and a 1/6 probability of the second dice being 5, but the throw 2:5 happens in both groups, so you remove to remove it. Add to this that you have to include the first dice being 5 and the second being 2, and life starts getting complicated! So there are two ways of calculating these or conditions.

The first is to work out the different probabilities, add them together, and remove the duplicates. I describe this method for calculating the probability of one number turning up in two dice, since there is only one duplicate. This will be 1/6 + 1/6 - 1/36, or 11/36. You could also use this method for either of two numbers on two dice. Here there are four duplicates, so the calculation becomes 1/6 + 1/6 + 1/6 + 1/6 - 4/36 = 20/36. You can see what's going on using the coloured squares on the chart, but it is a little hairy, so for this problem, on the webpage I have described the next method.

These and and or conditions are Boolean logic, and it is true that the opposite of or is not and (and vice versa). 'Opposite' means that any throw is either one or the other. The probabilities of opposites add up to one. Now the opposite of two dice showing 2 or 5 is two dice not showing a 2 and not showing a 5. This is an and condition, and is much easier to calculate. The chance of not having a 2 and not having a 5 on one dice is 4/6, so for two dice it's 4/6 x 4/6 = 16/36. So the probability of the opposite, two dice showing either 2 or 5 is 1 - 16/36 = 20/36.

It is possible to use this second method to calculate the probabilty of one number on two dice. The chance of not having a number on one dice is 5/6, so for two dice it will be 5/6 x 5/6 = 25/36. So the probability of having a particular number on two dice will be 1 - 25/36 = 11/36. It is satisfying that you get the same answer whichever method you use!



:thumbsup:

I'm the guy who scraped an O Level in maths and only did degree level maths after being threatened with being thrown out of Uni if I didn't pass the maths exam. :D

I think that Tim's right - it looks sensible to me but I'm not a statistician. Now, had it been a question about chaos or complexity theory then I would have proabbly been much happier!

Joe

I can't help with the maths but are we to substitute the word 'fish' for 'man / men' ?

My friend and I once 'caught' ( snogged) the same guy in a week of each other and the probability of us ever speaking to him again is not that good... but would need Tim-Rutter to do the maths

When the second guy comes along, there are 100 fish in the pond that have been caught before, out of a total of 950000 fish. So the chance of catching one of those fish is 100 in 950000 or 1/9500.

If he has 100 goes at it...thats 100*1 in 9500 or 1 in 95 as Tim says.

Well..thats how it looks to me...

he's just said that there is a catch to this type of problem and he needs to think about it. He also said you cant be sure you havent caught the same fish twice unless you mark them to show that they've been caught once before. With 950,00 fish that's quite a lot of work. Cant you x-ray them?

By the way he used to do all the stats for my published reports and spends his time doing mathematical formulae for fun !!! so if that's what he says I'd trust his judgment on this.

I have revised my answer!

With the new method it works out as about 98.9% as well with 950000 fish but thats coz there's a lot more fish involved


This makes sense of course. The more fish you catch or the less fish there are in the pond in the first place the lower the odds of not catching a caught fish....

Hmmm I think you'll all be just as confused now.

Maybe I'll draw a diagram

Look, Kristian, give it up...

...the guys are getting wet and fed-up, and it's not doing the fish any good either...

...just admit to yourself that we're never going to find the wretched button, and go buy a new Hawaiian shirt.










;) :D

PS: Sorry, I'm rubbish at this sort of maths and can't help.

This is what my neighbour dictated to me:

Unless each fish is different you will not know if it has been caught twice unless it has been marked, ie you have caught it once and marked it. Therefore the probability cant be calculated but it would possibly be 1/950,000.

If the fish are marked as they are caught and then returned the odds of picking a fish that has previously been caught are so large that it is not worth bothering to do the calculation.

Originally posted by wendygs
This is what my neighbour dictated to me:

Unless each fish is different you will not know if it has been caught twice unless it has been marked, ie you have caught it once and marked it. Therefore the probability cant be calculated but it would possibly be 1/950,000.

If the fish are marked as they are caught and then returned the odds of picking a fish that has previously been caught are so large that it is not worth bothering to do the calculation.

yeah but in this case they're catching 100 fish.

Originally posted by Kristian
I'm trying to work out the probability of two events coinciding. I don't work with fish, but will use this as an example otherwise I'm going to confuse you as much as me.

If there is a pool of water with 950,000 fish swimming in it, and I catch 1, the probability of any one fish being caught is 0.0001%. If I catch a hundred fish, this percentage increases to 0.01%

If the fish are then put back and someone else comes along and catches 100 fish, what is the probability that one of the fish will have been caught twice?

K x


Tim. I'm only a qualitative researcher. I just printed out the above, gave it to my neighbour and asked him for calculations based on the above info which I printed out in its entirety.

Based on what you wrote, I'm now wondering about Kristian's clarity of this problem. Is it a case of catching 100 fish and are they or are they not then marked from a stock of 950,000 and then returned to the same stock or is it a case of putting the marked stock of 100 fish in a completely different place. Either way its doing my head in just thinking about it.

I have assumed that 100 fish are caught, marked if you like with a tag of some kind, then thrown back.

Another man comes back and catches another 100 fish. THe question was what is the odds of no fish being caught twice?

I think.

Thats what I worked out anyway! :)

Originally posted by Kristian
If there is a pool of water with 950,000 fish swimming in it, and I catch 1, the probability of any one fish being caught is 0.0001%. If I catch a hundred fish, this percentage increases to 0.01%

If the fish are then put back and someone else comes along and catches 100 fish, what is the probability that one of the fish will have been caught twice?

It really is doing my head in. I've asked loads of folk here for help, but keep getting different replies.

Firstly forget using percentages - just use a simple probability figure 1.0 for "almost certain" and 0.0 for "probably impossible".

The scenario as I understand it.

you go and catch 100 out of the 950,000 fish in the lake and put them back.

Then someone else comes along, and catches 100 fish.
We need to know whether, when you catch your fish, you are putting them back after you catch them or not. If you are just catching 100 fish and then putting them all back in one go, you might as well leave yourself out of the problem, as all you are doing is randomly selecting a set of fish.

We also need to know whether the second fishing nut is putting them back after they catch each one or not. Not putting them back will increase the probability of hooking one of the selected fish with each cast of the line.

Now, do you want to know the probability of this second person catching one and only one of the fish that you caught, or at least one of the fish that you caught after they have caught 100 fish?

The problem is the problem...

Just double checked with my neighbour. He's happy with his calculations and took the 100 fish into account when working out the probabilities so doesnt mind if you want to challenge it.

The only useful contribution I've got to make on this one is I've always found his stats are sound and as I cant add up or get the same number twice there's nothing more I can usefully add.

Assuming no fish can be caught twice by the same person:


the probability of catching 100 fish at least one of which was one of the 100 fish that Kristian caught is :

100/950,000 or 1.05 e-4

the probability of catching exactly one of the 100 fish that kristian hooked, and no more plus an additional 99 uncaught fish is.......erm...anyone?

Kristian - maybe you should come clean and tell the forum about your fish fetish. I thought you'd agreed with your counsellor that you'd restrict your piscine ponderings to your 10-minute "fish-time" each morning? Keep snapping that rubber band! ;)

Originally posted by Phanerothyme
Assuming no fish can be caught twice by the same person:


the probability of catching at least one of the 100 fish that Kristian caught is 100/950,000 or 1.05 e-4

the probability of catching exactly one of the 100 fish that kristian hooked, and no more plus an additional 99 uncaught fish is.......erm...anyone?

Phan I dont think that's the problem and adding unnecessary layers of complication to a simple solution; it's a waste of time trying to calculate or rescue the pearl button from the fish.:D

Don't forget, the fish that have been caught already will now be very cautious in regards to hooks with bait floating around in the water.

That should decrease probability a bit ...

And dont forget that bad news travels fast. Fish are reputed to be hyper-sensitive to these sort of disturbances which might increase the bookies' stakes on winning out big style on this one :D

Originally posted by Phanerothyme
Assuming no fish can be caught twice by the same person:


the probability of catching 100 fish at least one of which was one of the 100 fish that Kristian caught is :

100/950,000 or 1.05 e-4

the probability of catching exactly one of the 100 fish that kristian hooked, and no more plus an additional 99 uncaught fish is.......erm...anyone?


Don't mean to be facitious but in situation one, are you not depleting the pool of one fish per fish caught and hence it is the sum of 100 in 950000, 100 in 949999, 100 in 949998 etc.

The second one needs a computer I think...

It's been a long time since I've done any statistics, so if there's a formula I've forgotten it, but here's the manual way of working it out:

There are 950,000 fish, 100 if which have been caught before. This means 949,900 fish have NOT been caught before. Taking each catch in turn, the odds change because the total number of fish in the pond changes.

Probability of NOT having been caught before:

First fish: 949,900/950,000 = 0.999894736842
Second fish 949,899/949,999 = 0.999894736731
Third fish: 949,898/949,998 = 0.999894736620
Fourth fish: 949,897/949,997 = 0.999894736509
Fifth fish: 949,896/949,996 = 0.999894736398
Sixth fish: 949,895/949,995 = 0.999894736288

And so on. To work this out manually will be a total pain in the neck. You'll have to multiply all 100 numbers together to obtain your final probability of NO fish having been caught before. For the inverse probability (of at least one fish having been caught before), just subtract the final figure from 1.

So if anyone knows a decent formula to help.... please feel free to tell us it :)

P.S. Each of the decimal numbers quoted are shortened from their maximum length for 'simplicity', which brings with it additional errors. Depending on the accuracy required, you may have to increase the number of significant figures used. For example:

949,900/950,000 = 0.99989473684210526315789473684211

Originally posted by wendygs
Phan I dont think that's the problem and adding unnecessary layers of complication to a simple solution; it's a waste of time trying to calculate or rescue the pearl button from the fish.:D

It seems to me to be ambiguous.

The probabilities are infinitesimal, and would not be borne out in the real world, because some fish are hungrier than others.

But the abstract problem, as Kristian outlined it is ambiguous.

The probabilities are different (minute differences in tiny values), but different nevertheless.


set A = 950,000 fish
set B = Kristian's 100 fish
Set C = 100 randomly selected fish

if both B and C are subsets of A -

in essence the two problems are -
what is the probability that the C intersects with B?

what is the probability that any intersection of B and C contains only 1 fish?

As my neighbour is a statistician he would have produced a formula if he thought it would help to resolve the problem. He left his calculations with me as follows:

Any one fish = 1/950,000
Any 100 fish = 1/950,000 x 1/949,000 x 1/ ... etc

this seems similar to some of the other stuff produced on this thread which is getting funnier and funnier by the post

Originally posted by tim_rutter
Don't mean to be facitious but in situation one, are you not depleting the pool of one fish per fish caught and hence it is the sum of 100 in 950000, 100 in 949999, 100 in 949998 etc.

The second one needs a computer I think...

What if all the fish are hooked simultaneously (like a big mackerel spinner or something).

I am not a statistician and I am trying to grasp this problem (like most stuff) intuitively, so you aren't being facetious.

Would the mass fish extraction method obviate the need for this additive calculation or not?

Kristian what is the actual objective of this exercise? ie do you need to recover your button or are you worried about the health of the fish?

Lets get this sorted, and I will help YOU sort it.

The original statement is flawed. If there were a million fish in the pond the chances would be 0.0001% chance. 950.000 fish does not give you the original equation.

However, translate this to the lottery, a number occuring from 49 balls is 49 x 48 x 47 etc. etc. this is diminished by having a line of 6 numbers. So the chances of winning are somewhere in the region of 14 million to 1.

Now translate to 950,000 fish and the chances of being caught twice.

I will contact the forum in 20 years to see if you have come up with an answer cos I dont think there will be enough paper in Sheffield to write the answer on.

I would work it out, but I like to get out now and again.

Originally posted by BoppinBruce
Lets get this sorted, and I will help YOU sort it.

I will contact the forum in 20 years to see if you have come up with an answer cos I dont think there will be enough paper in Sheffield to write the answer on.

I would work it out, but I like to get out now and again.

I'll donate ALL of my paper files which amounts to a very considerable storeroom here, in London, various Govt Departments etc, etc, etc to this very much more worthy cause so you resolve this unquantifiable problem for us once and for all. :D :D :D :D

I read a book once called the hitch hikers guide to the galaxy - you may have heard of it. In this book a computer is asked the answer to life the universe and everything. The answer comes after many years to 42.

They then had to postulate what the question actually was.

I think we need to clear that up before we go injecting any reems into this! And to be honest I don't think its that difficult.

I would like to know the reason behind kristians fish obsession?!?!

Originally posted by tim_rutter
...I would like to know the reason behind kristians fish obsession?!?!
I've already told you...

...it's that wretched button.

:D

Lets club together and buy him a new set of buttons, then we can all sleep tonight

Originally posted by valentine
Lets club together and buy him a new set of buttons, then we can all sleep tonight

:clap: :clap: :clap:

Excellent idea...

...I'll pledge 50p to the "good cause".

:thumbsup:

Originally posted by tim_rutter
it works out as about 98.9%

Yep if I've got my sums right, I get that too, as the chance of not catching any of the marked fish the second time around. I.e. the chance of recapturing at least one of them is around 1% or more accurately 1.05%.

[There's a bit of ambiguity in the question, but I'm guessing that you catch and mark 100 fish, put them back, then catch 100 fish again one after the other, putting them in a keepnet as you go to ensure you don't catch the same fish twice. Also have to assume everything is random and no fish get eaten by penguins, or whatever, in the interim.]

This question seems almost as nasty as one I saw once for the probability of wearing the same outfit as someone else and 'clashing' at an event, or wearing something you've worn before in front of the assembled crowd :confused:

Maybe that's what Kristian is working his way towards ? :)

As an absolute thicko in maths,I think that with 950,000 fish the odds of of catching a non specific fish are pretty damn good in your favour.The odds of catching Kristian's fish are 950,000/1 against and as far as I can see always will be,assuming the caught fish are returned.

Originally posted by Kristian
Now, regular users will recognise that I know more about designer shopping and general good taste than maths, but the good people who employ me unforunately don't want this area of my expertise; if someone can help me with the following maths problem you'll save me the embarrassment of having to remove my shoes and socks to count my toes in the office...

I'm trying to work out the probability of two events coinciding. I don't work with fish, but will use this as an example otherwise I'm going to confuse you as much as me.

If there is a pool of water with 950,000 fish swimming in it, and I catch 1, the probability of any one fish being caught is 0.0001%. If I catch a hundred fish, this percentage increases to 0.01%

If the fish are then put back and someone else comes along and catches 100 fish, what is the probability that one of the fish will have been caught twice?

It really is doing my head in. I've asked loads of folk here for help, but keep getting different replies. Please help before I show myself to be the proper numptie that I am! :D

Ta,

K x every possibility,fish have short memories so will forget that the bait had a hook in it last time,simple :banana:

Ok. I'll admit it. I've got a copy of Crawshaw and Chambers' - A Concise Course in A-Level Statistics open here.

After extensive reading, the problem is actually simpler than it first looks.

Just for arguments sake, and to make it all traceable, Kristian painted all his caught fish red :D

You now have a lake full of fish, 100/950,000 are red, and the problem becomes: What are the chances of selecting a red fish in a batch of 100, knowing 1/9500 are red?

The probability of catching 1 red fish on the first go is therefore 1/9500. But we are going to have 100 goes at this - so there is a better chance of getting a red one, and we therefore expect our answer to be a bigger number (or have a smaller number on the bottom, if that is easier to understand)

Quite simply, the chance of getting at least one of Kristian's red fish (1/9500) is 100 times more likely (x 100):

1/9500 x 100
1/95 :thumbsup:

Did anybody else get that? :suspect:

The very remote probability answers I read at the top of the thread are more relevant to htere being a fish called 'Bill', and both Kristian and his colleague both catch Bill in amongst their haul ;)

Guys, thanks everso for the help on this! :thumbsup: I can assure you all that I don't have a fish or button fetish, but if I had explained the situation I was in with the real situation of insurance policies you would have been even more confused - trust me!

The answer I used on the report I was writing was 1.05 % or 1/98, and everyone seems happy so far!

I really appreciate everyone that went out of their way to help, and everyone that gave me a giggle! :D

K x

EDIT: As I've sorted this, I'm going to close it now. Thanks again peeps.