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Canary Jedi

Calling all NCFC maths geniuses!

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3 minutes ago, Indy said:

True, that’s what I said from the start, probability is the measure likelihood is the question from the OP.

The OP did start off by asking what the probability was, so that's what most people gave him, though he did then ruin it by asking what the chances were, so yeah Emi is magic and the OP should stop trying to do his son's homework...

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I'd go with the following unless you want someone to look stupid in front of their maths teacher:

" 1/21 x 1/21 x 1/21 = 0.0001 approximately 

or 1 in 9,261"

 

Btw a nice explanation of the multiverse/ multi worlds theory and much more is in the book "in search of schroedingers cat" by John gribbon.   Accessible to all with a GCSE A level ish understanding of physics but enough to stretch anyone. 

Edited by Barbe bleu
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8 hours ago, Indy said:

I love all the debate which follows, the reality is every time you pick up that dice you have a 1 in 21 chance, it resets as soon as you pick it up. It’s just as likely as say 17, flowed by 3 then a 5!

Funny I posted what I though was a simple question this morning and now there are pages of answers! Mostly irrelevant Lakey beating it has to be said...

But on your reply Indy, I don’t think that’s right - I understand it’s 1 in 21 for an individual roll, but if I ask you to do it 3 times in a row surely that’s a different kettle of fish? But I’m not a mathematician, it’s a genuine question ☺️

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2 hours ago, Rock The Boat said:

Here you go Indy, I googled to find the difference between probability and likelihood. Apologies for the answer.

 

Discrete Random Variables

Suppose that you have a stochastic process that takes discrete values (e.g., outcomes of tossing a coin 10 times, number of customers who arrive at a store in 10 minutes etc). In such cases, we can calculate the probability of observing a particular set of outcomes by making suitable assumptions about the underlying stochastic process (e.g., probability of coin landing heads is pp and that coin tosses are independent).

Denote the observed outcomes by OO and the set of parameters that describe the stochastic process as θθ. Thus, when we speak of probability we want to calculate P(O|θ)P(O|θ). In other words, given specific values for θθ, P(O|θ)P(O|θ) is the probability that we would observe the outcomes represented by OO.

However, when we model a real life stochastic process, we often do not know θθ. We simply observe OO and the goal then is to arrive at an estimate for θθ that would be a plausible choice given the observed outcomes OO. We know that given a value of θθ the probability of observing OO is P(O|θ)P(O|θ). Thus, a 'natural' estimation process is to choose that value of θθ that would maximize the probability that we would actually observe OO. In other words, we find the parameter values θθ that maximize the following function:

L(θ|O)=P(O|θ)L(θ|O)=P(O|θ)

L(θ|O)L(θ|O) is called the likelihood function. Notice that by definition the likelihood function is conditioned on the observed OO and that it is a function of the unknown parameters θθ.

Continuous Random Variables

In the continuous case the situation is similar with one important difference. We can no longer talk about the probability that we observed OO given θθ because in the continuous case P(O|θ)=0P(O|θ)=0. Without getting into technicalities, the basic idea is as follows:

Denote the probability density function (pdf) associated with the outcomes OO as: f(O|θ)f(O|θ). Thus, in the continuous case we estimate θθ given observed outcomes OO by maximizing the following function:

L(θ|O)=f(O|θ)L(θ|O)=f(O|θ)

In this situation, we cannot technically assert that we are finding the parameter value that maximizes the probability that we observe OO as we maximize the PDF associated with the observed outcomes OO.

...you lost me at 'stochastic'

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The op asked can it come true, well the answer is yes.

It is improbable...but we could do it.  We were close in 92/93, 88/89, Forest, Derby.....Leicester......yes I know football has changed since the 80's, but the lines are closing.......we have been seeing for a number of seasons how the quality of player in teams throughout the premier league has risen...and in the championship too....the level has risen across the board.

So it is my strong opinion that it can be done, it requires a Leicester kind of roller coaster season to do it, everything would have to be absolutely right - hit the ground running, settled team, minimum injury problems, togetherness, belief, etc etc....and after all, we proved we could get it right last season..... 

Quality of player is an issue...can a team like us have the quality players that are good enough to do it?   Again the answer could be yes. Pukki may carry on where he left off last season, Aarons and Lewis too, Godfrey definitely has the quality to be a Company style centre back and Zimmermann/Klose are no fools.   We have good goalie options, we have some superb midfielders, or do people think Buendia Leitner and Vrancic are not top quality players??

The prognosis at this stage is that we have the equipment/mentality to do as well as any club in the league...so at the start of the season, the odds....probablilty....or whatever you want to call it...are not as bad as people/bookies etc are saying.

I would go as far as to say any club can win the title next season.  Liverpool and Man City will have to be at the tops of their game again to do what they did last season - and City have lost Company, who was a huge asset for them on and off the pitch. Liverpool will be good again, for sure, but then so can any number of clubs below them. 

This season in the PL is going to fascinating for many reasons, not least because we are in it.  I accept the probabilty is low that we will win it, but not that low - and we all start on the same number of points and nowhere in the rules does it say it can't be done....

 

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15 hours ago, Canary Jedi said:

Funny I posted what I though was a simple question this morning and now there are pages of answers! Mostly irrelevant Lakey beating it has to be said...

But on your reply Indy, I don’t think that’s right - I understand it’s 1 in 21 for an individual roll, but if I ask you to do it 3 times in a row surely that’s a different kettle of fish? But I’m not a mathematician, it’s a genuine question ☺️

The problem extends to this, if you throw that dice the result might be 17, as soon as you pick it up that result is insignificant as you have another i in 21 chance of rolling that 17! So each time you throw a single dice it reverts back to 1 in 21 probability.

In can see why the cubed route is used by some to try and work out the chance and say it’s 1 in 9261, but the reality is it resets to 1 in 21 as soon as you pick up that dice.

Just like tossing that coin 1 in 2 result.

By the way.......roll a 21 sided dice you have a 1 in 21 probability  of throwing a 17......pick it up again you have a 1 in 21 probability  of throwing a 17, pick it up again and you have a 1 in 21 probability of throwing that 17 again.

If you throw three dice at the same time to throw three 17’s it’s a 1 off 9261 combinations  😂

Edited by Indy

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It does. But that’s for an individual roll. The OP was talking about it happening 3 times in a row. So, the odds of that happening multiply. The odds for each individual event, as you state, remain constant and are in no way influenced by the previous event. So while the chance of rolling three in a row are cubed, once the first two have been rolled, the odds of rolling the third are the same as rolling the first one was originally. However, before the dice is touched the odds of rolling 3 consecutive same numbers are (21x21x21)/1. 

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Right here you go guys, how many combinations are there if you roll 2 x 6 sided dice?

Its not 36 unless take each dice separate, there are only 21 number combinations. So if you call dice 1 Fred, dice two Charlie and revere the number combinations it doesn’t work out as squared or cubed using three dice!

 

(you can roll.....1+1, 1+2, 1+3, 1+4, 1+5, 1+6, 2+2, 2+3, 2+4, 2+5, 2+6, 3+3, 3+4, 3+5, 3+6, 4+4, 4+5, 4+6, 5+5, 5+6 or 6+6) they are your number combinations.

So to consider the number combinations only the probability is less that cubed.

Edited by Indy

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3 minutes ago, Indy said:

Right here you go guys, how many combinations are there if you roll 2 x 6 sided dice?

Its not 36 unless take each dice separate, there are only 21 number combinations. So if you call dice 1 Fred, dice two Charlie and revere the number combinations it doesn’t work out as squared or cubed using three dice!

 

(you can roll.....1+1, 1+2, 1+3, 1+4, 1+5, 1+6, 2+2, 2+3, 2+4, 2+5, 2+6, 3+3, 3+4, 3+5, 3+6, 4+4, 4+5, 4+6, 5+5, 5+6 or 6+6) they are your number combinations.

So to consider the number combinations only the probability is less that cubed.

Those are unique combinations Indy

All combinations would include 1+2 and 2+1, for example

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9 minutes ago, Indy said:

Right here you go guys, how many combinations are there if you roll 2 x 6 sided dice?

Its not 36 unless take each dice separate, there are only 21 number combinations. So if you call dice 1 Fred, dice two Charlie and revere the number combinations it doesn’t work out as squared or cubed using three dice!

 

(you can roll.....1+1, 1+2, 1+3, 1+4, 1+5, 1+6, 2+2, 2+3, 2+4, 2+5, 2+6, 3+3, 3+4, 3+5, 3+6, 4+4, 4+5, 4+6, 5+5, 5+6 or 6+6) they are your number combinations.

So to consider the number combinations only the probability is less that cubed.

But the possibility of rolling the same specific number twice is 1 in 36. Which is 6².

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51 minutes ago, Rock The Boat said:

Those are unique combinations Indy

All combinations would include 1+2 and 2+1, for example

But it’s not is it! You’re rolling two dice. It doesn’t matter which one has 2 and which one has 1. That’s the point the number combinations aren’t squared or cubed.....not unless you count each dice individually, which you don’t, their just dice.

So rolling a 1+17+17 is the same as rolling 17+17+1 the same number combination. If you ask someone to roll 1+2+3 with three dice you don’t tell them to roll that specific way, they throw all three together! So the odds are not cubed for rolling three dice of any numbered sides!

 

Edited by Indy

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The cirrect answer was given within 7 minutes of the question being raised. 

36 hours and a lot of nonsense later this is still being discussed.

I suspect that this might be a reflection of a larger pattern

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Bit indy, your logic is flawed. Rolling 1+17+17 may be the same as rolling 17+17+1, but there are 3 permutations of rolling 2 17s and a 1, there is only one permutation of rolling 3 17s. In effect you are saying that you've got to roll the three numbers in order.

To use your 1&2 on a d6 example. The probability of rolling a 1 or 2 on the first roll is 1/3, the probability of rolling the other one on the second is then 1/6, giving a total probability of 1/18. However, the probability of rolling 2 6s is 1/6 then 1/6, which is total probability of 1/36. The dice being rolled together or individually doesn't make a difference when there is only one permutation of the result.

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48 minutes ago, cornish sam said:

As I said earlier, this needs to be treated as one unique event, not three sequential events.

And as I said earlier, it's an accumulator, not 3 separate events. As the challenge was to throw 17, three times in three throws. I don't get why some can't grasp this and have tried to cloud the issue with blather.

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22 minutes ago, wcorkcanary said:

And as I said earlier, it's an accumulator, not 3 separate events. As the challenge was to throw 17, three times in three throws. I don't get why some can't grasp this and have tried to cloud the issue with blather.

OK in an ideal world to throw 17 three times is indeed 1 in 9261.......

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I think we’re getting confused with probability, outcome, likelihood and Norwich wining the Premiership!

I stand corrected, I agree that you have one in 9261 chance in rolling three 17’s but that only applies if you are likely to throw every other 9260 sequence numbers!

 

The likelihood is far more random than 9261! 😂

 

I love statistics that’s why I did it for safety for thirteen years.....add variables and consequences it’s a damn interesting job for sad people!

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Theme the facts Indo. 

Somehow though, I doubt a bookie, if approached about this very scenario , would give anything like 9261/1.  Maybe half that ,say 5000/1.  Maybe less, that's why I don't generally gamble .

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So basically, if we go with 9261/1 and assume my son is not telling porkies, it’s quite an amazing result. 

I must interrogate him some more - I wonder if his teacher is a binner, that would be even better!

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6 hours ago, FenwayFrank said:

Ok, what are the chances of LDC being kidnapped by aliens tomorrow ?

Evens. 

 

He will or or he won’t. 

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13 hours ago, FenwayFrank said:

Ok, what are the chances of LDC being kidnapped by aliens tomorrow ?

 

7 hours ago, Duncan Edwards said:

Evens. 

 

He will or or he won’t. 

10.56 am and LDC hasn't appeared again on this thread. I'm going 4/6 kidnapped by aliens, 6/5 still on planet earth........

Earth to Lakey, Earth to Lakey...................

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14 hours ago, FenwayFrank said:

Ok, what are the chances of LDC being kidnapped by aliens tomorrow ?

He does live in Cumbria so quite high I reckon.

 

However the probability of them returning him within 5 minutes is much higher.

Edited by Drazen Muzinic
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11 hours ago, Canary Jedi said:

So basically, if we go with 9261/1 and assume my son is not telling porkies, it’s quite an amazing result. 

I must interrogate him some more - I wonder if his teacher is a binner, that would be even better!

Of course, retrospectively (at least in this universe) we now know that those numbers were certain to be rolled, although maybe it's more likely that he is making it up. In any case, each different combination (in mathematical terms) had the same probability of occurring and *one* of those combinations had to be rolled. But it is fair to say that, in a future repetition of the same exercise, it seems extremely unlikely that some one would declare the three numbers to be rolled, in the correct order before the event.

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16 hours ago, wcorkcanary said:

And as I said earlier, it's an accumulator, not 3 separate events. As the challenge was to throw 17, three times in three throws. I don't get why some can't grasp this and have tried to cloud the issue with blather.

Exactly. I really can't understand how this has been debated so much. It is, very, very simple.

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4 minutes ago, All the Germans said:

Exactly. I really can't understand how this has been debated so much. It is, very, very simple.

Simple for some perhaps!

Another question, does it make any difference to the accumulated odds the fact that it was 3 numbers the same, as opposed to say predicting rolling a 1 then 5 then 21?

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4 minutes ago, Canary Jedi said:

Simple for some perhaps!

Another question, does it make any difference to the accumulated odds the fact that it was 3 numbers the same, as opposed to say predicting rolling a 1 then 5 then 21?

Assuming you named the order then no, the odds are identical.

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9 minutes ago, Canary Jedi said:

Simple for some perhaps!

Another question, does it make any difference to the accumulated odds the fact that it was 3 numbers the same, as opposed to say predicting rolling a 1 then 5 then 21?

Apologies, it was not meant as a slight on you for asking the question, more I just can't understand how it has taken 4 pages of disagreements when the correct answer was given on the first page.

No. It makes no difference for the numbers differing or there being different combinations; they are all a 1 in 21 chance of occurring individually. The odds are then multipled to the power of the number events you are predicting (in this case 3).

Edit - Agreed, the order must be named (otherwise you have more options for the first two throws).

Edited by All the Germans

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