Calling all NCFC maths geniuses!

[Petriix 2,834]

Probability is only relevant in a purely mathematical world. As soon as you apply physics things become more complicated. In a physical world, probability is (at best) an approximation for the unknown factors involved. Over large sample sizes of relatively simple tasks (coin tosses, dice rolls etc) things tend towards the expected balance. However, something as complex as a football league is impossible to predict with any kind of mathematical accuracy.

Now, it is normal to go through life assuming that all things are possible (within reason), but there is a fairly compelling argument that every single future event is already predetermined by the current physical state of the universe and, by extrapolation, the Big Bang has already set in motion every single event until the end of time. In that sense, when someone says "what were the chances of that happening?" the answer is always "100%, definite, 1-in-1, certain". Retrospectively there is no point in probability because we already know the outcome. Perhaps a better question would be "what would the chance be of that happening again under similar circumstances?" to which I would reply "how similar?".

In any case, predicting outcomes reduces to how accurately you can measure and extrapolate the known factors and how they tend to interact based upon previous data. While Leicester were, in hindsight, 100% certain to win the league, Norwich are almost certainly (or are they?) not going to do so this season.

Edited July 8, 2019 by Petriix

[ricardo 7,347]

4 minutes ago, Petriix said:

Probability is only relevant in a purely mathematical world. As soon as you apply physics things become more complicated. In a physical world, probability is (at best) an approximation for the unknown factors involved. Over large sample sizes of relatively simple tasks (coin tosses, dice rolls etc) things tend towards the expected balance. However, something as complex as a football league is impossible to predict with any kind of mathematical accuracy.

Now, it is normal to go through life assuming that all things are possible (within reason), but there is a fairly compelling argument that every single future event is already predetermined by the current physical state of the universe and, by extrapolation, the Big Bang has already set in motion every single event until the end of time. In that sense, when someone says "what were the chances of that happening?" the answer is always "100%, definite, 1-in-1, certain". Retrospectively there is no point in probability because we already know the outcome. Perhaps a better question would be "what would the chance be of that happening again under similar circumstances?" to which I would reply "how similar?".

In any case, predicting outcomes reduces to how accurately you can measure and extrapolate the known factors and how they tend to interact based upon previous data. While Leicester were, in hindsight, 100% certain to win the league, Norwich are almost certainly (or are they?) not going to do so this season.

Since all possible futures already exist somewhere in the Multiverse, It just depends upon which Universe we happen to be in.

[Thirsty Lizard 3,155]

43 minutes ago, ricardo said:

Since all possible futures already exist somewhere in the Multiverse, It just depends upon which Universe we happen to be in.

Ricardo - You seem to be arguing that there is a universe somewhere where Ipswich Town aren't sh*te. Sorry, but that blows your Multiverse theory out of the water. 

[It's Character Forming 1,160]

According to Elon Musk, the chances are we're all  existing in a Matrix-type simulation rather than the real world

[Thirsty Lizard 3,155]

2 hours ago, Bill said:

err, that's because their odds reflect what money has been bet

not what they think will happen

the clue is in the name........... bookmaker

not predictors

They don't when the bookies first put their odds up. 

[lappinitup 629]

3 hours ago, dylanisabaddog said:

Of more interest to me is the 25-1 that Sky is offering on us at Liverpool.

2 hours ago, lappinitup said:

Where can you see this. Just joined up to place a bet but only 16-1 now. What am I missing?

 

Blimey, only 14-1 now, must be several people fancy our chances.

[wcorkcanary 4,329]

2 minutes ago, lappinitup said:

Blimey, only 14-1 now, must be several people fancy our chances.

Lakey has just put his Cumbrian Mansion House on City to win the league, that's why the odds have dropped. They know that he knows, you know.

[Thirsty Lizard 3,155]

13 minutes ago, lappinitup said:

Blimey, only 14-1 now, must be several people fancy our chances.

Betting exchanges nearly always the best place to go to get a good price. Betfair odds at the moment. 

[cornish sam 944]

1 hour ago, ricardo said:

Since all possible futures already exist somewhere in the Multiverse, It just depends upon which Universe we happen to be in.

The existence of futures surely only applies dependent on your frame of reference, and I fear you are mixing yours here. The future as not happened yet from within our frame of reference and as such those multiverses are mirrors of our own or yet to be. If you are saying they already exist then they then they are no longer  futures and are merely fact from the frame of reference where future and past have no meaning...

[dylanisabaddog 4,871]

1 hour ago, lappinitup said:

Blimey, only 14-1 now, must be several people fancy our chances.

That's strange. It can't be my 50p that has dropped the odds. You can still get 23-1 with William Hill 

https://www.oddschecker.com/football/english/premier-league/liverpool-v-norwich/winner

[All the Germans 1,075]

4 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!

I love those who cube it to try and gain the probability! Nah you can’t do that......it’s like tossing a coin it’s always 50 / 50.

This is not correct. Whilst I full agree that probability does not have a memory; each single flip of a coin / roll of a dice is equally likely to be any one of it's options as any other (assuming a fair test), regardless of any of the previous results. The probability of rolling three of the same in a row is less likely by ... three times (eg in this example - 1/21 x 1/21 x 1/21 = 1/9261). As you point out, the final roll would still be 1/21, as the probability does not remember or is any way influenced by previous rolls but that is very different to your inference above (unless I've misunderstood you?).

[Indy 3,283]

1 minute ago, All the Germans said:

This is not correct. Whilst I full agree that probability does not have a memory; each single flip of a coin / roll of a dice is equally likely to be any one of it's options as any other (assuming a fair test), regardless of any of the previous results. The probability of rolling three of the same in a row is less likely by ... three times (eg in this example - 1/21 x 1/21 x 1/21 = 1/9261). As you point out, the final roll would still be 1/21, as the probability does not remember or is any way influenced by previous rolls but that is very different to your inference above (unless I've misunderstood you?).

Probability is constant as I went on to explain, what you’re referring to is likelihood.

[Indy 3,283]

3 minutes ago, All the Germans said:

This is not correct. Whilst I full agree that probability does not have a memory; each single flip of a coin / roll of a dice is equally likely to be any one of it's options as any other (assuming a fair test), regardless of any of the previous results. The probability of rolling three of the same in a row is less likely by ... three times (eg in this example - 1/21 x 1/21 x 1/21 = 1/9261). As you point out, the final roll would still be 1/21, as the probability does not remember or is any way influenced by previous rolls but that is very different to your inference above (unless I've misunderstood you?).

 

Posted 4 hours ago

Don’t get probability and likelihood mixed up!

Probability is the certain occurrence of the event,  say that dice has 21 sides so the probability of that number being thrown is 1 in 21.

Likelihood is the condition of probability, so if you throw that dice three times and throw 17 three times is conditional so it’s calculated.

So if you ask about probability it’s static at 1 in 21 every time. Likelihood is much harder to calculate

[Hairy Canary 668]

4 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!

I love those who cube it to try and gain the probability! Nah you can’t do that......it’s like tossing a coin it’s always 50 / 50.

Well of course that's right. But if you are looking for the probability of throwing three consecutive 17's before you have thrown any dice then it is 21 cubed.

its like the probability of tossing a coin three heads on the trot from scratch. It's one chance in eight because there are seven other possible outcomes.

[Indy 3,283]

4 minutes ago, Hairy Canary said:

Well of course that's right. But if you are looking for the probability of throwing three consecutive 17's before you have thrown any dice then it is 21 cubed.

its like the probability of tossing a coin three heads on the trot from scratch. It's one chance in eight because there are seven other possible outcomes.

It’s not cubed, there’s a formula to work it out, I can’t be asked to troll through, but I used to be involved in risk assessments all the time, including probability and likelyhood.

[wcorkcanary 4,329]

Surely the odds of this three times getting seventeen on the 21 sided dice is the same as an accumulator with three bets at 21/1.

9261/1

[......and Smith must score. 1,336]

6 hours ago, lake district canary said:

Nope, I don't like flying

How do you know if you’ve never flown on one ?

[Hairy Canary 668]

5 minutes ago, wcorkcanary said:

Surely the odds of this three times getting seventeen on the 21 sided dice is the same as an accumulator with three bets at 21/1.

9261/1

That's exactly right. It is a straight forward 21cubed which is as you say 9261/1

[Rock The Boat 1,325]

The result is cubed because you are doing 3 rolls, if you were saying 17 four times in a row it would be 21 to the power of 4

 

A very simple example of the odds of flipping a coin twice and getting heads twice in a row, the possible outcomes are:

H-T

H-H

T-H

T-T

So both heads or both tails occur 1 in four times (1/2 x 1/2).  While a head OR a tail is 2 in four times (1/2 + 1/2)

Edited July 8, 2019 by Rock The Boat

[Indy 3,283]

6 minutes ago, Hairy Canary said:

That's exactly right. It is a straight forward 21cubed which is as you say 9261/1

Nope because that’s the calculated number, the likelihood takes into account that you can roll those three 21 sided dice 10,000 times and still not roll 17 three times in a row, then you could throw those dice just three times and roll 17 three times. That’s why probability is defined as 1 in 21 but likelihood is a far more complicated calculation.

[Indy 3,283]

6 minutes ago, Rock The Boat said:

The result is cubed because you are doing 3 rolls, if you were saying 17 four times in a row it would be 21 to the power of 4

 

A very simple example of the odds of flipping a coin twice and getting heads twice in a row, the possible outcomes are:

H-T

H-H

T-H

T-T

So both heads or both tails occur 1 in four times (1/2 x 1/2).  While a head OR a tail is 2 in four times (1/2 + 1/2)

Let’s put the next variable into this then, Chance, the fact you have a 1 in 2 probability doesn’t mean it’ll be heads then tails, Chance dictates that you could flip heads fifty times in a row, probability says it’s unlikely then the likelihood dictates this event is very rare. That’s why there’s a few calculations that can be used, I think you’ll find it’ll be a lot higher than 1 in 9261 to roll three 17’s in a row on a 17 sided dice.

But that’s my thoughts based on experience in applying probability in my industry.

[lappinitup 629]

 

[lappinitup 629]

Oops! Wrong thread.

[Wacky Waving Inflatable Arm Flailing Tube Man 3,780]

3 minutes ago, lappinitup said:

Oops! Wrong thread.

It's always the right thread for that clip.

Besides, maybe one of the geniuses could explain the physics behind how he did it.

[Rock The Boat 1,325]

8 minutes ago, Indy said:

Let’s put the next variable into this then, Chance, the fact you have a 1 in 2 probability doesn’t mean it’ll be heads then tails, Chance dictates that you could flip heads fifty times in a row, probability says it’s unlikely then the likelihood dictates this event is very rare. That’s why there’s a few calculations that can be used, I think you’ll find it’ll be a lot higher than 1 in 9261 to roll three 17’s in a row on a 17 sided dice.

But that’s my thoughts based on experience in applying probability in my industry.

I think that probability and likelihood are two different measures, are they not? In probability you know the set of parameters (something is either heads or tails) and the possible outcomes whereas in likelihood you know (observe) the possible outcomes but don't know the process that produces those outcomes

 

[Graham Paddons Beard 2,424]

3 hours ago, Thirsty Lizard said:

They don't when the bookies first put their odds up. 

Exactly . City 1st trots this one out from time to time. Amusing that he has completely missed the Book Maker part. 

[Rock The Boat 1,325]

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.

[Indy 3,283]

6 minutes ago, Rock The Boat said:

I think that probability and likelihood are two different measures, are they not? In probability you know the set of parameters (something is either heads or tails) and the possible outcomes whereas in likelihood you know (observe) the possible outcomes but don't know the process that produces those outcomes

 

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

[Indy 3,283]

3 minutes 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.

I’m glad you dragged this out! 😂 Sit back and watch this thread develop!

By the way Emi Beundia isa magician and his skill is beyond our comprehension! Nice one Lapps

[cornish sam 944]

No, it's not indy. The question is badly formed but it is a probability question. The probability of rolling a d21 and getting 17 the first three times is 1/9261. If it were being considered as an undefined subset of an infinite series then that would be where the likelihood came in. Essentially you have to treat the three rolls together as a discrete event, not three separate ones.