Basic Probability

[TheBramble]

So, based on what you think it currently is trendie, if you were offered the opportunity to take part in a high-stakes wager where you win on the first occurrence of two consecutive Heads in a maximum 4 coin tosses game, what would your Aw:Al ratio have to be for it to become marginally positive enough for you to consider participation?

 

[intradaybill]

Guys, probabilities make sense in your examples only as relative frequencies. There is a lot of debate about the relevance of of limiting probabilities to single or few step decision making processes.

 

[TheBramble]

I do understand what you're saying IDB, but where we don't get the luxury of deciding or debating or limiting probabilities to a single or few steps decision making process, we do need something to work with.

The entire purpose of this thread is to investigate what many have always (I'm guessing) used for their probabilistic decision making processes and compare that with what is currently believed to be more useful and which more closely mirrors the reality we encounter – especially when we trade.

 

[Hotch]

Next you're going to start spouting on about "probability pressure"

 

[TheBramble]

For any who have the vaguest hint of energy or interest left in this topic, have a shot at guessing the minimum coin tosses you’d need to make to, statistically, in our very real, but non-Gaussian Universe, generate all of the 16 possible strings of results from a nominal 4 coin toss sequence?

If you don’t have any idea, set a lower bound and an upper bound for the numbers you think are 90% likely to encapsulate the actual answer to this one. I’ll post the answer later.

It's 19.

2^N + N+1 (Havil, J. Gamma: Exploring Euler's Constant. Princeton, NJ: Princeton University Press, 2003. )

But who gives a toss. LOL












note to brain: don't bother with anyhting like this again. let's find the equivalent of page 3 for traders - they'll all like that....

 

[Calinor]

I don't think you have specified the question too well. That list of 16 isn't a sensible list of outcomes with regard to the experiment you're interested in. But before I get into this more

The Probability of two consecutive Heads in 4 coin tosses is derived from the formula (Gardner, Berlkamp below)

1-(F(n+2)/(2^n))
where F(n+2) is the (n+2)th Fibonacci number

In our example the value is 0.1875 (3/16).

For the last question asked, the probability of two consecutive Heads OR 2 consecutive Tails the result is a Probabilistic OR which yields 0.375 (6/16).

What exactly are you saying here? Your formula above has an 'n'. If you take the 'n' to be the coin tosses (you didn't define 'n') then you would not have the answer you gave, right?

 

[Calinor]

Got distracted on the FTSE closing the gap from the 1st April , so only just finished editing above

 

[TheBramble]

Got distracted on the FTSE closing the gap from the 1st April , so only just finished editing above

Yes, N is the number of coin tosses (4).

What did you get for 1-(F(n+2)/(2^n))
where F(n+2) is the (n+2)th Fibonacci number
and where N=4?

 

[TheHoneyMonster]

I dont know about basic probability, and i am no rocket scientist, but -

I dont think a trader need bog their heads down on endless issues regarding probability, luck, lack of luck, and so on.

For me, simply getting in early when the risk:reward is good, and holding until my target/s is met does and always will - work out well in the long run.

No need for random walk theories, cycles etc. just pick good entries, hold for your targets or stop losses, and take it from there.

For me, i only start to think about luck after i have made mistakes, such as entering too late, exiting too early, and then trying to enter or re-enter late. But if i trade correctly I know that the end result will be good, and that i will not need to concern myself with these types of issues.

Me i just relax, go with the flow and take the rough with the smooth. and most days i walk away with a big fat stash.
yes some days you might get a 60/40 split between favour and disfavour, yet other days you may get a 40/60 split between favour and disfavour (heads/tails or whatever). Some days you get the rub of the green, others you dont, just accept it and on you go.

 

[TheBramble]

N+2 = 4+2 = 6

6th Fib is 13.

2powerN where N=4 is 2power4 = 16.

THerefore 13/16.

1 - 13/16 = 3/16.

Happy to discuss any errors I may have made in the calculation.

 

[arabianights]

crikey, you don't open a thread on probability expecting to see handbags at dawn!

 

[TheBramble]

I dont know about basic probability, and i am no rocket scientist, but -

I dont think a trader need bog their heads down on endless issues regarding probability, luck, lack of luck, and so on.

Check your earlier post in this thread where you state "I'd be surprised if the split was more exagerrated than 55/45after 100 [tosses of a coin]".

Basic probability suggests we should expect a 50/50 split, yet you recognise even after an arbitrary number of coin tosses, there will be a discrepancy between what basic probability suggests we should expect and what we recognise as more likely to occur in reality.

The entire nub of this thread.

I’m attempting to bring more contemporary empirical views on the reality of probability to bear on what after all might be quite important to you as a trader if after 100 trades you had calculated a probabilistic 50/50 split of some kind – and found the actual results more clearly clung to the somewhat more skewed (and more believable) 45/55 you randomly suggest. Depending on what those numbers are and what they mean to you – it might make all the difference in the world.

 

[TheBramble]

crikey, you don't open a thread on probability expecting to see handbags at dawn!

Well, 14.34% of the time you do.....

You know how it works GJ. There’s always one (or two in this case) that have nothing to add to the topic, but take the opportunity to live out their fantasies, making personal attacks on those they most envy. In some ways I am flattered, but it is a distraction.

 

[Calinor]

1,1,2,3,5,8

My 6th Fib is 8 (or even 5 if you start at 0). But ok if you want to define the Fibs as you have, I won't be picky, that is just custom. As long as the formula uses the same custom as you

I am still concerned with what you're actually asking. It is my experience that most confusion arises when the problem isn't clearly stated. Are you looking for an exact 2 consecutive heads, (i.e. 3 consecs would not be a success) and are you looking for exclusive OR (i.e. HHTT in your 4 tosses would again not be a success). Be clear about the experiment and what is defined as a 'success' and what is not. One reason I ask is because you gave the chance of 2 heads to be 3/16, and the chance of two tails the same. But how do you know you are not counting double here when you sum the 2 probs? Whether you can do that or not depends on the problem.

Also your proof is...? I'm not dismissing it as wrong, I'm just naturally suspicious and I like to go through things myself.


Also

Basic probability suggests we should expect a 50/50 split, yet you recognise even after an arbitrary number of coin tosses, there will be a discrepancy between what basic probability suggests we should expect and what we recognise as more likely to occur in reality.

The entire nub of this thread.

I have to take issue here, because you are being loose with your words. Basic probability does NOT suggest it will be a 50/50 split, and maybe you used that word on purpose to highlight it. The problem is that people think it suggests things like this. Or suggests the dreaded 'law of averages'. It gives a distribution and that is all.

Also on tossing a coin 100 times I think in this case you'll also find a lot of people that are the opposite. On a fair coin toss there will be people that thing a 75-25 is quite likely. When it fact it is highly improbable.

 

[TheBramble]

1,1,2,3,5,8
My 6th Fib is 8 (or even 5 if you start at 0). But ok if you want to define the Fibs as you have, I won't be picky, that is just custom. As long as the formula uses the same custom as you

Good to get some quality feedback once again.

You’re correct, my Fib sequence was wrong. There seems to be a division between those who start with a zero and those with a 1 and as the majority of sources start it with a zero, we’ll do the same. Good spot.

1,1,2,3,5,8
I am still concerned with what you're actually asking. It is my experience that most confusion arises when the problem isn't clearly stated. Are you looking for an exact 2 consecutive heads, (i.e. 3 consecs would not be a success) and are you looking for exclusive OR (i.e. HHTT in your 4 tosses would again not be a success). Be clear about the experiment and what is defined as a 'success' and what is not. One reason I ask is because you gave the chance of 2 heads to be 3/16, and the chance of two tails the same. But how do you know you are not counting double here when you sum the 2 probs? Whether you can do that or not depends on the problem.

Confusion? Clearly stated? I am asking what is the probability within a 4 coin toss of getting two consecutive heads or two consecutive tails.

Counting double. Uh? What’s the probability of tossing a Head with a single toss of a fair coin? 1:2. What’s the probability of tossing a Tail with a single toss of a fair coin? 1:2. What’s the probability of tossing a Head or a Tail with a single toss of a fair coin? 1:1 of course. Double counting doesn’t come into it. The probability of event-A OR event-B is the sum of Event-A and event-B.

1,1,2,3,5,8
Also your proof is...? I'm not dismissing it as wrong, I'm just naturally suspicious and I like to go through things myself.

Proof of what? The formula using the Fib sequence? I’ve cited my reference source in the post where I introduced it.

1,1,2,3,5,8
Also
I have to take issue here, because you are being loose with your words. Basic probability does NOT suggest it will be a 50/50 split, and maybe you used that word on purpose to highlight it. The problem is that people think it suggests things like this. Or suggests the dreaded 'law of averages'. It gives a distribution and that is all.

Loose in what respect? It absolutely does make that the outcome. If you push the binary outcome of either H or T and the number 100 through any standard probability formula you will come out with a 1:2 and a 50/50. If you can find a basic probability formula that yields anything other than result from those parameters you need to post it – with the appropriate references of course. LOL.

1,1,2,3,5,8
Also on tossing a coin 100 times I think in this case you'll also find a lot of people that are the opposite. On a fair coin toss there will be people that thing a 75-25 is quite likely. When it fact it is highly improbable.

That’s the point. I wouldn’t argue with those people, but they are not quoting standard probability are they. And their guess is a lot less improbable than basic probability theory allows for. My whole point.

 

[Calinor]

the probability of two consecutive Heads OR 2 consecutive Tails the result is a Probabilistic OR which yields 0.375 (6/16).

Ok, so if your Fib sequence was wrong with respect to the formula, then was your answer wrong, yes? I'm not sure you even mean what you wrote above. The probability of getting either 2 heads or 2 tails consecutively is pretty high theBramble (do a simulation and get an approximation).

The reason I asked for clarification on the question is that there are two separate problems. There is the question where you toss the coin 4 times, you have the 16 outcomes, each equally likely, and you can get the probability as Gecko did of there being two heads or two tails. 7/8 was correct for that.

There is a second problem dealing with runs/sequences. And that looks at a sequence of heads and tails (imagine an infinite string of them if it helps). You then focus in on a group of 4 and ask the question what is the probability of having two heads or two tails. For that you have a formula like the one you quoted. Why do you assume this is the same question Bramble?

The probability of event-A OR event-B is the sum of Event-A and event-B.

This is certainly NOT true in general. It is only true for disjoint events.

Loose in what respect? It absolutely does make that the outcome. If you push the binary outcome of either H or T and the number 100 through any standard probability formula you will come out with a 1:2 and a 50/50. If you can find a basic probability formula that yields anything other than result from those parameters you need to post it – with the appropriate references of course. LOL.

You've misunderstood what I am saying. I am not saying the mean won't be 50:50, I'm saying probability doesn't tell you what the outcome will be. That is an important distinction, getting 99 heads and 1 tail is an outcome. If I have tossed a coin 99 times and it is 49:50, probability does not tell me it will be 50:50 after the next coin toss. It is not saying what the outcome is. It is giving you a distribution, and that's all. From that you can have a confidence interval. But what is the probability of you actually getting 50 tails and 50 heads from 100 coin tosses? Is probability telling us to be confident of that outcome? How confident? Which is why i commmented. For probability to suggest something to us about an outcome, then you would want the probability to be pretty high of that outcome or event.

That’s the point. I wouldn’t argue with those people, but they are not quoting standard probability are they. And their guess is a lot less improbable than basic probability theory allows for. My whole point.

You haven't made any sense here. When you say it is a lot less probable than probability allows for, it is self-contradictory. Probability allows for a lot, but it is a model for the real world. It is up to the individual to decide whether that is a good model or a bad one. But that doesn't mean that probability doesn't allow an accurate model.

 

[TheBramble]

Yes, my answer was wrong.

What's the probability of getting 2 consecutive Heads in 4 tosses of a coin?

 

[Calinor]

0.5

 

[TheBramble]

What's the probability of getting 2 consecutive Tails in 4 tosses of a coin?

 

[TheBramble]

The reason I asked for clarification on the question is that there are two separate problems. There is the question where you toss the coin 4 times, you have the 16 outcomes, each equally likely, and you can get the probability as Gecko did of there being two heads or two tails. 7/8 was correct for that.

There is a second problem dealing with runs/sequences. And that looks at a sequence of heads and tails (imagine an infinite string of them if it helps). You then focus in on a group of 4 and ask the question what is the probability of having two heads or two tails. For that you have a formula like the one you quoted. Why do you assume this is the same question Bramble?

Just realised what you've said. Yes. They are one and the same question.

An infinite string of random (LOL) Heads and Tails taken ANY four at a time is exactly the same as any discrret set of 4 coin tosses.

If you don't agree then I need proof of why this isn't the case.