Basic Probability

[TheHoneyMonster]

A coin has 2 sides to it.

Every time it is tossed, is like dipping your hand into a bag and being asked to pull out the red Ball or the blue Ball.

Two tosses, requires 2 reds and 2 blues. Dipping into the bag for 2 balls presents the following chances -

2 reds
2 blues
1 red 1 blue

So theres a chance you will get 100% of one, or 0% of it. Statistically, the average is a 50% chance for picking out a red or a blue.

 

[TheHoneyMonster]

Siblings share 50% of their genes on average.

However, the actual relatedness could theoretically range from 0% to 100%.

Children inherit 50% from their mother & 50% from their father.

Think of your mothers genes being the red cards in a deck, and your father's the black cards.

A child is asked to pick out 26 cards from the combined 52 red/black.

They could pick out all the same ones of their sibling, or none of the same as their sibling. But the likelyhood is they will on average select 50% of the same "cards" (genes) as their sibling - hence 50% shared genes by siblings.

 

[TheBramble]

TTTT
TTTH
TTHT
TTHH
THTT
THTH
THHT
THHH
HTTT
HTTH
HTHT
HTHH
HHTT
HHTH
HHHT
HHHH

These are the 16 permutations possible from a 4 coin toss.

Count the number of consecutive 'TT's (or 'HH's for that matter).

Bear in mind the string 'TTTT' has 3 'TT' strings in it.

 

[MrGecko]

Dcraig kind of hit on what I’m driving at here.

In any series of 4 coin flips, while the probability of getting 2 consecutive heads is indeed 1 in 4 and the probability of getting 2 consecutive tails is 1 in 4, statistically, the probability of getting two consecutive tails OR two consecutive heads is 1 in 2.

But of the 16 possible permutations from a 4 coin flip, there are 12 in 16 occurrences of two consecutive Heads, and 12 in 16 occurrences of two consecutive Tails. On that basis, the probability of two consecutive Heads in a 4 coin toss exercise is 3 in 4. Same for two consecutive Tails.

So if two consecutive tails has a 3 in 4 and two consecutive heads has a 3 in 4, what it the probability of two consecutive heads OR two consecutive tails?

Haven't really got time to write a detailed response, but my hunch is that you should investigate:

a) Permutations and Combinations
b) Central Limit Theorem

with the emphasis on a). My suspicion is that as you increase your sample size, the number of permutations / combinations rises exponentially (factorially), which is where you run into trouble.

In your example, you increase the permutations exponentially, then translate this into a liner sample space.

Just my 1st thoughts.

 

[blackcab]

Dcraig kind of hit on what I’m driving at here.

In any series of 4 coin flips, while the probability of getting 2 consecutive heads is indeed 1 in 4 and the probability of getting 2 consecutive tails is 1 in 4, statistically, the probability of getting two consecutive tails OR two consecutive heads is 1 in 2.

But of the 16 possible permutations from a 4 coin flip, there are 12 in 16 occurrences of two consecutive Heads, and 12 in 16 occurrences of two consecutive Tails. On that basis, the probability of two consecutive Heads in a 4 coin toss exercise is 3 in 4. Same for two consecutive Tails.

So if two consecutive tails has a 3 in 4 and two consecutive heads has a 3 in 4, what it the probability of two consecutive heads OR two consecutive tails?

It's equivalent to expressing a random integer 0-15 in binary (0000...1111). The no. of sequences that do not contain 00 or 11 is small:
Starting with 0 you have to follow with 1, then 0 then 1: 0101
Starting with 1 you have to follow with 0, then 1 then 0: 1010
So only 2 of the 16 sequences do not contain 00 or 11, therefore you'd expect 14/16 (87.5%) of 4-coin flips to contain at least one instance of consecutive heads or consecutive tails.

 

[TheBramble]

Haven't really got time to write a detailed response

I don’t really have time to read a detailed response so that works out quite nicely.

Nothing to do with Combinations. Where the sequence is key, it is Permutations that interest us. And while coin toss data would be expected to be Gaussian in nature (and thus a potential candidate) the number of data we are discussing here falls outside the scope of CLT.

Post #16 was the sleight of hand.

Without any challenge, I segued from getting ‘two consecutive heads is a 1 in 4 probability’ to ‘in any series of 4 coin flips’.

Of course, there’s no direct relationship between a ‘1 in 4’ probability and 4 coin flips. The 1 in 4 probability was the result of 2 coin flips. With 4 coin flips you have 3 chances of a 1 in 4. Which, when using the additive property of probabilistic OR you get 3 in 4 which is the 12 in 16 I referred to in that same post.

So, given you have 4 coin flips you indeed do have a 3 in 4 chance of getting 2 consecutive Heads. You also have a 3 in 4 chance of getting 2 consecutive Tails. So what’s the probability of you getting two consecutive Heads or two consecutive Tails with 4 flips of the coin?

 

[MrGecko]

Jesus, what a ****ing waste of time.

 

[Joey25]

factorials innit

 

[Hotch]

What exactly is the question?

Bramble, when you came up with 12 chances of getting 2 consecutive heads you were wrong.

HHHH still counts as only one consecutive HH. This is because HHHH is ONE outcome of the 16.

The chance of getting a HH is still 1/2, even though there are 12 occurrences of HH in the sequences.

Let's take a football match, if you are betting on team A winning, they can win 1-0, 2-0, 37-5, but they've still won, it is all encompassed within "winning".

Hope that example helps.

Maths isn't broken.

 

[TheBramble]

What exactly is the question?

The current question is:-

“what’s the probability of you getting two consecutive Heads or two consecutive Tails with 4 flips of the coin?”

Bramble, when you came up with 12 chances of getting 2 consecutive heads you were wrong.

HHHH still counts as only one consecutive HH. This is because HHHH is ONE outcome of the 16.

Of course it doesn’t. You commit to making 4 flips of the coin. It gives you ‘HHHH’. There are 3 sets of consecutive ‘HH’ strings in there.

The chance of getting a HH is still 1/2, even though there are 12 occurrences of HH in the sequences.

What? Chance of getting ‘HH’ in what?

Let's take a football match, if you are betting on team A winning, they can win 1-0, 2-0, 37-5, but they've still won, it is all encompassed within "winning".

For goodness sake. You went off on a bloody tangent with the 747 as well. LOL.

OK, as you’re keen to discuss in an area you prefer…

How many sets of two consecutive wins in a row has your team A had using your data?

 

[Hotch]

7/8 Is the answer.

The point I was trying to make, which you missed, as usual, is that the question you are asking is basicly "count how many of these sets of 4 coin flips have a consecutive head or tails in them, and divide by 16", what you are doing, is "count the number of consecutive head or tails in them and count by 16". These are different things.

Mathematics still works, you're a fool. The end.

 

[TheBramble]

7/8 Is the answer.

To what?

Your hat size?

 

[MrGecko]

Really, what is the point?

P(Two consecutive heads or tails in a series of 4 flips)

=[1 - P(you don't get two consecutive heads or tails)]

= [1 - P(alternating heads and tails)]

= 1 - [P(THTH) + P(HTHT)]
= 1 - [2 * 1/(2^4)]
= 7/8

 

[MrGecko]

For my sins I am going to say that the question needs clarification; either you are asking the probability of:

a) EXACTLY TWO consecutive H's or T's
b) AT LEAST TWO consecutive H's or T's
c) ANY TWO consecutive events in the series of four being the same

But really, what is the point of asking a purposefully simple question, and using it as bait to intentionally mislead, use "smoke and mirrors" to confuse, and couple it with an arrogant narrative?

It must play havoc with your PnL, having to prove that you are right all the time. Particularly if you need to decieve those around you in order for it to be possible.

Can't you compete?

 

[TheBramble]

Mr. G, try and keep the response on topic. It’s nothing to do with my trading, it’s a question on basic probability. And try and keep it civilised. It’s not my fault you’re confused. That’s the whole point of this thread. Confusion often precedes learning.

My point was ands still is it seems, to examine the elementary blunders most make with even the most basic of probability questions. 'Purposefully simple' indeed...LOL

I asked what was the probability of two consecutive heads being tossed in a two coin toss sequence. 1 in 4 came the response. Statistically correct.

I then asked what was the probability of two consecutive heads being tossed in a 4 coin toss sequence.

I then asked what was the probability of two consecutive heads or two consecutive tails being tossed in a 4 coin sequence.

So far, for all your bluster, neither you nor anyone else has actually come up with the correct answer?

Hint: Look at the 16 Permutations I provided in my earlier post .

 

[TheBramble]

The underlying purpose of this thread is to examine the psychology of defence. I am primarily interested in the mechanisms that are employed to defend established dogma.

Not the longer term members who know me well enough to have seen me coming, but there’s a new crop of ‘stars’ out there who have yet to be put through the mill and whose insights and fresh angles could be potentially significant. Even more so, on a personal level, their approaches and styles of response.

Although I appreciated the majority, the vast majority would not for a second consider the possibility that a review of any nature on this topic would be appropriate or necessary, there are, as always, two camps.

Of those who felt it was beyond question, few, if any, would offer a logical basis of proof for supporting existing beliefs other than to simply regurgitate the self same laws which are apparently being scrutinised. Although it was to be expected that the majority would be those resistant to any form of challenge, I am fairly sure the few who did stop to consider the question afresh, however briefly, would do so in anonymity and would be unlikely to stick their head above the parapet to disclose their momentary heresy. There is a visceral biological imperative to staying small and keeping quiet in the early stages of any major shift.

Of those who considered my thinly disguised intentions (post #16) impertinence, there would be the usual crop of blowhards who would feel it was an attack on them personally and respond in kind. I could have nailed their names to the mast ahead of time. LOL. You weren’t targeted, you arrived here of your own free will and you participated in your own way. No blame.

I’m not sure whether it is the sense of being personally attacked (questioning current knowledge and beliefs which, for all of us, are potentially traumatic) or of having an external belief (that which we believe ‘everyone’ believes) brought into question. Which do we defend most strongly?

It’s a measure of intellectual development the extent to which we are not only willing, but eager to have the opportunity to robustly review that which we currently hold to be Fact or Truth – however that opportunity presents itself to us. The more solid, long-standing, obvious or sacrosanct the belief the more powerfully should we seize these opportunities to re-examine the fundamentals upon which they are thought to be based, the proofs and processes.

For those still up for it, stick the 16 Permutations in a word processing document and string count the occurrences of ‘HH’ and forget for a moment you know anything about probability.

Of 4 coin flips, what is the probability you will encounter a string of two consecutive Heads?

 

[trendie]

I think you have established that there are 12 consec Heads (and Tails).
I think that sometimes we misread the intention of a question, and start answering it before we reach the end of the question.

The effort to manually go through a small exmaple, and then deduce a general property is an exercise conducted by those who are bored, or love puzzles.
(you may find puzzle-addicts are more open-minded to new ideas, as its the "Aha!" moment, or the discovery that is the prize, not re-inforcement of pre-concieved ideas.)

Of the 16 perms;
4 Hs. (HHHH) positions (1,2), (2,3) (3,4) = 3
3 Hs. positions (123, 124, 134, 234) (12, 23, 12, 23, 34, 34) = 6
2 Hs. positions (12, 13, 14, 23, 24, 34) (12,23,34) = 3
1 Hs. positions (1,2,3,4) = no consecs.

Where do we go from here?

 

[TheBramble]

you may find puzzle-addicts are more open-minded to new ideas, as its the "Aha!" moment, or the discovery that is the prize, not re-inforcement of pre-concieved ideas.

Ah. Nail on the Head. Not inappropriately either for a student of Maslow…

Not sure if it’s that evenly divided between just two camps, but you may be right.

I think you have established that there are 12 consec Heads (and Tails)
Where do we go from here?

OK, bearing that figure in mind and the formal statistical probability for the same, 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?

 

[TheBramble]

OK. Enough is enough.

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).

These are ‘real life’ probabilities. Ones that actually occur. The ones YOU WILL experience regardless of what the textbooks say. These apply to us, to our trades and our trading performance statistics and all the other ratios associated with trading performance over time.

That’s why I kept referring you to the 16 Permutations in my earlier post. They are the least likely set of strings you’ll ever find in 16 sets of 4 coin toss results. They are idealised outputs from a Gaussian Universe that will never be encountered in any reality you inhabit, accidentally or otherwise. Yet they look ‘correct’. That’s what you’d expect from elementary probability theory – and you come to expect it and feel comfortable with it. There’s the real danger.

If you carry on trotting out, or even worse believing, and even worse still, financially depending upon all the other textbook stuff on Probability which applies to a Gaussian Universe you will potentially come a cropper. Just like LTCM did. The real Universe of Coin tosses (and trading) has fat tails AND skew.

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.









Gardner, M. "Nontransitive Paradoxes." Time Travel and Other Mathematical Bewilderments. New York: W. H. Freeman, pp. 64-66, 1988.

Berlekamp, E. R.; Conway, J. H; and Guy, R. K. Winning Ways for Your Mathematical Plays, Vol. 1: Adding Games, 2nd ed. Wellesley, MA: A K Peters, p. 777, 2001.

 

[trendie]

TTTT
TTTH
TTHT
TTHH
THTT
THTH
THHT
THHH
HTTT
HTTH
HTHT
HTHH
HHTT
HHTH
HHHT
HHHH

I get 8 out of 16 will give 2 consec heads!

EDIT: actually, with the exception of HTHT and THTH, ALL other options will give at least one consec, be it Heads or Tails.
So, 2/16 will not give a consec, thus 14/16 options will give a consec Head or Tail.