Basic Probability

This is a discussion on Basic Probability within the The Foyer forums, part of the Off the Grid category; Dcraig kind of hit on what I’m driving at here. In any series of 4 coin flips, while the probability ...

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Old Mar 27, 2009, 12:18pm   #16
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TheBramble started this thread 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?
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Old Mar 27, 2009, 12:25pm   #17
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Tony,

Try coming at it from a different angle and determine what the chance is of not getting two heads. I know that Scripophilist is an expert in this area and he has used that approach that often gives surprising answers to probabilities.

One question that may have been posted here before but had a curios answer was:
How may people would you need to have in one room for the chances being even that two people share the same birthday ?


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Old Mar 27, 2009, 12:38pm   #18
 
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Probability Theory

is a site created by an old friend and contains interesting stuff.
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Old Mar 27, 2009, 12:40pm   #19
 
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Ah !
Just seen Paul's post.
The author of the aforementioned site is indeed Scripophilist.
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Old Mar 27, 2009, 12:52pm   #20
 
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Quote:
Originally Posted by TheBramble View Post
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?
ok. still on first page, and I am now confused.

if you flip a coin 4 times, the permutations of consecutive Heads positions are: (1,2), (2,3) or (3,4).
the 4 heads in a row option gives you the above all in one go, so, say 3 again,
total 6.
but, strictly speaking, only 4 of those throws results in a 2-consec result.

dont know where you get the 12 possibilities from.

from dcraigs .75; you ask how many times would you need to throw to get a Head. If you get a Head on the first throw, you stop throwing.
You only flip the coin again if the first wasnt a Head. this negates one of dcraigs options, so we are back to 0.5.

EDIT: writing before thinking. yes, the possibility of (1,2) and (2,3) also exists, so HHHT shows 2 hits. so, yes maybe 12. havent checked it.
I was assuming finishing once you got your consec.

EDIT2: yes, there are 12 instances of consecs. (Pascals Triangle)
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Old Mar 27, 2009, 1:59pm   #21
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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.
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Old Mar 27, 2009, 2:07pm   #22
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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.

Last edited by TheHoneyMonster; Mar 27, 2009 at 9:44pm.
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Old Mar 27, 2009, 2:08pm   #23
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TheBramble started this thread 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.
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Old Mar 27, 2009, 2:13pm   #24
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Quote:
Originally Posted by TheBramble View Post
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.
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Old Mar 27, 2009, 4:39pm   #25
 
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Quote:
Originally Posted by TheBramble View Post
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.
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Old Mar 27, 2009, 4:40pm   #26
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TheBramble started this thread
Quote:
Originally Posted by MrGecko View Post
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?
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Old Mar 27, 2009, 6:00pm   #27
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Jesus, what a ****ing waste of time.
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Old Mar 27, 2009, 6:28pm   #28
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factorials innit
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Old Mar 27, 2009, 6:55pm   #29
 
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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.
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Old Mar 27, 2009, 7:09pm   #30
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TheBramble started this thread
Quote:
Originally Posted by Hotch View Post
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?”

Quote:
Originally Posted by Hotch View Post
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.

Quote:
Originally Posted by Hotch View Post
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?

Quote:
Originally Posted by Hotch View Post
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?
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