# heads or tails ........

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#### Deleted member 115935

Joined Dec 31, 1969
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#### Deleted member 115935

Joined Dec 31, 1969
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The solution to pollution is dilution.
Thank you @wayneh,
may be you could expand on your theory please

#### bogosort

Joined Sep 24, 2011
696
so for 999 tosses if were not at 50:50
then by 100000 tosses we should be closer to 50:50

Hence the question, if were at 999 tosses, and have already 490 H,
to get nearer to 50:50 by 100000 tosses, must there be more H than T ?
The answer to the bolded question is no. After 999 trials, we have 490/999 = 49.05% heads. Suppose that we perform another 10,000 flips and observe 5,000 heads and 5,000 tails. We've added the same amount of H and T, yet the total heads percentage has increased: (490 + 5000) / (999 + 10000) = 49.91% heads.

If we keep adding a perfectly fifty-fifty split of heads and tails, the heads percentage will increasingly and monotonically approach 50%, regardless of how big a "head start" the tails had. This is the law of large numbers at work: any random difference from the expected value gets swamped in the long term. A run of a million consecutive tails is utterly insignificant to the heads percentage in 10^1000 trials. A run of 10^1000 tails is utterly insignificant in 10^(10^1000) trials. And so on.

#### 402DF855

Joined Feb 9, 2013
271
I think the answer is along the lines of this: If at some point T=H+10, it is likely (certain?) a subsequent series will have H=T+10.

#### Wendy

Joined Mar 24, 2008
23,477
My take is if it isn't 50/50 then there is something biasing the system you haven't accounted for.

#### joeyd999

Joined Jun 6, 2011
5,410
My take is if it isn't 50/50 then there is something biasing the system you haven't accounted for.
So, if I flip two heads in a row, my coin is biased?

#### BobTPH

Joined Jun 5, 2013
9,278
By the way, your conjecture is known as “the gambler’s fallacy.”

Bob

#### wayneh

Joined Sep 9, 2010
17,505
Thank you @wayneh,
may be you could expand on your theory please
Anomalies disappear with more data. If your sample includes the entire population, sampling error disappears.

#### bogosort

Joined Sep 24, 2011
696
Anomalies disappear with more data. If your sample includes the entire population, sampling error disappears.
What's the population of coin flips? I'm not sure the question is meaningful, though perhaps the answer exposes whether one has a frequentist interpretation of probability or a Bayesian interpretation.

I suppose a frequentist might say, "The coin, whether fair or not, models some distribution, and the underlying distribution is the population. Coin flips are samples, and if we could flip it forever we would find the exact distribution."

I suppose a Bayesian might say, "Flipping a coin is not sampling from a population, there are no anomalies. My prior is that the coin is fair; if I flip it a few times and see a run, I'll update my prior accordingly."

#### wayneh

Joined Sep 9, 2010
17,505
What's the population of coin flips? I'm not sure the question is meaningful...
Right, it's not. But suppose you wanted to estimate the average height of females in your town. You could, in theory, measure every one of them and eliminate sampling error. In reality you'd probably stop after 100 or so, once the mean and standard deviation suggest you have enough.

#### joeyd999

Joined Jun 6, 2011
5,410
What's the population of coin flips?
More importantly: how many are left?

#### Deleted member 115935

Joined Dec 31, 1969
0
The answer to the bolded question is no. After 999 trials, we have 490/999 = 49.05% heads. Suppose that we perform another 10,000 flips and observe 5,000 heads and 5,000 tails. We've added the same amount of H and T, yet the total heads percentage has increased: (490 + 5000) / (999 + 10000) = 49.91% heads.

If we keep adding a perfectly fifty-fifty split of heads and tails, the heads percentage will increasingly and monotonically approach 50%, regardless of how big a "head start" the tails had. This is the law of large numbers at work: any random difference from the expected value gets swamped in the long term. A run of a million consecutive tails is utterly insignificant to the heads percentage in 10^1000 trials. A run of 10^1000 tails is utterly insignificant in 10^(10^1000) trials. And so on.
@bogosort

Thank you so much
this is exactly the sort of answer that was hopping for,
I can understand that as an idea,

Its exactly the great kind of response I was hoping for,
when ever I have asked around this question
we seem to get the almost rude answers of

a) the coin has no memory ( Dah, did say that at the beginning, are they implying Im stupid enough to think the coin doe shave memory )

b) Its the "gamblers fallacy", I highlighted the two knowns at the beginning, each toss is 50:50

Its amazing over the decades, this has cropped up in lectures every now and then,
I used to give the bland answer 50:50 or gamblers theory, but now days I get at mathematician in to cover it ,
and they tend to use the same 50:50 or gamblers theory answer. !!!!

Its been great to have a proper conversation,
not the many one line answers that are so easy to recant with no understanding.

Thank you again,

#### Deleted member 115935

Joined Dec 31, 1969
0
So, if I flip two heads in a row, my coin is biased?
@joeyd999
thank you for your one line answer
not certain what it adds to the conversation though ,

#### Deleted member 115935

Joined Dec 31, 1969
0
My take is if it isn't 50/50 then there is something biasing the system you haven't accounted for.
Thank you @Wendy

This is a thought experiment,
Even so , with a 50:50
there is always going to be a probability of say 40 heads in a row,
not much of one, but its what one would expect,

no need for a bias in the system,

#### Deleted member 115935

Joined Dec 31, 1969
0
You can request that this thread be closed but do we really need to start another thread on "heads or tails"?
@MrChips
I was more saying that this thread was about coin tossing
And as I said at the start and a few times since,
I do not want this thread to get into a general chat

and that if people want to have a general chat about the mods they should may be consider opening a new thread.

#### visionofast

Joined Oct 17, 2018
106
consider that most of the nature's phenomenons intrinsically observe Normal Distribution of statistics.maybe due to harmony and isochronism...
so,as far as your experiment's source is belong to the Nature you can expect 50-50 in total results.

#### MrChips

Joined Oct 2, 2009
31,087
And has been repeatedly stated, the probability of heads in the flip of a coin is 0.5. End of discussion.
Any deviation from 50:50 split on n-trials is explainable by statistical variance.

#### Deleted member 115935

Joined Dec 31, 1969
0
And has been repeatedly stated, the probability of heads in the flip of a coin is 0.5. End of discussion.
Any deviation from 50:50 split on n-trials is explainable by statistical variance.
Thank you @MrChips

I think you have given that one line answer to the question you think is asked,
Still not certain what it adds, may be you could refer back to original post, and add a few more lines of answer )

but thank you for your contribution,

#### Deleted member 115935

Joined Dec 31, 1969
0
consider that most of the nature's phenomenons intrinsically observe Normal Distribution of statistics.maybe due to harmony and isochronism...
so,as far as your experiment's source is belong to the Nature you can expect 50-50 in total results.
Thank you @visionofast

The original thought experiment is outlined in the first post, where we state that each toss has a 50% probability of being a head.

The question was regarding the apparent ambiguity,
when after a large number of tosses (A) , you have more of say H than T,
We agree the odds on the next toss is 50:50,
and we agree that after a sufficiently large number of tosses (B), then there will be the same number of H as T.
but at any time you will have more H than T
and the apparent argument that thus in the next number of tosses ( B - A )
there must be more of one than the other.

The argument we now have,
is that its all down to percentages.

Between A and B , there could be the same number of H and T, but at B, the number of H and T will be more equal due to the difference still being the but the number of samples being more,

No one has come back on this , and I think its a good enough answer.

unless you know a more succinct answer

thank you

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