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Guest

Tue Jan 22, 2019 5:45 pm

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results â€“ say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails. So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH". Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have. "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way? I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

Guest

Tue Jan 22, 2019 5:45 pm

On Tuesday, January 22, 2019 at 11:21:31 AM UTC-5, Jamie M wrote:

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results â€“ say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results â€“ say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

Not correct. The probability of those two sequences are exactly the same. Any combination of four tosses has exactly a 1 in 16 probability and will occur with the same frequency. To test this you would need to do 20 coin tosses a fair number of times, say 10 times or even more. So I'm not doing it. This can't really be automated since computer code isn't really random..

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.

The nature of probability in these situations is based on the fact that any one coin toss does NOT depend on the previous coin tosses.

So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH". Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have. "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH". Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have. "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

These are all different in the eyes of probability. If you want to say instead "three heads and one tails" then that is very different and will be VERY much more likely than all heads.

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way? I have always

just assumed a coin flip is 50/50, but too lazy to check.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way? I have always

just assumed a coin flip is 50/50, but too lazy to check.

I don't know what you mean by "this way". The probability of a coin coming up heads or tails is always 50/50. No? How would the previous results impact how you flip the coin??? Is there a string tied to the coin? Does the coin get sticky on one side? Spooky action at a distance?

Rick C.

- Get 6 months of free supercharging

- Tesla referral code - https://ts.la/richard11209

Guest

Tue Jan 22, 2019 5:45 pm

On 1/22/19 11:21 AM, Jamie M wrote:

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results â€“ say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.Â So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH".Â Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have.Â "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way?Â I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results â€“ say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.Â So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH".Â Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have.Â "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way?Â I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

Psychologists are notoriously hopeless at statistics.

Cheers

Phil Hobbs

Guest

Wed Jan 23, 2019 2:45 am

On Tue, 22 Jan 2019 08:21:26 -0800, Jamie M <jmorken_at_shaw.ca> wrote:

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results – say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails. So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH". Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have. "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way? I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results – say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails. So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH". Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have. "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way? I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

Dosen't viewing the outcome of a coin-toss affect the outcome? ;)

Cheers

Guest

Wed Jan 23, 2019 3:45 am

Martin Riddle <martin_ridd_at_verizon.net> wrote in

news:2fhf4e5ehq2i9kqpc3hs1k2df0c6ouq4j0_at_4ax.com:

Dosen't viewing the outcome of a coin-toss affect the outcome? ;)

Cheers

Cheers

No.

Maybe... IF you had flapping butterfly wings on at the time.

Here is the stat... The outcome of a coin toss is 50/50.

The outcome of ANY subsequent toss will have exactly nothing to do

with the result of ANY previous toss, and is exactly 50/50.

So 100 tosses could result in 100 events with the same outcome.

Now, the PROBABILITY of that happening is the question when one is

examining a set or group of like events.

Guest

Wed Jan 23, 2019 5:45 am

On Tuesday, January 22, 2019 at 8:40:30 PM UTC-5, Martin Riddle wrote:

On Tue, 22 Jan 2019 08:21:26 -0800, Jamie M <jmorken_at_shaw.ca> wrote:

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results â€“ say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails. So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH". Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have. "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way? I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

Dosen't viewing the outcome of a coin-toss affect the outcome?

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results â€“ say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails. So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH". Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have. "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way? I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

Dosen't viewing the outcome of a coin-toss affect the outcome?

Yes in that it resolves the outcome to a single state rather than a superposition of all possible states.

Rick C.

+ Get 6 months of free supercharging

+ Tesla referral code - https://ts.la/richard11209

Guest

Wed Jan 23, 2019 7:45 am

On Wed, 23 Jan 2019 02:37:00 +0000 (UTC),

DecadentLinuxUserNumeroUno_at_decadence.org wrote:

Martin Riddle <martin_ridd_at_verizon.net> wrote in

news:2fhf4e5ehq2i9kqpc3hs1k2df0c6ouq4j0_at_4ax.com:

Dosen't viewing the outcome of a coin-toss affect the outcome? ;)

Cheers

No.

Maybe... IF you had flapping butterfly wings on at the time.

Here is the stat... The outcome of a coin toss is 50/50.

The outcome of ANY subsequent toss will have exactly nothing to do

with the result of ANY previous toss, and is exactly 50/50.

So 100 tosses could result in 100 events with the same outcome.

Now, the PROBABILITY of that happening is the question when one is

examining a set or group of like events.

news:2fhf4e5ehq2i9kqpc3hs1k2df0c6ouq4j0_at_4ax.com:

Dosen't viewing the outcome of a coin-toss affect the outcome? ;)

Cheers

No.

Maybe... IF you had flapping butterfly wings on at the time.

Here is the stat... The outcome of a coin toss is 50/50.

The outcome of ANY subsequent toss will have exactly nothing to do

with the result of ANY previous toss, and is exactly 50/50.

So 100 tosses could result in 100 events with the same outcome.

Now, the PROBABILITY of that happening is the question when one is

examining a set or group of like events.

I can imagine a small correllation between successive tosses.

--

John Larkin Highland Technology, Inc trk

jlarkin att highlandtechnology dott com

http://www.highlandtechnology.com

Guest

Wed Jan 23, 2019 8:45 am

Jamie M wrote:

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results â€“ say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.Â So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH".Â Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have.Â "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way?Â I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results â€“ say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.Â So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH".Â Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have.Â "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way?Â I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

The 50/50 probability of heads(tails) is not exactly true, and

depends on the coin (assuming no outside interference like magnetic

fields, air movement, etc).

The design of the image heads side must balance with the design of

the image tails side, for an "accurate" 50/50 result.

Guest

Wed Jan 23, 2019 10:45 am

On 23/01/2019 07:01, John Larkin wrote:

On Wed, 23 Jan 2019 02:37:00 +0000 (UTC),

DecadentLinuxUserNumeroUno_at_decadence.org wrote:

Martin Riddle <martin_ridd_at_verizon.net> wrote in

news:2fhf4e5ehq2i9kqpc3hs1k2df0c6ouq4j0_at_4ax.com:

Dosen't viewing the outcome of a coin-toss affect the outcome? ;)

Cheers

No.

Maybe... IF you had flapping butterfly wings on at the time.

Here is the stat... The outcome of a coin toss is 50/50.

The outcome of ANY subsequent toss will have exactly nothing to do

with the result of ANY previous toss, and is exactly 50/50.

So 100 tosses could result in 100 events with the same outcome.

Now, the PROBABILITY of that happening is the question when one is

examining a set or group of like events.

I can imagine a small correllation between successive tosses.

DecadentLinuxUserNumeroUno_at_decadence.org wrote:

Martin Riddle <martin_ridd_at_verizon.net> wrote in

news:2fhf4e5ehq2i9kqpc3hs1k2df0c6ouq4j0_at_4ax.com:

Dosen't viewing the outcome of a coin-toss affect the outcome? ;)

Cheers

No.

Maybe... IF you had flapping butterfly wings on at the time.

Here is the stat... The outcome of a coin toss is 50/50.

The outcome of ANY subsequent toss will have exactly nothing to do

with the result of ANY previous toss, and is exactly 50/50.

So 100 tosses could result in 100 events with the same outcome.

Now, the PROBABILITY of that happening is the question when one is

examining a set or group of like events.

I can imagine a small correllation between successive tosses.

That would only be if it were not a fair toss, or not a fair coin.

In reality, of course, few things are /really/ fair. Once you have done

your first hundred tosses, you might have a certain pattern of how you

pick up the coin and put it on your fingers, and how hard you toss it -

that could certainly give a slight correlation. And the coin itself

might not be perfectly balanced, giving a slight balance.

But usually these effects are too small to be noticed.

Guest

Wed Jan 23, 2019 10:45 am

In article <2fhf4e5ehq2i9kqpc3hs1k2df0c6ouq4j0_at_4ax.com>,

martin_ridd_at_verizon.net says...

On Tue, 22 Jan 2019 08:21:26 -0800, Jamie M <jmorken_at_shaw.ca> wrote:

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results ? say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails. So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH". Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have. "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way? I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

Dosen't viewing the outcome of a coin-toss affect the outcome? ;)

Cheers

That must depend on the mental ability of the viewer (or perhaps their

dose!)...

Mike.

Guest

Wed Jan 23, 2019 10:45 am

On 23/01/2019 06:01, John Larkin wrote:

On Wed, 23 Jan 2019 02:37:00 +0000 (UTC),

DecadentLinuxUserNumeroUno_at_decadence.org wrote:

Martin Riddle <martin_ridd_at_verizon.net> wrote in

news:2fhf4e5ehq2i9kqpc3hs1k2df0c6ouq4j0_at_4ax.com:

Dosen't viewing the outcome of a coin-toss affect the outcome? ;)

Cheers

No.

Maybe... IF you had flapping butterfly wings on at the time.

Here is the stat... The outcome of a coin toss is 50/50.

The outcome of ANY subsequent toss will have exactly nothing to do

with the result of ANY previous toss, and is exactly 50/50.

So 100 tosses could result in 100 events with the same outcome.

Now, the PROBABILITY of that happening is the question when one is

examining a set or group of like events.

I can imagine a small correllation between successive tosses.

DecadentLinuxUserNumeroUno_at_decadence.org wrote:

Martin Riddle <martin_ridd_at_verizon.net> wrote in

news:2fhf4e5ehq2i9kqpc3hs1k2df0c6ouq4j0_at_4ax.com:

Dosen't viewing the outcome of a coin-toss affect the outcome? ;)

Cheers

No.

Maybe... IF you had flapping butterfly wings on at the time.

Here is the stat... The outcome of a coin toss is 50/50.

The outcome of ANY subsequent toss will have exactly nothing to do

with the result of ANY previous toss, and is exactly 50/50.

So 100 tosses could result in 100 events with the same outcome.

Now, the PROBABILITY of that happening is the question when one is

examining a set or group of like events.

I can imagine a small correllation between successive tosses.

That is highly unlikely. It is much more likely that the exact ratio is

not exactly 50:50 and varies gradually with time as the coin edge wears.

Manufacturing defects, centre of gravity, the patterns on the surface

and edge wear will very slightly bias the results so that it is more

like 5000000:5000001 - but few people have the patience to throw that

many times. One guy who did but with a casino grade dice was the Swiss

astronomer Wolf who threw the dice 20000 times in the late 1800's

recording the outcomes and got a mean of 3.5983 instead of 3.5. He is

the same Wolf who gives his name to the Zurich sunspot number still in

use today. He was very interested in random physical processes.

https://en.wikipedia.org/wiki/Rudolf_Wolf

Ed Jaynes famously analysed the likely manufacturing imperfections in

his Brandeis series of lectures on Maximum Entropy (heavy going advanced

mathematics but clearly explained).

http://www.sns.ias.edu/~tlusty/courses/InfoInBio/Papers/JaynesStandOnEntropy.pdf

Classic Maxent treatment on p34 and then see "Wolf's Dice Data" p48-55

Real dice and coins have real physical defects. If the die is drilled

and the paint less dense than the base material then 6 will come up

slightly more often than 1 because of the CoG location. The same is also

true of 5 vs 2 and 4 vs 3 but to a lesser extent.

--

Regards,

Martin Brown

Guest

Wed Jan 23, 2019 12:45 pm

On 2019-01-22 8:21 a.m., Jamie M wrote:

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results â€“ say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.Â So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH".Â Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have.Â "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way?Â I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a time

(roughly the size of our memory capacity) while going through a series

of results â€“ say from 20 coin tosses. The mathematics show that the

contents of that window will hold "HHHT" more often than "HHHH" ("H" and

"T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.Â So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH".Â Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have.Â "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way?Â I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

A simple test that a computer program could do based on the above to see

if it is possible to guess higher than 50% what the next pseudorandom

coin will be:

Program is a loop that does sets of 3 random flips, heads or tails, then

a guess is made. The guess is based on a lookup table looking at the

three previous flips:

lookup table:

------------

coin1 coin2 coin3 guess

HHHT

HHTT

HTHT

HTTH

THHT

THTH

TTHH

TTTH

I just made the 4th coin guess by trying to balance out each group of 4

flips to give the closest count of heads to tails for each set of 4 coin

tosses. This of course only covers half of the 2^4 possibilities of 4

coin flips. This lookup table has overall 16 H and 16 T guesses, and

6 of the 8 groups of 4 flips have equal number of heads and tails flips.

The opposite guess lookup table:

---------------------------------

coin1 coin2 coin3 guess

HHHH

HHTH

HTHH

HTTT

THHH

THTT

TTHT

TTTT

This lookup table which is the opposite of the first regarding the

guesses, has overall 16 H and 16 T guesses. However only 2 of the 8

groups have equal number of heads and tails flips.

So to me it seems the first example is more balanced but not sure if

it is more random.

cheers,

Jamie

Guest

Wed Jan 23, 2019 12:45 pm

On 23/01/2019 12:13, Jamie M wrote:

On 2019-01-22 8:21 a.m., Jamie M wrote:

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a

time (roughly the size of our memory capacity) while going through a

series of results â€“ say from 20 coin tosses. The mathematics show that

the contents of that window will hold "HHHT" more often than "HHHH"

("H" and "T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.Â So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH".Â Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have.Â "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way?Â I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

A simple test that a computer program could do based on the above to see

if it is possible to guess higher than 50% what the next pseudorandom

coin will be:

Program is a loop that does sets of 3 random flips, heads or tails, then

a guess is made.Â The guess is based on a lookup table looking at the

three previous flips:

lookup table:

------------

coin1 coin2 coin3 guess

HHHT

HHTT

HTHT

HTTH

THHT

THTH

TTHH

TTTH

I just made the 4th coin guess by trying to balance out each group of 4

flips to give the closest count of heads to tails for each set of 4 coin

tosses.Â This of course only covers half of the 2^4 possibilities of 4

coin flips.Â This lookup table has overall 16 H and 16 T guesses, and

6 of the 8 groups of 4 flips have equal number of heads and tails flips.

The opposite guess lookup table:

---------------------------------

coin1 coin2 coin3 guess

HHHH

HHTH

HTHH

HTTT

THHH

THTT

TTHT

TTTT

This lookup table which is the opposite of the first regarding the

guesses, has overall 16 H and 16 T guesses.Â However only 2 of the 8

groups have equal number of heads and tails flips.

So to me it seems the first example is more balanced but not sure if

it is more random.

cheers,

Jamie

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a

time (roughly the size of our memory capacity) while going through a

series of results â€“ say from 20 coin tosses. The mathematics show that

the contents of that window will hold "HHHT" more often than "HHHH"

("H" and "T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.Â So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH".Â Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have.Â "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way?Â I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

cheers,

Jamie

A simple test that a computer program could do based on the above to see

if it is possible to guess higher than 50% what the next pseudorandom

coin will be:

Program is a loop that does sets of 3 random flips, heads or tails, then

a guess is made.Â The guess is based on a lookup table looking at the

three previous flips:

lookup table:

------------

coin1 coin2 coin3 guess

HHHT

HHTT

HTHT

HTTH

THHT

THTH

TTHH

TTTH

I just made the 4th coin guess by trying to balance out each group of 4

flips to give the closest count of heads to tails for each set of 4 coin

tosses.Â This of course only covers half of the 2^4 possibilities of 4

coin flips.Â This lookup table has overall 16 H and 16 T guesses, and

6 of the 8 groups of 4 flips have equal number of heads and tails flips.

The opposite guess lookup table:

---------------------------------

coin1 coin2 coin3 guess

HHHH

HHTH

HTHH

HTTT

THHH

THTT

TTHT

TTTT

This lookup table which is the opposite of the first regarding the

guesses, has overall 16 H and 16 T guesses.Â However only 2 of the 8

groups have equal number of heads and tails flips.

So to me it seems the first example is more balanced but not sure if

it is more random.

cheers,

Jamie

<https://en.wikipedia.org/wiki/Law_of_averages>

<https://en.wikipedia.org/wiki/Gambler%27s_fallacy>

Guest

Wed Jan 23, 2019 1:45 pm

On 22/01/2019 16:41, Phil Hobbs wrote:

On 1/22/19 11:21 AM, Jamie M wrote:

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a

time (roughly the size of our memory capacity) while going through a

series of results â€“ say from 20 coin tosses. The mathematics show that

the contents of that window will hold "HHHT" more often than "HHHH"

("H" and "T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.Â So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH".Â Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have.Â "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way?Â I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

Psychologists are notoriously hopeless at statistics.

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel to

be less random are precisely the ones that are least likely to occur.

Imagine a sliding window that can only "see" four coin tosses at a

time (roughly the size of our memory capacity) while going through a

series of results â€“ say from 20 coin tosses. The mathematics show that

the contents of that window will hold "HHHT" more often than "HHHH"

("H" and "T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.Â So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH".Â Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have.Â "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way?Â I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

Psychologists are notoriously hopeless at statistics.

Indeed. They seem to subscribe to the Gamblers fallacy :(

I wonder if they could be persuaded to accept the two people with the

same birthday bet in a room with 23 people in it ?

https://en.wikipedia.org/wiki/Birthday_problem

However I thought it would be fun to compute the probability of finding

the exact sequence of 2 bits "HT" at least once in a finite length

bitstream. Turns out it is *much* easier to enumerate the states not

containing this specific bit pattern. The reasoning is simple once there

has been an H in the sequence there cannot be any T's following it.

For a specific N there are N+1 states without "HT" in and 2^N states.

n p 1-p Not containing "HT"

2 1/4 3/4 HH TH TT

3 1/2 4/8 HHH THH TTH TTT

4 11/16 5/16 HHHH THHH TTHH TTTH TTTT

5 13/16 6/32 HHHHH THHHH TTHHH TTTHH TTTTH TTTTT

6 57/64 7/64

7 7/8 8/64

....

N (2^N-1-N)/2^N (N+1)/2^N

A much more interesting problem involving understanding prior and

posterior probabilities is the three door problem beloved of a certain

game show which even some otherwise astute mathematicians struggle with:

https://www.theproblemsite.com/games/treasure-hunt/door-hint

Bayesians find the answer rather more easily than others.

--

Regards,

Martin Brown

Guest

Wed Jan 23, 2019 1:45 pm

On 2019-01-23 4:07 a.m., Martin Brown wrote:

On 22/01/2019 16:41, Phil Hobbs wrote:

On 1/22/19 11:21 AM, Jamie M wrote:

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel

to be less random are precisely the ones that are least likely to

occur. Imagine a sliding window that can only "see" four coin tosses

at a time (roughly the size of our memory capacity) while going

through a series of results â€“ say from 20 coin tosses. The

mathematics show that the contents of that window will hold "HHHT"

more often than "HHHH" ("H" and "T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.Â So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH".Â Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have.Â "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way?Â I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

Psychologists are notoriously hopeless at statistics.

Indeed. They seem to subscribe to the Gamblers fallacy :(

I wonder if they could be persuaded to accept the two people with the

same birthday bet in a room with 23 people in it ?

https://en.wikipedia.org/wiki/Birthday_problem

However I thought it would be fun to compute the probability of finding

the exact sequence of 2 bits "HT" at least once in a finite length

bitstream. Turns out it is *much* easier to enumerate the states not

containing this specific bit pattern. The reasoning is simple once there

has been an H in the sequence there cannot be any T's following it.

For a specific N there are N+1 states without "HT" in and 2^N states.

nÂ Â Â pÂ Â Â Â Â Â Â 1-pÂ Â Â Not containing "HT"

2Â Â Â 1/4Â Â Â Â Â Â Â 3/4Â Â Â HH TH TT

3Â Â Â 1/2Â Â Â Â Â Â Â 4/8Â Â Â HHH THH TTH TTT

4Â Â Â 11/16Â Â Â Â Â Â Â 5/16Â Â Â HHHH THHH TTHH TTTH TTTT

5Â Â Â 13/16Â Â Â Â Â Â Â 6/32Â Â Â HHHHH THHHH TTHHH TTTHH TTTTH TTTTT

6Â Â Â 57/64Â Â Â Â Â Â Â 7/64

7Â Â Â 7/8Â Â Â Â Â Â Â 8/64

...

NÂ Â Â (2^N-1-N)/2^NÂ Â Â (N+1)/2^N

A much more interesting problem involving understanding prior and

posterior probabilities is the three door problem beloved of a certain

game show which even some otherwise astute mathematicians struggle with:

https://www.theproblemsite.com/games/treasure-hunt/door-hint

Bayesians find the answer rather more easily than others.

On 1/22/19 11:21 AM, Jamie M wrote:

Hi,

Reading this paper on coin toss probabilities:

https://medicalxpress.com/news/2019-01-rationality-stupid.html

The article states that:

"for a finite set of coin tosses, the sequences we intuitively feel

to be less random are precisely the ones that are least likely to

occur. Imagine a sliding window that can only "see" four coin tosses

at a time (roughly the size of our memory capacity) while going

through a series of results â€“ say from 20 coin tosses. The

mathematics show that the contents of that window will hold "HHHT"

more often than "HHHH" ("H" and "T" stands for for heads and tails)."

I think if there were three coin tosses and then you pick what the next

coin will be, you should consider all 4 coins as already tossed, with

the fourth coin as either heads or tails.Â So in the case of the first

three already tossed as "HHH" then depending on the fourth pick it will

be "HHHT" or "HHHH".Â Then also the order of those coins doesn't really

matter, so it is good to shuffle them to see how many configurations

they have.Â "HHHT" obviously has more configurations than "HHHH", ie:

HHHT, HHTH, HTHH, THHH = 4 configurations

while "HHHH" has only one configuration.

The total configurations for four heads or tails coins, is 2^4 = 16,

so "HHHT" (in all 4 configurations) has a probability of 4/16=25%

while "HHHH" (in its only configuration) has a probability of 1/16=6.25%

So, if there are four coin tosses, could the probability of the fourth

coin being heads or tails could be determined this way?Â I have always

just assumed a coin flip is 50/50, but too lazy to check. :)

Psychologists are notoriously hopeless at statistics.

Indeed. They seem to subscribe to the Gamblers fallacy :(

I wonder if they could be persuaded to accept the two people with the

same birthday bet in a room with 23 people in it ?

https://en.wikipedia.org/wiki/Birthday_problem

However I thought it would be fun to compute the probability of finding

the exact sequence of 2 bits "HT" at least once in a finite length

bitstream. Turns out it is *much* easier to enumerate the states not

containing this specific bit pattern. The reasoning is simple once there

has been an H in the sequence there cannot be any T's following it.

For a specific N there are N+1 states without "HT" in and 2^N states.

nÂ Â Â pÂ Â Â Â Â Â Â 1-pÂ Â Â Not containing "HT"

2Â Â Â 1/4Â Â Â Â Â Â Â 3/4Â Â Â HH TH TT

3Â Â Â 1/2Â Â Â Â Â Â Â 4/8Â Â Â HHH THH TTH TTT

4Â Â Â 11/16Â Â Â Â Â Â Â 5/16Â Â Â HHHH THHH TTHH TTTH TTTT

5Â Â Â 13/16Â Â Â Â Â Â Â 6/32Â Â Â HHHHH THHHH TTHHH TTTHH TTTTH TTTTT

6Â Â Â 57/64Â Â Â Â Â Â Â 7/64

7Â Â Â 7/8Â Â Â Â Â Â Â 8/64

...

NÂ Â Â (2^N-1-N)/2^NÂ Â Â (N+1)/2^N

A much more interesting problem involving understanding prior and

posterior probabilities is the three door problem beloved of a certain

game show which even some otherwise astute mathematicians struggle with:

https://www.theproblemsite.com/games/treasure-hunt/door-hint

Bayesians find the answer rather more easily than others.

Hi,

There is one option missing in that, which would be to guess again

between the two remaining closed doors, which would give a 50%

probability of being correct. Considering that, then the probability of

1/3 or 2/3 for keeping the guess or changing the guess makes more sense

(maybe!)

cheers,

Jamie

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