Flipped coin, 3 possibilities?

[Capiert]

I once flipped a coin

& it finally landed

on edge.

 

I was just wondering

how you would calculate

the probability

of that (happening,

because it never happened again).

 

That (edge_stand)

would be the (undecided)

maybe answer,

between yes & no.

 

Assuming we know

the diameter D

thickness T

density (rho)

& leaving you free

to choose your own

flipping method (calculation,

e.g. height h, twist rate, etc

what(ever) you need,

for the normal way people flip coins

& coin contour):

is there a way to calculate

that probability (for edge standing)?

 

E.g. (100%=)

1=H+E+T

heads, edge, tails (H=1-T-E, E=1-H-T, T=-H-E),

E=1/10000000.

 

Signed

 

SheerLuck Homes.

[swansont]

http://www.scienceforums.net/topic/101934-the-odds-of-a-coin-landing-vertically-5149-theory/

[Capiert]

http://www.scienceforums.net/topic/101934-the-odds-of-a-coin-landing-vertically-5149-theory/

Thanks. Fascinating stuff.

But I'm interested in something more practical.

Although Janus (#11, #16, #18)

is doing a good job

at getting us started,

(~5% does NOT seem real or correct to me);

I would expect (with this probabilty stuff)

after flipping the coin 20 times

(that) I would get an edge_stander e.g. 5%=0.05=1/20.

 

But in fact the statistics

(of everybody trying

in the whole world)

say an edge stander

is much rarer

than 1/20.

 

The bar's case (Acme #15)

is the 2nd example

I (now) know;

but I suspect

there are many more examples

I have not heard of.

 

Simply put,

if I flipped the coin

100 times,

I still would NOT get an edge_stander

i.e. NOT even 1/100.

Edited May 8, 2017 by Capiert

[DrKrettin]

Simply put,

if I flipped the coin

100 times,

I still would NOT get an edge_stander

i.e. NOT even 1/100.

 

You don't know that - it might happen on the first toss. But probably not.

[Capiert]

You don't know that - it might happen on the first toss. But probably not.

It won't happen on my 1st toss;

I've already experienced that!

(=History for my statistic.

Anything that happens after

can NOT be my 1st toss;

because it has a different number

of toss.)

 

Sorry if that sounds harsh;

but that is what I mean

by more practical.

 

E.g. How many times

must a coin be tossed

till an edge_stander happens,

guaranteed (=practically)?

 

Generally, (maybe) once

in a lifetime

for (only) some (people).

 

The rest never experience it (= an edge_stander),

no matter

how long they toss.

 

The odds are (for me) useless

if I don't see practical results

that can be verified

experimentally.

 

That's science:

experimentally reproduceable,

independently.

Edited May 8, 2017 by Capiert

[DrKrettin]

It won't happen on the 1st toss;

I've already experienced that!

 

If you say it will not happen on the first toss, you are saying that the probability is zero. If the probability is not zero, then you cannot say it will not happen on the first toss. Well, you could, but you would be wrong.

Edited May 8, 2017 by DrKrettin

[Capiert]

If you say it will not happen on the first toss, you are saying that the probability is zero.

The possibility of getting an edge_stander

on the very 1st toss

(in life (=a lifetime))

is even more rarer

than to get 1 (=edge_stander)

at all.

 

That (1st toss edge_stander)

is a (completely) different statistic

& probabability.

 

I do NOT say it is impossible,

but I've NEVER seen it (happen).

 

If the probability is not zero, then you cannot say it will not happen on the first toss. Well, you could, but you would be wrong.

I think we agree (there). ?

(#5 is edited.)

Edited May 8, 2017 by Capiert

[MigL]

That's not the way probability works...

The chance of the 1st toss landing on edge is exactly the same as any other toss.

Furthermore, a chance of 1/20 doesn't mean that after 20 tosses you should get an edge-stander.

It means that if repeated an infinite number of times, the chance of an edge-stander approaches 1/20.

[Capiert]

But probably not.

Phrases like that indicate

the fraction is smaller

than expected;

because the tosses needed

to get success

are (really) (much) larger,

which the math

should reveal

(for me)

but did NOT.

 

When we add an opposing opinion

it seems to me

(that) the math is more complicated

than what we have done (=formulated).

Edited May 8, 2017 by Capiert

[Lord Antares]

If you read the thread Swansont linked, you would have seen that the researchers empirically got the odds of around 1:6000, which is lower than I would expect, to be honest.

 

I'm not sure why you're complicating the issue. I don't understand your last post. Do you need clarification with something regarding probability?

[Capiert]

That's not the way probability works...

The chance of the 1st toss landing on edge is exactly the same as any other toss.

Furthermore, a chance of 1/20 doesn't mean that after 20 tosses you should get an edge-stander.

It means that if repeated an infinite number of times, the chance of an edge-stander approaches 1/20.

Ok let's start counting to infinity.

The sooner we start,

the sooner we'll be finished.

 

A 1st toss (stander) is rarer, than any other toss stander.

(I'm (talking about) looking at world statistics, historically.)

Edited May 8, 2017 by Capiert

[Lord Antares]

 

A 1st toss (stander) is rarer, than any other toss stander.

 

???

 

Are you trying to say that getting it to land on the rim on the first toss has a lower chance than on any other toss? What on earth would make you say that?

[Capiert]

???

 

Are you trying to say that getting it to land on the rim on the first toss has a lower chance than on any other toss?

Yes.

What on earth would make you say that?

Experience.

(I didn't get 1 on the 1st try (in my life)

& I don't know anybody else, either.)

 

You've put an extra limitation

on the math,

so if you collected the data

(of the whole earth, let's say a million to keep the math simple although it's wrong)

& found

1000 people had an edge_stander toss;

& asked how many had it (=the edge_stander)

on their (very) 1st toss

& only 1 person reported (=1/1000)

that they did,

then you would know what I mean.

E.g. ~(1/1000)^2=1/1000000.

 

 

The statistics of being 1st

versus not 1st (=simply being any)

depend on

the square of being (at all).

Edited May 8, 2017 by Capiert

[Lord Antares]

No, you are very wrong. When have you seen it land vertically on exactly the 9th toss? What about 22nd? What about 741st? Never, right?

 

Your problem is that you consider the first toss as ''special'', whereas you see all subsequent tosses as ''other tosses''. So, you take note when it lands vertically on the ''special'' toss and you don't take note of the exact number of tosses it took to land vertically on the ''other tosses''. This is something akin to the gambler's fallacy.

 

To say that the probability is reduced for the 1st toss because you haven't seen it happen is a VERY flippant comment. Probability doesn't work that way.

Also, as I mentioned, the odds were empirically shown to be about 1:6000 so that 1 person got it on the first toss is not only not unusual, but it is above expectation.

[Capiert]

No, you are very wrong. When have you seen it land vertically on exactly the 9th toss? What about 22nd? What about 741st? Never, right?

 

As soon as you limit to a single number (of sequence)

you have increased the complexity.

Let us say the edge_stander probability for any toss is E=1/6000, =1/N, N=6000.

Now if you want E for only the 1st toss, then that is 1/(N^2)

but it does not matter,

as you said,

it could be 1 of any other than that, e.g. 9th, 22nd, 741st, .. or 6000th, etc.

But let's say it did not matter

whether we received a stander on either the 1st or 2nd toss, instead: then E=2/(N^2).

How about a toss on either the 1st or 2nd or 3rd toss, then E=3/(N^2).

I think you get the idea how it's going.

If it didn't matter which toss, then any is E=N/(N^2), =1/N.

Above, I don't have to use 1st, 2nd, 3rd;

instead other numbers could be used e.g. 8th, 21st, 103rd, for E=3/(N^3).

Am I wrong?

 

Your problem is that you consider the first toss as ''special'', whereas you see all subsequent tosses as ''other tosses''. So, you take note when it lands vertically on the ''special'' toss and you don't take note of the exact number of tosses it took to land vertically on the ''other tosses''. This is something akin to the gambler's fallacy.

 

To say that the probability is reduced for the 1st toss because you haven't seen it happen is a VERY flippant comment.

 

I'd say real comment, because I experienced it so;

but that's only my choice of expression.

Probability doesn't work that way.

Fine. I used the word possibility (instead, in the header)

maybe there is a difference

(even in the math?)

Also, as I mentioned, the odds were empirically shown to be about 1:6000 so that 1 person got it on the first toss is not only not unusual, but it is above expectation.

Does that mean they had something like at least 6000^2=36000000 tosses

to display that each toss order (of sequence)

was ruffly distributed?

(e.g. stander on 1st toss: 1, 2nd toss: 1, 3rd toss: 1 etc

for the ideal case, which doesn't really happen so perfect,

but I can't show it too well, forgive me please.)

Caution: special is distinguished, from the group.

Special is only 1 of the group('s number).

But "1st" & "any" are NOT the same.

It is more difficult

to get a stander

only on the 1st toss.

 

If you color 1 ball (of many) in buckshot

& shoot at a fly

expecting the coloured ball to hit it (=the fly);

& repeat that experiment

over & over,

you will find

much less color balls hits

than the other balls,

because there are more of them (other non_coloured balls).

 

Please, don't forget,

I was asking for something

more practical

than the typical probability.

E.g. what we really find.

1/6000 sounds good.

(I guess I was lucky,

I didn't need so many attempts.)

 

I still can't imagine

an exclusive 1st toss

is just as easy to obtain

(but you guys & gals

have your methods).

Edited May 8, 2017 by Capiert

[Lord Antares]

Fine. I used the word possibility (instead, in the header)

maybe there is a difference

(even in the math?)

 

No, we're talking about the same thing.

 

 

Does that mean they had something like at least 6000^2=36000000 tosses

to display that each toss order (of sequence)

was ruffly distributed

 

??

Why would that be necessary?

They could have flipped it 18 000 times (6000 x 3) and got 3 vertical landings, hence 1 in 6000.

 

To assume that they have to be evenly distributed would be another fallacy. There is no reason they couldn't have gotten, for example, the 3577th, 15 321st and 15 328th tosses as vertical. If they all landed in perfect distribution, it would be remarkable coincidence! Of course, you would need more flips for more accurate results. I don't know how many they flipped. I'm just giving you an example.

 

 

Special is only 1 of the group('s number).

But "1st" & "any" are NOT the same.

It is more difficult

to get a stander

only on the 1st toss.

If you color 1 ball (of many) in buckshot

& shoot at a fly

expecting the coloured ball to hit it (=the fly);

& repeat that experiment

over & over,

you will find

much less color balls hits

than the other balls,

because there are more of them (other non_coloured balls).

 

Of course there is lower probability of it occuring on toss 1 than any other toss COMBINED. Why is it even necessary to mention that? If you coloured every 7th ball instead, you will find that they will also come up more rarely than all others combined. There's nothing special about that. It goes without saying.

 

 

I still can't imagine

an exclusive 1st toss

is just as easy to obtain

(but you guys & gals

have your methods).

 

Have you seen a coin being tossed around 6000 times? I will guess that you haven't, hence your comments about the first toss coming up more rarely. I repeat, there is NOTHING special about the first toss. There are equal odds for any toss individually.

[Capiert]

No, we're talking about the same thing.

 

??

Why would that be necessary?

They could have flipped it 18 000 times (6000 x 3) and got 3 vertical landings, hence 1 in 6000.

Please reread #15 again, I've added some math & explaination, & then tell me what you think.

To assume that they have to be evenly distributed would be another fallacy.

I agree.

There is no reason they couldn't have gotten, for example, the 3577th, 15 321st and 15 328th tosses as vertical. If they all landed in perfect distribution, it would be remarkable coincidence!

Yes (I agree).

Of course, you would need more flips for more accurate results. I don't know how many they flipped. I'm just giving you an example.

Yes, I'm with you.

 

Buckshot:

Of course there is lower probability of it occuring on toss 1 than any other toss COMBINED. Why is it even necessary to mention that?

In an attempt to persuade you above.

If you coloured every 7th ball instead, you will find that they will also come up more rarely than all others combined.

Yes, it does not matter which ball we have chosen;

the important thing

is whether we have selected "only 1" (or NOT).

 

There's nothing special about that. It goes without saying.

Good, then maybe now you can see a parallel similarity, in example.

 

Have you seen a coin being tossed around 6000 times?

No I haven't.

I will guess that you haven't,

You are right.

hence your comments about the first toss coming up more rarely. I repeat, there is NOTHING special about the first toss.

But there is something different

about excluding

all other tosses.

There are equal odds for any toss individually.

Yes, but there is a difference between those (any)

versus specifying that it must be only 1 specific toss from all possibilities (of any).

Edited May 8, 2017 by Capiert

[studiot]

I didn't see the December '16 thread referred to in the outset posts, but you have subtitled this thread ' 3 possibilities'.

 

That brings me to an interesting spin off point about this.

 

The coin toss is about the simplest random process we have.

 

We usually accept that we have equal probabilities of 0.5 and 0.5 for each side.

 

If we step up one level of complexity to the standard die of six sides we have equal probabilities for each of six sides.

 

But introducing the die introduces something not available witht the simple coin toss.

 

We can conside the probability of a number less than or equal to 3 (other other combination).

 

This probability is also 0.5.

 

Now that we have understood that we can create new categories of event by combining outcomes we can look again at the coin toss.

 

We can now consider the probability of 'anything else happening at all'

 

This is a very useful step forward as we can now have 3 outcomes Heads, Tails, Anything Else.

 

If we can show that the probability of anything else is below the radar, ie insignificant, we can go safely return to our unbiased coin.

 

Evidence has been presented to suggest that the probability of anything else is less than .0002.

 

Do you think this is significant?

Edited May 8, 2017 by studiot

[Capiert]

 

Evidence has been presented to suggest that the probability of anything else is less than .0002.

 

Do you think this is significant?

Yes, looking at Lord Antares 1/6000

0.0002=1/5000

comes into that range!

 

Please proceed Studiot.

Edited May 8, 2017 by Capiert

[Strange]

Let us say the edge_stander probability for any toss is E=1/6000, =1/N, N=6000.

Now if you want E for only the 1st toss, then that is 1/(N^2)

 

 

If the probability for any toss is 1/N then the probability for the first toss is (by definition) 1/N. Because the first toss is just one of "any".

[Capiert]

If the probability for any toss is 1/N then the probability for the first toss is (by definition) 1/N. Because the first toss is just one of "any".

"one of any" is the key there.

(It should become obvious to you.)

If "any"=N possibilities,

you forgot

you are (also) excluding N-1 other possibilities (from any)

when you want either (but) "only" the 1st or 7th or 216th,..

 

(See #15 again please.)

 

E=(1/N)*(1/N).

 

The extra (1/N)

is needed for the further restriction (of only 1st (instead of any)).

Edited May 8, 2017 by Capiert

[Lord Antares]

Please reread #15 again, I've added some math & explaination, & then tell me what you think.

 

The math is wrong. I don't see how you arrived at N^2. It's certainly incorrect. The odds that the coin will land vertically for ANY specific toss is (according to the statistic) 1 in 6000. Saying that the first toss is biased is unscientific and wrong.

 

 

I agree.

Yes (I agree).

Yes, I'm with you.

 

 

I'm slightly confused because I don't agree with anything you said. It should be reciprocal here.

 

 

Buckshot:In an attempt to persuade you above.

Yes, it does not matter which ball we have chosen;

the important thing

is whether we have selected "only 1" (or NOT).

 

 

It's hard for me to understand this 100% but maybe you are talking about the odds of landing vertically UP TO a specific number? For example, the odds for the 1st toss are smaller than for UP TO 10 tosses. This is obvious, so I'm not sure whether you tried to include this or not.

 

 

But there is something different

about excluding

all other tosses.Yes, but there is a difference between those (any)

versus specifying that it must be only 1 specific toss from all possibilities (of any).

 

No there isn't. The odds for an unspecified number are the same as picking a specific number. I'm talking about single numbers individually, as mentioned before. Of course the odds that it will land vertically on any number of tosses other than 1 are substantially higher than it landing on the first toss, but the odds are the same for every individual number of tosses.

[Capiert]

The math is wrong.

I disagree (but I'm only saying that

to prevent confusion).

 

I don't see how you arrived at N^2.

It's made of 2 decisions: a toss, & then specifying which.

It's certainly incorrect.

I can't say I agree

(but that's nothing new).

The odds that the coin will land vertically for ANY specific toss is (according to the statistic) 1 in 6000.

I agree.

That is the 1st decision.

It is biased (so to speak)

to find "any" stander.

If we collect data

that's what we get.

(Let's say we've tossed for multi_billion times,

as rediculous as it sounds,

so we have enough stander counts.)

However if we become more biased

 

(please remember

we have (biasedly) selected

for (any) edge_standers;

because heads & tails have been excluded (by our bias=decision(s) rules),

 

we can take that "same data"

& ask how many standers

happened only on the 1st tosses.

(=That is more bias, &)

we will find

that new number is much less.

 

How do you explain

that difference

in those 2 numbers?

It's quite significant.

Saying that the first toss is biased is unscientific and wrong.

Why?

If you toss 10 times

& have no selection rules

(=preferences of head, tails, edge, i.e biases)

then you only know you tossed,

but you do not know

what received the larger counts.

The bias (results) is which data "you" want (=preferred to see).

If "you" want "only" the 1st tosses (count, wrt to the total=all)

then you have (biasedly) selected (only) that (set of data)

to plot your statistics.

The percentages

will show that

in the plot.

 

I'm slightly confused because I don't agree with anything you said.

It's ok,

(maybe my wrong (informal) vocabulary?)

you just need a bit of time

(to let things settle & clear)

to understand

because you are logical

you will find the answer.

It should be reciprocal here.

But as you see it is not

because I have a different perspective

than you.

But I'm confident you will get it.

 

It's hard for me to understand this 100%

but maybe you are talking about the odds

of landing vertically UP TO a specific number?

 

For example,

the odds for the 1st toss

are smaller

than for UP TO 10 tosses.

 

This is obvious, so I'm not sure whether you tried to include this or not.

I don't think so.

 

But I'm sorry, because I'm not sure what (=how) you mean

with that context.

"Up to"

is a limiting expression

meaning

"not more than".

 

If you mean:

if I tossed only once,

versus

if I tossed ten times,

then naturally

10 tosses

has better results.

 

But those probability fractions (e.g. 1/N=1/6000)

can be used as multiplying factors

with the total number of tosses (e.g. 1 or 10)

to give you

the expected results.

E.g.

1*1/6000, versus

10*1/6000.

 

In that sense,

you can see

it wasn't

why we have

1/6000 for any, versus

(1/6000)*(1/6000) for "only" the 1st toss.

 

 

1/6000 can be the 1st toss,

but does NOT have to be (the 1st toss).

 

(1/6000)*(1/6000) is for a selected toss

of "your" choice;

& you can choose

from all possibilities

within the 6000 (virtually speaking).

 

Have you at least understood

that the results

are from a selection (=choice, bias)

(of (the total) data)?

I.e. a specific percent

(that had selection rules).

No selection rules means

all the data (E.g. multi_billion 100%;

but (uselessly) NOT sorted

into H, T, E.)

Completely (=100%) neutral

means you weren't looking

for anything,

& found everything

that told you nothing

but the tossing's grand total.

 

No there isn't. The odds for an unspecified number are the same as picking a specific number.

I'm talking about single numbers individually, as mentioned before.

Of course the odds that it will land vertically

on any number of tosses

other than 1

are substantially higher than it landing on the first toss,

Doesn't that ring a bell?

How do you show that mathematically?

Answer: with the extra *(1/N).

but the odds are the same for every individual number of tosses.

I agree.

Good. It looks like you have got it. ?

[Laughablestuff]

Enter quantum physics. Classical physics can't figure that one out. Its not likely to happen, possibly even in a lifetime. Personally, I can't throw the math down, but one must always explore all possibilities. Like one person posted, it could b a edge stand on toss 1. And also 2,3,4,5,6,7,8 and so forth. What gets calculated at that point is the deviation of the likelihood. I think they use sigmas to explore that stuff. Like the higgs boson was thrown around with the term sigma 5 or whatever, when regular people wouldn't even know what that means. Even further, they gave the wrong number for what it represents from Some sources, but that's beside the point. The equation is definitely already out there and solvable. I'm lazy tho

[Capiert]

Thanks for your tip.

I'm lazy tho

Try green coffee extract (400 mg/day before your meal),

it (=the chlorogenic acid, .. helps, &)

burns fat for more energy (Joules, but that's biochem).

 

(You'll have to toss a few (=many) coins (away) for it's price. )

Edited May 9, 2017 by Capiert