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Hillary Clinton Has The Most Statistically Improbable Coin-Toss Luck Ever

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posted on Feb, 5 2016 @ 02:12 PM
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originally posted by: eNumbra
6 coins with 2 outcomes each means the probability of getting 6 heads is at bare minimum 1 in 7. (6h-0t, 5h-1t, 4h-2t... etc.)

Each toss has a 1:2 probability but as you add the tosses together (which you have to do in this case) the odds begin to stack against the outcome.


6 50/50 shots is .5^6 (chance of not happening raised to the power of the number of attempts) which is 0.015625% or you can take 1/0.015625 which gives you 64. Alternatively you could use (1/.5)^6 which is 2^6, which again gives you 64 or... 1 in 64. It's not all that rare really.




posted on Feb, 5 2016 @ 02:27 PM
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a reply to: preezy120

This is incorrect. Each flip is 50/50, but you're not constructing a set of 1. Even though all of the flips happened in different locations it's still a set of 6 flips. There are 7 possible outcomes, here they are along with the odds of each happening

Clinton 6, Sanders 0 - 1 in 64
Clinton 5, Sanders 1 - 6 in 64
Clinton 4, Sanders 2 - 15 in 64
Clinton 3, Sanders 3 - 20 in 64
Clinton 2, Sanders 4 - 15 in 64
Clinton 1, Sanders 5 - 6 in 64
Clinton 0, Sanders 6 - 1 in 64



posted on Feb, 5 2016 @ 03:46 PM
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originally posted by: Aazadan
a reply to: preezy120

This is incorrect. Each flip is 50/50, but you're not constructing a set of 1. Even though all of the flips happened in different locations it's still a set of 6 flips. There are 7 possible outcomes, here they are along with the odds of each happening

Clinton 6, Sanders 0 - 1 in 64
Clinton 5, Sanders 1 - 6 in 64
Clinton 4, Sanders 2 - 15 in 64
Clinton 3, Sanders 3 - 20 in 64
Clinton 2, Sanders 4 - 15 in 64
Clinton 1, Sanders 5 - 6 in 64
Clinton 0, Sanders 6 - 1 in 64


Sure hope you don't clam to be a Mathematician . . . Tell us; just how does 6 binary events have 7 results?



posted on Feb, 5 2016 @ 05:19 PM
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originally posted by: tanka418

originally posted by: Aazadan
a reply to: preezy120

This is incorrect. Each flip is 50/50, but you're not constructing a set of 1. Even though all of the flips happened in different locations it's still a set of 6 flips. There are 7 possible outcomes, here they are along with the odds of each happening

Clinton 6, Sanders 0 - 1 in 64
Clinton 5, Sanders 1 - 6 in 64
Clinton 4, Sanders 2 - 15 in 64
Clinton 3, Sanders 3 - 20 in 64
Clinton 2, Sanders 4 - 15 in 64
Clinton 1, Sanders 5 - 6 in 64
Clinton 0, Sanders 6 - 1 in 64


Sure hope you don't clam to be a Mathematician . . . Tell us; just how does 6 binary events have 7 results?


Because 0 is also a possible result. Therefore the range is 0 to 6 which is 7 different numbers. There are 64 distinct outcomes which sum to 7 possibilities.

In your response to me you even quoted the 7 possibilities as I listed them. Count it for yourself.
edit on 5-2-2016 by Aazadan because: (no reason given)



posted on Feb, 5 2016 @ 06:54 PM
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originally posted by: Aazadan

originally posted by: tanka418

originally posted by: Aazadan
a reply to: preezy120

This is incorrect. Each flip is 50/50, but you're not constructing a set of 1. Even though all of the flips happened in different locations it's still a set of 6 flips. There are 7 possible outcomes, here they are along with the odds of each happening

Clinton 6, Sanders 0 - 1 in 64
Clinton 5, Sanders 1 - 6 in 64
Clinton 4, Sanders 2 - 15 in 64
Clinton 3, Sanders 3 - 20 in 64
Clinton 2, Sanders 4 - 15 in 64
Clinton 1, Sanders 5 - 6 in 64
Clinton 0, Sanders 6 - 1 in 64


Sure hope you don't clam to be a Mathematician . . . Tell us; just how does 6 binary events have 7 results?


Because 0 is also a possible result. Therefore the range is 0 to 6 which is 7 different numbers. There are 64 distinct outcomes which sum to 7 possibilities.

In your response to me you even quoted the 7 possibilities as I listed them. Count it for yourself.


I was afraid of that...

You are mistaken. There are only 6 possibilities from 6 events.

The addition of the seventh "outcome" is the result of broken logic, though, understandable I suppose.

Think of it like this; you have 6 individual binary events, thus there can only be 6 outcomes.

However, IF you still have an issue with this; please ask a high school math teacher.

edit on 5-2-2016 by tanka418 because: (no reason given)



posted on Feb, 5 2016 @ 08:44 PM
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a reply to: tanka418

There are 7 events. Lets scale it down to a single event, 1 or 0. It either happens or it doesn't happen, that's two possibilities. Now lets add an event, you have 4 outcomes
00
01
10
11

Sum those up and your possible outcomes are 0 (25%), 1 (50%), and 2 (25%).

Repeating what I said before
Clinton 6, Sanders 0 - 1 in 64
Clinton 5, Sanders 1 - 6 in 64
Clinton 4, Sanders 2 - 15 in 64
Clinton 3, Sanders 3 - 20 in 64
Clinton 2, Sanders 4 - 15 in 64
Clinton 1, Sanders 5 - 6 in 64
Clinton 0, Sanders 6 - 1 in 64

Those are the outcomes along with their odds of happening. Which of these do you not agree is possible?

I have now shown this mathematically 3 times, and your only response has been "that's wrong". Prove it.



posted on Feb, 5 2016 @ 09:43 PM
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originally posted by: Aazadan
a reply to: tanka418

There are 7 events. Lets scale it down to a single event, 1 or 0. It either happens or it doesn't happen, that's two possibilities. Now lets add an event, you have 4 outcomes
00
01
10
11

Sum those up and your possible outcomes are 0 (25%), 1 (50%), and 2 (25%).

Repeating what I said before
Clinton 6, Sanders 0 - 1 in 64
Clinton 5, Sanders 1 - 6 in 64
Clinton 4, Sanders 2 - 15 in 64
Clinton 3, Sanders 3 - 20 in 64
Clinton 2, Sanders 4 - 15 in 64
Clinton 1, Sanders 5 - 6 in 64
Clinton 0, Sanders 6 - 1 in 64

Those are the outcomes along with their odds of happening. Which of these do you not agree is possible?

I have now shown this mathematically 3 times, and your only response has been "that's wrong". Prove it.


Referencing the OP; there were 6 events reported, not seven (7), six (6)...

The probabilities are calculated by multiplying the several instance values (in this case 0.5) by each other. So, we have 0.5^6; that's 0.5 raised to the sixth (6) power.

I invite you to check it our with a math teacher your trust.

Probabilities:
(0) 1 . . . 0.5 1 in 2
(1) 2 . . . 0.25 1 in 4
(2) 3 . . . 0.125 1 in 8
(3) 4 . . . 0.0625 1 in 16
(4) 5 . . . 0.3125 1 in 32
(5) 6 . . . 0.0156 1 in 64

Again, take this to a math teacher!

edit on 5-2-2016 by tanka418 because: (no reason given)



posted on Feb, 5 2016 @ 09:52 PM
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a reply to: Aazadan



Out of six throws at 50/50 probability should gets you three. So to get a run of six, the probability should drop 50% after each throw. First throw 2/1 second throw 4/1 third throw 8/1 fourth throw 16/1 fifth throw 32/1 six throw 62/1 So her chances of winning all six are 62/1. Highly unlikely but still possible. But you can see from then on it gets worse seventh throw 124/1 etc .etc.



posted on Feb, 5 2016 @ 11:05 PM
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a reply to: tanka418

I hope you're not a math teacher then, because the probability of 5 heads out of 6 flips is not 3.125% or 1 in 32, it's 6 in 64 or 3 in 32, and again you are missing 0.

To go into detail on this 5 out of 6 for example, here are the configurations (1 being heads, 0 being tails) that result in 5/6 being heads
011111
101111
110111
111011
111101
111110

As you can see, there are 6 of them out of the 64 total possibilities. 6 flips gives 7 total outcomes, 0 heads, 1 heads, 2 heads, 3 heads, 4 heads, 5 heads, and 6 heads.


originally posted by: anonentity
a reply to: Aazadan



Out of six throws at 50/50 probability should gets you three. So to get a run of six, the probability should drop 50% after each throw. First throw 2/1 second throw 4/1 third throw 8/1 fourth throw 16/1 fifth throw 32/1 six throw 62/1 So her chances of winning all six are 62/1. Highly unlikely but still possible. But you can see from then on it gets worse seventh throw 124/1 etc .etc.


But she didn't win 7 throws, she won 6 which is really not all that unlikely. 1% is a tiny number psychologically but very large number in mathematics and she was well over that.

If I read/remember correctly, Bernie needed to win 3 coin tosses to win the state, that means Hillary needed to win atleast 4 of them. I listed the odds above, so it's pretty easy to calculate. Hillary was 22/64 or 35.375% to win on the coin tosses and she hit (that she hit on the most unlikely combination isn't really relevant, but it's what everyone is focusing on).
edit on 5-2-2016 by Aazadan because: (no reason given)

edit on 5-2-2016 by Aazadan because: (no reason given)



posted on Feb, 5 2016 @ 11:32 PM
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originally posted by: Aazadan
a reply to: tanka418

I hope you're not a math teacher then, because the probability of 5 heads out of 6 flips is not 3.125% or 1 in 32, it's 6 in 64 or 3 in 32, and again you are missing 0.



No, I'm not a mat teacher, however mathematics is one area where I do have a degree, And you so obviously do not! So, before you embarrass yourself any more, ask a math teacher!!

Where did you get"5"? We are not talking about 5 of 6 here, we are talking about 6 of 6.

To reiterate: Ask a math teacher!.



posted on Feb, 5 2016 @ 11:49 PM
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Context! There's a situation which can be contextualized however one pleases. If neither of you are trolling and honestly can't see the other individuals valid reasoning, then this is just a sad unfolding.

I'm going to throw something else in the mix and walk off...

So 6 of6 is 1/64, but I find it kinda interesting that if you're making a larger number of tosses, the probability of course increases towards getting 6 in a row one perspective or the other.

Makes me feel like there's a group for me out there... somehow... somewhere outside my own ramblings.

Then I remember, I'm not stuck on either or. Damned, I'll stay a loner then lol
edit on 6-2-2016 by pl3bscheese because: (aaz perspective is more inclusive, btw)



posted on Feb, 6 2016 @ 02:46 AM
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originally posted by: tanka418
No, I'm not a mat teacher, however mathematics is one area where I do have a degree, And you so obviously do not! So, before you embarrass yourself any more, ask a math teacher!!

Where did you get"5"? We are not talking about 5 of 6 here, we are talking about 6 of 6.

To reiterate: Ask a math teacher!.



I'll try this once more. There are 64 possible outcomes, this is determined by there being 6 flips with 2 outcomes each, that is 2^6. Since you clearly do not understand this through formulas, I'm going to list every outcome and it's sum (sorry, this is a little long) which is the number of flips that come up heads (not to be confused with the numerical value of those digits)

000000 = 0
000001 = 1
000010 = 1
000011 = 2
000100 = 1
000101 = 2
000110 = 2
000111 = 3
001000 = 1
001001 = 2
001010 = 2
001011 = 3
001100 = 2
001101 = 3
001110 = 3
001111 = 4
010000 = 1
010001 = 2
010010 = 2
010011 = 3
010100 = 2
010101 = 3
010110 = 3
010111 = 4
011000 = 2
011001 = 3
011010 = 3
011011 = 4
011100 = 3
011101 = 4
011110 = 4
011111 = 5
100000 = 1
100001 = 2
100010 = 2
100011 = 3
100100 = 2
100101 = 3
100110 = 3
100111 = 4
101000 = 2
101001 = 3
101010 = 3
101011 = 4
101100 = 3
101101 = 4
101110 = 4
101111 = 5
110000 = 2
110001 = 3
110010 = 3
110011 = 4
110100 = 3
110101 = 4
110110 = 4
110111 = 5
111000 = 3
111001 = 4
111010 = 4
111011 = 5
111100 = 4
111101 = 5
111110 = 5
111111 = 6

Count those values up and you will get the amounts I listed before. There are 7 outcomes. 0 1's, 1 1's, 2 1's, 3 1's, 4 1's, 5 1's, and 6 1's (you can treat 1's as heads or tails, it doesn't matter which). You can add those up to verify my previous numbers if you wish. If you know base 2, which I assume you do since you have a degree in mathematics, you should also be able to easily look through those numbers and confirm they are the values 0 through 63 (that is 64 outcomes, those pesky zero's) in base 2, listed in order so it will be easy to confirm that I listed all possible outcomes and that it isn't a random selection of numbers.

I got 5 because you claimed 5 heads is a 3.125% chance. It is not. The chance for 5 heads in 6 flips is 6 in 64 or 9.375%. If you were to flip a coin 5 times there would be 2^5 outcomes or 32, and one of those would be all heads just as another would be all tails, and those respective outcomes would each be a 3.125% chance, but that is not the situation we are dealing with.


originally posted by: pl3bscheese
Context! There's a situation which can be contextualized however one pleases. If neither of you are trolling and honestly can't see the other individuals valid reasoning, then this is just a sad unfolding.

I'm going to throw something else in the mix and walk off...

So 6 of6 is 1/64, but I find it kinda interesting that if you're making a larger number of tosses, the probability of course increases towards getting 6 in a row one perspective or the other.

Makes me feel like there's a group for me out there... somehow... somewhere outside my own ramblings.

Then I remember, I'm not stuck on either or. Damned, I'll stay a loner then lol


I am not trolling, I cannot say the same for tanka418, but this is my last post on the subject either way, because if I am not getting through then there's no point in continuing, and if I am getting through then there's no need to continue.

The answer to your statement is that as you make an increasingly larger number of tosses everything drifts towards statistical averages but also that everything becomes more likely to happen (or more specifically, less likely to have not happened). It gets a little weirder with infinity, because if you make an an infinite number of tosses then you will also have an infinite series of outcomes and every finite sequence no matter how improbable will be contained in infinity, it will also happen an infinite number of times. Infinity gets really weird.

This is why it's best to stick to finite sequences, although that's not always possible.
edit on 6-2-2016 by Aazadan because: (no reason given)



posted on Feb, 6 2016 @ 02:51 AM
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Just because something is statistically improbable does not make it impossible. I have lost 20+ games at a blackjack table. Statistically, should not happen; but it did. I think honing in on the coin flip is the wrong thing to be looking at. We should be wondering, why are we flipping a coin in a caucus?!
edit on 6-2-2016 by ownbestenemy1 because: (no reason given)



posted on Feb, 6 2016 @ 02:55 AM
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a reply to: Aazadan

It comes off as either trolling, or not really being capable of entering the other person's mindset all too well.

He doesn't have it, you should have realized that long ago.

Not sure why you care.

There is no "answer" to a statement, because it's not a question. Just weird you want to be so precise and correct on matters that are just not so, and in doing get it all wrong for the right reasons.

Just weird.
edit on 6-2-2016 by pl3bscheese because: (no reason given)



posted on Feb, 6 2016 @ 10:23 AM
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a reply to: Aazadan



I am not trolling, I cannot say the same for tanka418, but this is my last post on the subject either way, because if I am not getting through then there's no point in continuing, and if I am getting through then there's no need to continue.



The reason you are not "getting through" is because you are very wrong, mathematically speaking that is.

I suggested that you check it out with a math teacher, why haven't you? By asking a math teacher you can resolve this is less than 5 minutes, or about as long as it will take him to explain this to you.

One would think that this would be the logical course for someone like yourself with few math skills.

Oh, and by the way; there are only 6 possible outcomes for 6 binary events, as contrasted to the 64 outcomes you seem to have amplified this into....that's one single outcome per event. the value 64 is a part of the probability that all six are the same...hence 1 in 64.

Now, please do yourself a favor and find someone with real math skills, and ask! Stop arguing something you have no understanding of!

There is an ATS member; "Harte" who claims to be a math teacher...perhaps you could ask him. Or maybe Phage...

So...anyway; please deny ignorance


@ownbestenemy1: The reason a coin toss is used is because it is a fast, inexpensive, fair/random way of breaking a tie. I mean; what would you have them do...spend $10,000's to do it over? Or less than 5 minutes, with no expense, flipping a coin.

edit on 6-2-2016 by tanka418 because: (no reason given)



posted on Feb, 6 2016 @ 11:56 AM
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a reply to: tanka418

I'm not really sure what you're arguing about, but you're actually sort of both right. Strictly speaking, there are 64 unique combinations of results, but functionally, there are only seven possible outcomes.

C/S
6/0
5/1
4/2
3/3
2/4
1/5
0/6

Think of it this way. If we take 1 as a Clinton win, and 0 as a Sanders win, then to obtain the C/S score of 1/5, the possible arrangements are.

100000
010000
001000
000100
000010
000001

However, which individual position (or state) the flip is won in is ultimately irrelevant. There's six ways of this happening out of 64 possible outcomes, which is why the chance of Clinton winning EXACTLY once and Sanders winning five times is 6/64. If you want to ask what the chance of Clinton winning at least once is, however, that chance is 63/64. A 6-win streak only has one possible combination, which is

111111.

There's formulae for this, but it really shouldn't require this much discussion. The probability of either candidate winning 6/6 coin flips, under the assumption that each coin flip has an exactly 50/50 chance of landing on either side, is 1/64. THAT IS A FACT.



posted on Feb, 6 2016 @ 12:00 PM
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Sorry to break in, but one of my blogs does a week in pictures post and I thought I'd share these:





They seem relevant to the discussion.



posted on Feb, 6 2016 @ 12:02 PM
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a reply to: tanka418


Oh, and by the way; there are only 6 possible outcomes for 6 binary events, as contrasted to the 64 outcomes you seem to have amplified this into.


I'm responding to this part specifically in another post since it seems to be the root cause of the issue.

There are seven possible outcomes because there are six binary events. There would be thirteen outcomes if you had twelve binary events. It's [No. of Binary Events + 1]=Total Outcomes.

If you have one binary event, how many outcomes are there?
Two.

The 64 is coming from the number of unique combinations of events that lead to each score. That's where the higher probability of a 3/3 score comes from, as there are many more ways to end up with 3/3 than there is to end up with 6/0.



posted on Feb, 6 2016 @ 12:24 PM
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Thanks for all of the great math.

I posted this earlier...I think it's the simplest chart:

Hillary had a:

50% chance of winning the first toss
25% chance of winning the second toss in a row
12.5% chance of winning the third toss in a row
6.25% chance of winning the fourth toss in a row
3.125% chance of winning the fifth toss in a row
1.56% chance of winning the sixth toss in a row

It still gets us to a 1 in 64 chance of six in a row...with her consecutively overcoming some increasingly impressive odds.
edit on 6-2-2016 by IAMTAT because: (no reason given)



posted on Feb, 6 2016 @ 12:44 PM
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originally posted by: Eilasvaleleyn
There are seven possible outcomes because there are six binary events. There would be thirteen outcomes if you had twelve binary events. It's [No. of Binary Events + 1]=Total Outcomes.

If you have one binary event, how many outcomes are there?
Two.


One...one event one result. The results produced by this simple math are not a "set", but a single result instance.


here...solve this simple method...

integer I = 0;
double d = 0.5;
while(I lessthanorequalto 6)
[
d = d *0.5;
I++; // increments i
]

When you work this out you will notice very quickly that the result is always wrong...in that it wont be 0.0156 (1 in 64).

The main reason for this is that there are only 6 events. That seventh "thing" you ae trying to insert doesn't belong, and is in fact incorrect, and based on broken logic.

I'm rather tired trying to educate those who apparently don't wish to deny ignorance and would rather embrace it instead...So...go ask a math teacher, please! Then you will have a better idea of who is correct here. And, I won't demand an apology.


edit on 6-2-2016 by tanka418 because: Great difficulty representing "C" language code here, due to poor quality of text editor[/editby

edit on 6-2-2016 by tanka418 because: (no reason given)
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