probability of coin toss

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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 t osses 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 .

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.

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

%

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 im pact how you flip the coin??? Is there a string tied to the coin? Does th e coin get sticky on one side? Spooky action at a distance?

Rick C.

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Psychologists are notoriously hopeless at statistics.

Cheers

Phil Hobbs

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

Cheers

Martin Riddle wrote in news: snipped-for-privacy@4ax.com:

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.

the

5%

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

Rick C.

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I can imagine a small correllation between successive tosses.

--

John Larkin   Highland Technology, Inc   trk 

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com

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.

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.

That must depend on the mental ability of the viewer (or perhaps their dose!)...

Mike.

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.

Ed Jaynes famously analysed the likely manufacturing imperfections in his Brandeis series of lectures on Maximum Entropy (heavy going advanced mathematics but clearly explained).

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

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

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 ?

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:

Bayesians find the answer rather more easily than others.

--
Regards, 
Martin Brown

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

Actually I see this is incorrect, nevermind lol

Robert Baer wrote in news:wQU1E.41849$ snipped-for-privacy@fx20.iad:

Absolute bullshit. Perpendicular edge margin == 50/50 toss EVERY TIME EVERY COIN.

The difference, if ever, if any is so negligible as to barely be measureable in 10000 tosses.

AND each toss strikes a different point of the edge of the coin as well. 100% 50/50 result.

Even coin spin results are 50/50 and anyone would have a hard time creating a biased coin.

Mike Coon wrote in news: snipped-for-privacy@news.plus.net:

Like giving a goat a heart attack? Yeah, right.

Martin Brown wrote in news:q29bhc$1edb$ snipped-for-privacy@gioia.aioe.org:

Bullshitzky.

Attempt at figuring this out:

Case1: 2/3 probability

Pick incorrect door, then 100% probability host also picks incorrect door. Remaining doors are the correct door and an incorrect door.

option1: keep original pick, probability of winning is 0%

option2: change door, probability of winning is 100%

Case2: 1/3 probability

Pick correct door, then 100% probability host picks incorrect door. Remaining doors are the correct door and an incorrect door.

option1: keep original pick, probability of winning is 100%

option2: change door, then probability of winning is 0%

Case1 option1 will occur 2/3 of the time, since there 2 out of 3 doors that are incorrect, so adding up the probabilities:

Change doors no matter what:

Case1+Case2 = (2/3*100%)+(1/3*0%)=2/3

Keep original pick no matter what:

Case1+Case2 = (2/3*0%)+(1/3*100%)=1/3

Makes sense! :D

cheers, Jamie