probability of coin toss

Might as well correct the incorrect correction:

Randomly re-guessing the second door will give 50% probability, Keeping the original pick is 1/3, reguessing is 1/2, and changing the original pick is 2/3.

cheers, Jamie

I have to say that when I first came across that, I couldn't see the problem.

Three doors A, B, C. Say the prize is in A. Three possibilities:

You choose A - then B or C is opened - you would be wrong to change. You choose B - then C is opened - you would be right to change. You choose C - then B is opened - you would be right to change.

That's it, simple. Two out of three times you would be right to change.

Intuition, opinions, feelings, common sense, none of that matters.

Cheers

--
Clive

LOL!

Rick C.

-- Get 6 months of free supercharging -- Tesla referral code -

Yes, I was thinking that a person will pick up and toss a coin in a fairly repeatable manner, so the way it lies can affect the way it lands next throw. I expect that people could learn - consciously or unconsciously - to do that

Maybe some one has quantified that by experiment. "Too small to be noticed" would be zero statistical effect. A few tents of a per cent wouldn't surprise me.

--

John Larkin   Highland Technology, Inc   trk 

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

Knife throwers can be pretty impressive that way.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com

The key to the N-door Monty Hall problem is that there's no way for you to know in advance which of the doors will be left besides the one you picked, and so when you get down to two doors, the problem is not nearly so symmetrical as it appears.

Your original door is always 1/N, but when N-2 doors have been opened, the (initially unknown) remaining door picks up the remaining 1-1/N. So switch. ;)

Any two doors chosen at random at the beginning will have the same probability, but that other door hasn't been chosen randomly.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com

h:

ties:

hange.

I've always had problem with this argument. I think it needs to be stated as (game show host must know where the prize is):

You choose A - then B is opened - you would be wrong to change. You choose A - then C is opened - you would be wrong to change. You choose B - then C is opened - you would be right to change. You choose C - then B is opened - you would be right to change.

Seems to me the odds of changing thereby improving your situation would be 50|50.

John :-#)#

Hi,

I think you are missing one detail, the choosing of A is 1/3 of the time not 1/2, even though there are two possibilities of B or C being opened it is still 1/3 not 1/2.

cheers, Jamie

I am not going to try to guess the level here, but I expect it to be very small. I am just agreeing with you on reasoning why it might be a non-zero effect, even if it may in practice be a tiny effect.

only if you guess A twice as often as you guess B or C

--
  When I tried casting out nines I made a hash of it.

They have to not only aim, but get exactly the right number of rotations. Hitting handle-first won't do.

So one could learn at coin tossing, and maybe even dice tossing.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

In practice it would be infinitesimally small and I can't see how it could be measurable. The only possibility would be a poor measurement coming up with a false positive result.

I think the point is that there is virtually nothing repeatable about the coin toss. The person can think they did it exactly the same, but only a very tiny difference would produce a different result.

I'm really surprised Larkin is saying there would be a measurable result since this is the penultimate chaotic experiment and Larkin is such a believer that chaotic systems can't result in predictability.

LOL!

Rick C.

-+ Get 6 months of free supercharging -+ Tesla referral code -

o s

one

u -

Knife throwing is trivial compared to coin tossing. In knife throwing you just have to hit your mark with the right number of rotations. In coin tos sing you have to do both of the above, but to an accuracy many times higher . If the rotation is a fraction of a degree off or the trajectory is a fra ction of an inch off the coin bounce will magnify the difference and result in a potentially different outcome.

It would be interesting to create a simulation with the two dimensions of r otation angle and speed of impact and plot the resulting heads/tails as bla ck vs.white pixels in 2D. It might result in something like the Mandelbrot set but without the color.

LOL!

Rick C.

+- Get 6 months of free supercharging +- Tesla referral code -

Martin Brown wrote in news:q29oci $19ho$ snipped-for-privacy@gioia.aioe.org:

And to think I was touting you as one of the more intelligent in this group.

And then you had to go and jump on the RETARDED KRW bandwagon.

That pretty much negates any modicum of intelligence you may have ever used here.

Fuck you, Brown. Even your name is generic.

But stats? You don't know shit if you are pissing and moaning about coin face protrusion 'differences' causing a coin toss result invariance.

That statement by you, pretty much means you ain't real bright, boy.

The 'always wrong', krw childish crack, pretty much means that you are dumber than Donald J. Trump.

John Larkin wrote in news: snipped-for-privacy@4ax.com:

Then you are a true idiot. There are no folks out there tossing yours, mine or even their own coin in ANY repeatable manner.

It just ain't there.

And neither is you guys attempts at nit picking the physics here.

There is NO coin toss that has ANY influence on ANY subsequent toss, EVER!

And no toss is predictable, even if a machine loaded with the same coin at the same idexing and same drop height did it over and over.

It would STILL NOT BE PREDICTABLE.

YOU DIG YET???

Phil Hobbs wrote in news: snipped-for-privacy@supernews.com:

A knife throw is almost more predictable and IS more doable than a coin toss.

It is like spot shots on a pool table or free throws at the foul line.

The best expert can still miss.

Funny... there are exactly ZERO "coin tossers" out there. Nope... not a one.

John Larkin wrote in news: snipped-for-privacy@4ax.com:

Nope. Dice roll is ALSO 100% random outcome, EVERY TIME.

Coin toss: 100% random outcome, 100% of the time.

PERIOD, boys.

It is like claiming you can roll a pool ball off two rails and make it stop with the number pointing straight up.

I suggested that a toss outcome could be correlated with the previous state, not that a toss affects the future.

I could pick up a coin and drop it flat: a zero-order toss. The end position would be highly correlated to the previous state.

Give it a half-flip in the air and it would still be correlated.

If a person were reasonably consistent, the number of flips would have a statistical peak.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

*I* named you Always Wrong. KRW just acknowledged my judgement.

You called yourself Massive Prong, which I evolved to Mini Thong and then settled on the more descriptive Always Wrong.

If you had a real name, we could use that instead.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics