wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> hard >> A wannabe fair coin
        
(Message started by: BenVitale on Feb 4th, 2008, 10:11pm)
Title: A wannabe fair coin
Post by BenVitale on Feb 4th, 2008, 10:11pm
Suppose we have a coin that "tries" to be fair. To be more specific, if we flip the coin n times, and have X heads, then the probability of getting a heads in the (n+1)st toss is 1-(X/n).

[The first time we flip the coin, it truly is fair, with p=1/2.]

A short example is in order. Each row here represents the i-th coin toss, with associated probability and its outcome:

p=1/2: H
p=0: T
p=1/2: H
p=1/3: T
p=1/2: T
p=3/5: T

It certainly would seem that the expected value of X should be n/2. Is this the case? If so (and even if not), then think of this coin as following a sort of altered binomial distribution.

How does its variance compare to that of a binomial [=np(1-p)]?

Does the coin that tries to be fair become unfair in the process? Or does it quicken the convergence to fairness?

N.B. If the algebra becomes intractable, computer solutions are allowable.
Title: Re: A wannabe fair coin
Post by Eigenray on Feb 4th, 2008, 11:06pm
The expected value of X is n/2.  One way to see this is that given any sequence of n flips, with k heads, the sequence obtained by negating each flip is equally probable, and has n-k heads.  It looks like Var(X) = n/12 for n > 2, which is exactly one third that for a truly fair coin.  Attached is a plot for the probability distribution for X heads out of 100 (fair coin in blue, 'wannabe fair' in red).
Title: Re: A wannabe fair coin
Post by Eigenray on Feb 4th, 2008, 11:30pm
Actually the proof for the variance is not difficult.  Let p(n,k) be the probability of getting k heads out of n throws.  Define the moment

M(r,n) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif kr p(n,k).

Since p(n,k) = p(n,n-k), we have

M(1,n) = 1/2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif [k + (n-k)] p(n,k) = n/2,
M(3,n) = 1/2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif [k3 + (n-k)3] p(n,k) = 1/2[ n3 - 3n2(n/2) + 3n M(2,n) ].

We also have

p(n+1,k) = k/n p(n,k) + (1-(k-1)/n) p(n,k-1).

Multiplying by k2, and summing over all k, gives an expression for M(2,n+1) in terms of M(r,n), r=0,...,3, and inductively we can show M(2,n) = n2/4 + n/12.  Thus Var(X) = M(2,n) - M(1,n)2 = n/12.
Title: Re: A wannabe fair coin
Post by BenVitale on Feb 5th, 2008, 2:54pm
Thanks.
Title: Re: A wannabe fair coin
Post by BenVitale on May 29th, 2008, 4:02pm
Dice can be loaded -- that is, one can easily alter a die so that the probabilities of landing on the six sides are dramatically unequal.

Is it possible to bias a coin-flip, that is to weight a coin so that it is substantially more likely to land “heads” than “tails” when flipped and caught in the hand in the usual manner?

[Conditions of the experiment: the coin is not allowed to bounce or be spun but rather simply flipped in the air.]
Title: Re: A wannabe fair coin
Post by william wu on May 29th, 2008, 7:16pm

on 05/29/08 at 16:02:44, BenVitale wrote:
Dice can be loaded -- that is, one can easily alter a die so that the probabilities of landing on the six sides are dramatically unequal.

Is it possible to bias a coin-flip, that is to weight a coin so that it is substantially more likely to land “heads” than “tails” when flipped and caught in the hand in the usual manner?


This might not be the kind of answer you want, but one way of biasing coin-flips is to train your hand. Mathematician/magician Persi Diaconis has allegedly trained himself to flip a coin so that it comes up heads 10 out of 10 times. He has also done research on whether the outcomes of coin flips are really equally likely, even when you're not deliberately trying to bias it. His research suggests that coins are more likely to come up the same way they started. A reference can be found here: http://comptop.stanford.edu/preprints/heads.pdf
Title: Re: A wannabe fair coin
Post by BenVitale on May 30th, 2008, 10:21am

on 05/29/08 at 19:16:41, william wu wrote:
This might not be the kind of answer you want, but one way of biasing coin-flips is to train your hand. Mathematician/magician Persi Diaconis has allegedly trained himself to flip a coin so that it comes up heads 10 out of 10 times. He has also done research on whether the outcomes of coin flips are really equally likely, even when you're not deliberately trying to bias it. His research suggests that coins are more likely to come up the same way they started. A reference can be found here: http://comptop.stanford.edu/preprints/heads.pdf



Thank you very much William. That does help indeed.

This is the other part of the problem I was given:

You have a coin.
The coin is biased.
You don't know which way it's biased or how much it's biased. All I can say is that the coin is biased.
This is all the information I'm willing to give you.

So, you draw the coin, flip it, and slap it down on a table before you.

Now, before you look at the result, I ask you, "Are you willing to assign a 1/2 probability to the coin having come up heads?

So, we can answer 'yes', we can assign a 1/2 probability to the coin.

Your thoughts, please.
Title: Re: A wannabe fair coin
Post by towr on May 30th, 2008, 12:17pm

on 05/30/08 at 10:21:09, BenVitale wrote:
Your thoughts, please.
I'd adjust the probability I'd assign the coin with each toss. Of course for the first toss, without any information about how it may be biased, I may as well assign 50% probability.
Title: Re: A wannabe fair coin
Post by BenVitale on May 30th, 2008, 2:56pm
We don't know which way the coin is biased on this one occasion. Does it matter?

A Bayesian accepts uncertainty. A coin has to come either heads or tails. A Bayesian plans for 50% state of uncertainty where he doesn't weigh outcomes conditional on heads any more heavily in my mind than outcomes conditional on tails.

But, if you are not a Bayesian, you could argue that if we say that flipping a coin has a probability of 50% means that that the coin has an inherent propensity to come up heads as often as tails. But we know that the coin is biased, so it cannot have a probability of 50%.

But the thing is, even if you're holding a fair coin in your hand, if you say that the coin has inherently 50% probability of coming up heads is wrong.
Title: Re: A wannabe fair coin
Post by towr on May 30th, 2008, 3:09pm

on 05/30/08 at 14:56:29, BenVitale wrote:
We don't know which way the coin is biased on this one occasion. Does it matter?
Yes, by symmetry I should be willing to pay as much to bet one way as for the other. So 2:1 odds are fair, in that sense.
(Of course this is a case of uncertainty vs probability; in a sense, knowledge of reality vs reality.)


Quote:
But the thing is, even if you're holding a fair coin in your hand, if you say that the coin has inherently 50% probability of coming up heads is wrong.
That may depend on what you mean by probability; although I'm not quite sure under what interpretation you can reconcile 'fair' with "not having 'inherently' 50% probability of coming up heads".
Title: Re: A wannabe fair coin
Post by BenVitale on May 30th, 2008, 3:48pm
It depends on how you're holding the coin, maybe the way you're holding the coin gives you heads more often than tails, given the force at which you flip it, and the air currents around you.

It's a different matter with an automated coin-flipper or using a computer flipping coin.
Title: Re: A wannabe fair coin
Post by BenVitale on May 31st, 2008, 11:34am

on 05/30/08 at 12:17:52, towr wrote:
I'd adjust the probability I'd assign the coin with each toss. Of course for the first toss, without any information about how it may be biased, I may as well assign 50% probability.


And how many trials do you need to determine the probability, to assign a probability? With a fair coin we have no trouble assigning a 50% probability even though if I flip 10, 100 or 1000 times I won't get exactly a 50% probability.
Title: Re: A wannabe fair coin
Post by towr on May 31st, 2008, 12:27pm

on 05/31/08 at 11:34:51, BenVitale wrote:
And how many trials do you need to determine the probability, to assign a probability?
You can never, ever, determine the actual probability. But you can make a good estimation at every step.
With one toss, and given its a biased coin, best estimate would be you got the most likely side, so I'd go with 100% for that one; but only until the next toss.
Title: Re: A wannabe fair coin
Post by BenVitale on May 31st, 2008, 1:55pm
Suppose that you are not satisfied with 2 outcomes, heads and tails, and you want that coin to land on its edge, with a probability of x%.

So, we could have:

probability for heads h1%, probability for tails h2% and probability to land on its edge x%

(1)
What should then the thickness [minimum] of the coin be?

(2)
And what should the thickness [minimum] be to have h1 = h2 = x = 1/3?
Title: Re: A wannabe fair coin
Post by towr on May 31st, 2008, 3:02pm
That's very hard to say without finding out empirically. It's not as simple as, say, the taking the fractions of surface area.
Title: Re: A wannabe fair coin
Post by BenVitale on May 31st, 2008, 4:02pm
How about:

Source: http://en.wikipedia.org/wiki/Spherical_cap


coin's radius = a
2*pi*rh = (the coin-circumfering sphere's surface area) / 3
= 4*pi*r^2 / 3
h = 2r / 3

thickness of coin
= 2r - 2h
= 2r (1 - 2/3)
= 2r / 3

but that's in r (sphere's radius). we'll try to change the term into a.

a^2 + (r/3)^2 = r^2
a^2 = 8r^2 / 9
a = 2*sqrt.2 r / 3
r = 3a / 2*sqrt.2

thickness (in a)
= 2r / 3
= 2 [3a / 2*sqrt.2] / 3
= a / sqrt.2

this is all is by the assumption that the point on the sphere's that touches the landing surface first will determine the landing face on a coin. since we want all 3 surfaces (1 head, 1 tail, and 1 edge) to have a fair chance, we divide by 3.
Title: Re: A wannabe fair coin
Post by towr on Jun 1st, 2008, 6:59am
I'm not sure what you're trying to do there; are you approximating a coin by a sphere?
Title: Re: A wannabe fair coin
Post by BenVitale on Jun 1st, 2008, 11:32am

on 06/01/08 at 06:59:00, towr wrote:
I'm not sure what you're trying to do there; are you approximating a coin by a sphere?


Yes, i'm using approximations. But i'm not sure whether or not this makes sense. Last night, i used different sizes of small cylinder-shaped objects that one can find at a Hardware store. It works with some surfaces. With other surfaces, the chock is not well absorbed, then you have that object bouncing and rolling or falling on its side. And it depends on the altitude, on the height of the fall.

I was trying to think of a math and physics problem, here. Do you know if this kind of problem has been looked at, documented?
Title: Re: A wannabe fair coin
Post by Grimbal on Jun 2nd, 2008, 5:23am
This reminds me of a trick to favour one of the side while "flipping" a coin.

Instead of flipping the coin in the air, spin it on the edge until it falls on either side.

Here you have to see that a coin is a cylinder that has 2 edges, one around each face.  When you spin the coin, it spins on the one or the other edge.  And the edge that touches the table is likely to end up down.

So when you spin the coin, be careful to do so with the face you want slightly oriented up.  That works well on a very flat surface.


Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board