Tricky Coin Flipping Probability Problem

gAr · 2011-08-16 17:33:04

Hi all,

I checked George,Y's solution and it matches mine.

He actually means this:

which matches exactly!

Thanks George!

Anyway, I haven't understood completely yet. I am still trying.

Last edited by gAr (2011-08-16 17:37:05)

bobbym · 2011-08-16 18:22:34

Hi gAr;

I think you found another way to do it.

Possibly he means this:

Left out any simplification that might hide his method. This is what I was looking for, for two days. A formula! No one had it.

gAr · 2011-08-16 18:40:48

Hi bobbym,

Okay.
His formula is spectacular, much less calculation's required!

I don't know stochastic calculus yet. Is stochastic calculus mostly used for fincance? I can't find out how it can be used here.

bobbym · 2011-08-16 18:45:02

Hi;

I do not agree. I favor your solution and here is why.

1) It was done using experimental ideas. This establishes the relationship between the Catalan numbers and the solution. Linking the problem to a lattice.

2) It provides a generating function, from which we might determine the moments, unimodality an asymptotic form and a recurrence.

The stochastic calculus is used for brownian motion. The Feller books covers it. But did not provide that formula.

gAr · 2011-08-16 19:00:09

Hi,

Hmmm, okay.

Anyway, I'm happy that there's another method.
Can't we have a simpler generating function from that?

?

Last edited by gAr (2011-08-16 19:39:54)

bobbym · 2011-08-16 19:53:55

Hi gAr;

Can there be two of them? There is a theorem for Taylor series that says there is only one Taylor series for a given sequence.

George,Y · 2011-08-16 19:55:35

Hi,

The reason why I deduce only half of Prob(X=50) is that there are 101 possible outcomes for X in a symmetry fashion.

P(X=0)=P(X=100) P(X=1)=P(X=99) ... P(X=49)=P(X=51)

P(X=0)+P(X=1)+P(X=2)+...+P(X=49) = P(X=51)+...+P(X=98)+P(X=99)+P(X=100)= 1-P(X=50)

So 1/2=P(X<50)+P(X=50)/2

George,Y · 2011-08-16 20:05:38

The reason why we need to double the P(R(100)<15) to get P(R(t)<15) is this:

Set R(t) as the total gains after t trials.

The probability of R(100)-R(t)>0 or R(100)-R(t)<0 is the same if R(t)=-15 for the first time and is nearly 0.5
(considering the case R(100)=R(t)).

-----E[ R(100)-R(t) ]=0

when the player losses all the 15 bucks some time <100, the game seems over.
However, if we let the player continue to play until time=100, the chances of winning back some r bucks and losing r bucks more are the same in the coin flip case. That approximately constitute the whole sample space and the total conditional probability is almost 1 (neglecting the r=0 case)

So

P{R(100)<-15} = P{R(t)<-15,R(100)-R(t)<0} = P{R(t)<-15}/2

gAr · 2011-08-16 21:36:52

Hi George,Y,

I'm finding it hard to understand your reasoning.
Your random variable is "number of wins", mine's "number of dollars". Maybe that's making a difference here.
I'll try to think your way and see if it makes sense to me.

Hi bobbym,

I modified my previous g.f to a probability g.f, now it's "decent"!

We need the sum of the probabilities, so it's simply:

bobbym · 2011-08-17 00:13:57

Hi gAr;

Very good! I am still in the testing phase for all these formulas.

gAr · 2011-08-17 00:43:37

Hi bobbym,

Okay!

bobbym · 2011-08-17 00:46:45

Hi gAr;

I forgot you already did that for some smaller problems. I think this one is done.

Thanks to gAr and George,Y the OP's question is answered.

gAr · 2011-08-17 01:03:51

Hi,

And thanks for your simulation, which gave some confidence with the result!

Now, we do not know whether the OP will ever see this?

bobbym · 2011-08-17 02:55:23

Not only is MathsIsFun it is also lonely. he may never see the answer.

gAr · 2011-08-17 03:06:52

Yes, that's true!

bobbym · 2011-08-17 03:09:17

Hi;

Sometimes posters never return because they post on another forum and get another answer. Although not as good. But it is strange to me why some would just forget about the problem they wanted an answer to.

gAr · 2011-08-17 03:17:57

Appears strange to me too.
Maybe they do not note down the sites where they've asked questions.

bobbym · 2011-08-17 03:27:30

Hi;

That is an idea. Maybe they decide that mathematics is too tough and they become bored. Most likely someone else gave the wrong answer but answered first.

gAr · 2011-08-17 03:40:17

Hi,

I searched a bit for this question.
Doesn't appear to be posted anywhere else?

Whatever... I'll note down the result! After a few days, this thread drowns down the horizon, with the formulas!

bobbym · 2011-08-17 03:43:40

Hi gAr;

I usually copy them. Later you will need this result. You never know when.

gAr · 2011-08-17 04:26:28

Hi bobbym,

Yes, that's right!

George,Y · 2011-08-17 14:58:31

gAr wrote:

Hi George,Y,
I'm finding it hard to understand your reasoning.
Your random variable is "number of wins", mine's "number of dollars". Maybe that's making a difference here.
I'll try to think your way and see if it makes sense to me.

Hi gAr
I used two variables:
One is number of wins as X in the total 100 trials.
One is the profit/loss R(t) after t times of trial.

Therefore R(100) = X-(100-X) = 2X-100, which is the link between the two.

gAr · 2011-08-17 16:43:52

Hi George,Y,

From #18, I understood it like this:
We are calculating P ( losses - wins > 15 ), which is the probability of going broke.

russellhq · 2011-08-18 04:58:25

I'm here!! I'm trying to digest all your formulae and get it into some programme code to run the calcs.

gAr · 2011-08-18 05:04:32

Hi russellhq,

Glad that you are back!

Math Is Fun Forum

#26 2011-08-16 17:33:04

Re: Tricky Coin Flipping Probability Problem

#27 2011-08-16 18:22:34

Re: Tricky Coin Flipping Probability Problem

#28 2011-08-16 18:40:48

Re: Tricky Coin Flipping Probability Problem

#29 2011-08-16 18:45:02

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#30 2011-08-16 19:00:09

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#31 2011-08-16 19:53:55

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#46 2011-08-17 04:26:28

Re: Tricky Coin Flipping Probability Problem

#47 2011-08-17 14:58:31

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#48 2011-08-17 16:43:52

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