Is this cool with you?

gAr · 2011-03-03 15:03:13

Hi bobbym,

Yes, I too like that proof, good one!

This prob came up somewhere else:
A coin with probability p of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is {2,4,6,8,10...} is 2/5, then find p:
B solves it like this:
The chance that a head comes up on an even toss is:
Now sum that series:
Set the sum to 2/5 and solve for p.
Solve the equation any way you can, to get p = 1/3
Has B solved the problem?

The solution is correct, isn't it?

And please check post #300 too

bobbym · 2011-03-03 18:50:12

Hi gAr;

bobbym · 2011-03-11 16:45:46

A has had it with B. All that group theory has turned A into a meanie. A decides to prove that B is overrated. A poses this problem:

The points (2,2),(6,8),(10,9),(15,12),(18,6),(14,3),(8,1) form a 7gon. (8,1)->(2,2) closes it. What is the area enclosed by this 7gon?

A says) Solve that you blowhard, you fake, you fraud!

B chuckles, he knows he can use geogebra and get it quickly thus delivering a hard slap to A. He fires it up and to his horror, it is not working. B thinks, I will just reinstall it. Woops he does not have the installation program on his hard drive. Well then B will go download it, but unfortunately his internet is knocked out. To make matters worse the power goes out and B cannot use his beloved Mathematica. Now B is sweating, then he remembers a method that only requires a pen and paper. He feverishy calculates and delivers an answer to the benumbed A.

B says) 94 + 1 / 2 square units.
C says) I got that too.
D says) Yep!

How did all 3 of them get the answer with just a pencil and paper and so quickly?

gAr · 2011-03-11 18:43:26

Hi bobbym,

bobbym · 2011-03-11 19:25:00

Hi gAr;

gAr · 2011-03-11 20:05:56

Hi bobbym,

That's a nice method too.

bobbym · 2011-03-12 00:08:44

Hi gAr;

Here is Ramanujan formula I just came across.

Where C is the circumference of an ellipse and a,b are the major and minor semiaxes. When I was kid coming up with an approximation of the circumference of an ellipse was one of the first problems I worked on. I do not remember what I came up with but it sure was not that!

gAr · 2011-03-12 01:16:13

Hi bobbym,

That's a nice formula, haven't seen till now.
I first thought about arc length of the ellipse when taking a course in calculus. I couldn't find a formula, of course!

bobbym · 2011-03-12 01:22:13

Not everybody is a Ramanujan. Most of us go around never finding anything.

gAr · 2011-03-12 23:17:58

Hi bobbym,

Yes, but there's a moment just before finding a solution. So anyone who's trying shouldn't give up too soon.

bobbym · 2011-03-12 23:23:57

Hi gAr;

Whenever I have one that I cannot solve I just put it away for the future. Some of them have been in there for more than 12 years.

gAr · 2011-03-12 23:35:35

Cool, I note down and forget where I noted it down!
Can you post the ones you tried, if you don't mind?

bobbym · 2011-03-13 00:07:03

Hi gAr;

Here are some of them that stand out.

1) Get the integral to 100 digits.(solved)

Would be nice to have another way and need to identify the value in terms of other known constants.

2) A regular deck is spread face up on a table. What is the chance that 3 adjacent cards are of the same suit?
( Solution shown to me but I could not understand it )

3) The Alias Smith and Jones problem. One week a character on the show arranged the following bet. He was to be dealt 25 playing cards and he was to arrange them into 5 pat hands. He must use all 25 cards. A pat hand is one in which you would not change and consists of either

a straight (3,4,5,6,7) of mixed suits
full house 2,2,2,3,3
flush all 5 card the same suit.
four of a kind 5,5,5,5,j or 5,5,5,5,3 or 7,7,7,7,2... 5 th card could be anything. The character in the show said it worked more than 9 times out of ten. What is the actual probability?

4) What is the front digit of:

5) 3 cards are drawn from the unit interval {1,2,3,4,...,n} with replacement and added together. Then 2 more cards are drawn with replacement and added what is the probablity that the 2 card sum is greater than the 3 card sum as n approaches infinity. (Solved but would like to see another answer )

6) Six cards are drawn from a deck. what is the mean of the 3 highest cards. ( I have a simulation solution.)

gAr · 2011-03-13 21:49:23

Hi bobbym,

That's a nice set of problems!

I haven't programmed for numerical methods, which method do you consider good at controlling errors ( for integration ) ?
Is it RungeKutta method?

bobbym · 2011-03-13 21:56:09

Hi gAr;

Runge Kutta can be viewed as an integration technique but it is primarily used for the numerical solution of DE's

Numerical integration is as much an art as a science. The control of errors or Numerical Analysis is a huge field with thousands of books. For numerical integration no one method will suffice. I think the preferred method today is Romberg Integration or some Adaptive scheme, I do not use it much. I prefer Gaussian integration.

gAr · 2011-03-13 22:05:54

Hi bobbym,

Good to know about it. I'll check about the methods.

bobbym · 2011-03-14 00:28:30

Hi gAr;

Those are not challenge problems, I only posted them because you asked. They are very tough.

gAr · 2011-03-14 00:42:39

Hi bobbym,

I can see that they are tough. Still trying to figure out how to start.
Thanks for the problems.

bobbym · 2011-03-14 01:14:05

Perhaps with a fresh insight... Maybe you could.

New problem!

10 pigeons settle on a seesaw. Seven of them are on left end, A and three of them are on the right end, B. Every minute all ten pigeons make a decision to go to either A or B. What is the chance that all 10 pigeons end up at A at the same time?

A says) Clearly a gamblers ruin problem. The chance they all end up at A is 7 / 10.
B says) That is impressive but incorrect, the chance is 1 / 2.
C says) I ran a simulation and A is right.
D says) B is definitely correct.

gAr · 2011-03-14 02:02:44

Hi bobbym,

bobbym · 2011-03-14 02:07:36

Hi gAr;

gAr · 2011-03-14 02:11:31

Oh, I see. I'll try again.

gAr · 2011-03-14 06:16:18

Hi bobbym,

Last edited by gAr (2011-03-14 06:19:01)

bobbym · 2011-03-14 09:58:07

Hi gAr;

gAr · 2011-03-14 15:03:00

Hi bobbym,

Good answer!
If we assume game ends only when all the birds are on side A, there'll be only one absorbing state and the probability of absorption is 1.

Math Is Fun Forum

#301 2011-03-03 15:03:13

Re: Is this cool with you?

#302 2011-03-03 18:50:12

Re: Is this cool with you?

#303 2011-03-11 16:45:46

Re: Is this cool with you?

#304 2011-03-11 18:43:26

Re: Is this cool with you?

#305 2011-03-11 19:25:00

Re: Is this cool with you?

#306 2011-03-11 20:05:56

Re: Is this cool with you?

#307 2011-03-12 00:08:44

Re: Is this cool with you?

#308 2011-03-12 01:16:13

Re: Is this cool with you?

#309 2011-03-12 01:22:13

Re: Is this cool with you?

#310 2011-03-12 23:17:58

Re: Is this cool with you?

#311 2011-03-12 23:23:57

Re: Is this cool with you?

#312 2011-03-12 23:35:35

Re: Is this cool with you?

#313 2011-03-13 00:07:03

Re: Is this cool with you?

#314 2011-03-13 21:49:23

Re: Is this cool with you?

#315 2011-03-13 21:56:09

Re: Is this cool with you?

#316 2011-03-13 22:05:54

Re: Is this cool with you?

#317 2011-03-14 00:28:30

Re: Is this cool with you?

#318 2011-03-14 00:42:39

Re: Is this cool with you?

#319 2011-03-14 01:14:05

Re: Is this cool with you?

#320 2011-03-14 02:02:44

Re: Is this cool with you?

#321 2011-03-14 02:07:36

Re: Is this cool with you?

#322 2011-03-14 02:11:31

Re: Is this cool with you?

#323 2011-03-14 06:16:18

Re: Is this cool with you?

#324 2011-03-14 09:58:07

Re: Is this cool with you?

#325 2011-03-14 15:03:00

Re: Is this cool with you?

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