Is this cool with you?

gAr · 2011-03-29 02:19:10

Hi bobbym,

Ok, I understand! Thanks.
Can we extend it to more than 2 players?

bobbym · 2011-03-29 02:21:53

Hi gAr;

I do not think so. We can have the positive exponents represent one player and the negative for the other. There just is not any other choices.

gAr · 2011-03-29 02:29:54

Hi bobbym,

Ok, thank you.
I too could not find a way.

bobbym · 2011-03-29 02:38:03

I guess if you had 3 you might be able to do the problem 2 at a time.

gAr · 2011-03-29 02:45:04

Or can we have 3 different variables in the generating functions, and then add the coefficients as we require?

bobbym · 2011-03-29 02:51:56

Yes, we could. But that would get complicated. The beauty of the powers and the reciprocals is that they cancel out automatically.

Take this one for example:

This says that B loses by exactly 11 strokes 1975 / 308367 percent of the time. We do not have to think about the answer, the GF's do it for us.

gAr · 2011-03-29 02:59:20

Yes, I too like it.
By the way, nice problem!

bobbym · 2011-03-29 03:08:58

Thanks glad you looked at it.

bobbym · 2011-03-31 19:00:52

New Problem!

D poses this one. A guy has 5 bananas, 5 pears, 5 apples and 5 oranges. How many different arrangements of 4 groups can he make?

A says) Using Sylow p subgroup I come up with 1460.
B says) I remember them but I do not see how they apply here. Anyway that answer is wrong. The correct answer is...
C says) Hold on B, do not just blurt it out, I am still working on it.
D says) Hold it guys, I got hungry and ate the fruit. Now there are 0 ways to arrange the fruit.

Ignoring D's answer what do you get?

gAr · 2011-04-05 21:12:47

Hi bobbym,

Did not see your post.

How many different arrangements of 4 groups can he make?

Does this mean arrangements in groups of 4?
We can use exponential generating function which you taught me!

bobbym · 2011-04-06 03:10:06

Hi gAr;

Very good! Yes, wanted to see if you remembered that.

gAr · 2011-04-06 03:45:08

Hi bobbym,

Thanks.

Anyway, the

bobbym · 2011-04-06 03:58:38

Kayrect!

Generating Functions,Generating Functions,Generating Functions,Generating Functions,Generating Functions,Generating Functions,Generating Functions,Generating Functions!!!!

Hip hip hooray
Use them
And all our combinatorics problems go away.

gAr · 2011-04-06 04:08:12

Yep, they are sooooo coooooooooooool!!!

bobbym · 2011-04-06 13:31:00

New Problem!

C says if two people play russian roulette they have an even chance to die. He explains the rules that 1 bullet is put into the cylinder that can hold six shells. They spin the cylinder each time before firing. They take turns pointing the gun at there own head and pulling the trigger! They are obviously imbeciles.

A says) Sounds reasonable to me. Not even much of a problem to pose. Yes they have an even chance.

B says) Nope, the first player, meaning the first guy to point the gun at his own head is a big underdog.

C says) I posed it and they have an even chance.

D says) If I have to play I want to go first you know to get it out of the way.

How big is the second players advantage?

gAr · 2011-04-06 17:33:16

Hi bobbym,

gAr · 2011-04-06 18:14:26

bobbym · 2011-04-06 18:44:27

Hi gAr;

That is correct! Very good! It seems there is always another way.

gAr · 2011-04-06 18:48:54

Hi bobbym,

Yes.
These are the only methods I know so far!

bobbym · 2011-04-06 18:53:40

Hi;

That is all I can think of too.

gAr · 2011-04-06 21:57:07

Hi bobbym,

I thought of a variation of your question: (Actually, this is how I misunderstood your question before reading it again!)

There are 2 players with one pistol each.
The cylinders are fully loaded in the beginning.
Both of them have an accuracy of 50 percent.
They spin the cylinder each time before firing. They take turns to shoot the other person. ( After pulling the trigger, other player gets the chance )
i) What is the probability that second player survives?
ii) What is the probability that both survive?

I thought markov chain would be easy here too, but it's not!

bobbym · 2011-04-06 23:17:27

Hi gAr;

Markov chains have some problems without replacement and this is what your problem is. For both to have a chance to live they must run out of bullets.

I solved it like this:

gAr · 2011-04-07 00:10:23

Hi bobbym,

Thanks for the solution!
You know what, the markov chain which I thought was incorrect gave the same answer!

Three cheers to Markov Chain!

bobbym · 2011-04-07 00:18:31

Hi gAr;

That is two ways then. Thanks for working on it.

gAr · 2011-04-07 00:27:43

Your answer helped me to verify, thanks.
Though, I'm not fully convinced with the transitions in the chain!

Math Is Fun Forum

#426 2011-03-29 02:19:10

Re: Is this cool with you?

#427 2011-03-29 02:21:53

Re: Is this cool with you?

#428 2011-03-29 02:29:54

Re: Is this cool with you?

#429 2011-03-29 02:38:03

Re: Is this cool with you?

#430 2011-03-29 02:45:04

Re: Is this cool with you?

#431 2011-03-29 02:51:56

Re: Is this cool with you?

#432 2011-03-29 02:59:20

Re: Is this cool with you?

#433 2011-03-29 03:08:58

Re: Is this cool with you?

#434 2011-03-31 19:00:52

Re: Is this cool with you?

#435 2011-04-05 21:12:47

Re: Is this cool with you?

#436 2011-04-06 03:10:06

Re: Is this cool with you?

#437 2011-04-06 03:45:08

Re: Is this cool with you?

#438 2011-04-06 03:58:38

Re: Is this cool with you?

#439 2011-04-06 04:08:12

Re: Is this cool with you?

#440 2011-04-06 13:31:00

Re: Is this cool with you?

#441 2011-04-06 17:33:16

Re: Is this cool with you?

#442 2011-04-06 18:14:26

Re: Is this cool with you?

#443 2011-04-06 18:44:27

Re: Is this cool with you?

#444 2011-04-06 18:48:54

Re: Is this cool with you?

#445 2011-04-06 18:53:40

Re: Is this cool with you?

#446 2011-04-06 21:57:07

Re: Is this cool with you?

#447 2011-04-06 23:17:27

Re: Is this cool with you?

#448 2011-04-07 00:10:23

Re: Is this cool with you?

#449 2011-04-07 00:18:31

Re: Is this cool with you?

#450 2011-04-07 00:27:43

Re: Is this cool with you?

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