What do you think?

bobbym · 2011-02-20 08:06:47

Hi gAr;

My solution is different than yours, so I am having trouble understanding. Can you explain why 3 ... 24?

gAr · 2011-02-20 21:39:12

Hi bobbym,

We start by taking one diagonal :

Then we compute the other diagonal q, using that formula.
We take p=24, 23,22 ... 3.
Down till p=3, because for that value the parallelogram degenerates to a line segment.

On a second thought, I guess it is more accurate to start from p = 35. Again, I don't get an integral value for q !

bobbym · 2011-02-21 09:01:10

Okay, thank you for the explanation. I did it like this:

The parallelogram law says that the sum of the squares of the 4 sides must be equal to to sum of the squares of the diagonals:

There is only one way to represent the LHS as a sum of two squares and that is 3^2 + 35^2. By the triangle inequality 35 is impossible so there are no solutions.

gAr · 2011-02-21 16:49:27

Hi bobbym,

Yes, that's a nice way!

bobbym · 2011-02-21 22:39:28

Thanks gAr;

New problem:

A standard deck is dealt out one card at a time. Each time the card is replaced and the deck reshuffled. When the 4 different aces are dealt the game stops. What is the probability that it happens on the nth card?

A says) The answer is

B says) The answer is much more complicated than that.

C says) Use the hypergeometric distribution.

D says) I have seen B's approach he constructed a 44 x 44 absorbing chain. Pure genius! By the way A's approach is garbage.

E says) A is right!

gAr · 2011-02-21 23:35:29

Hi bobbym,

bobbym · 2011-02-22 08:08:56

Hi gAr;

gAr · 2011-02-22 15:26:15

Hi bobbym,

For n=5, what I think of the number of ways is:

|*|*|*| |*|
|*|*| |*|*|
|*| |*|*|*|
| |*|*|*|*|

where space represents any card except A's, and *'s are 4 different A's. They can be chosen as shown above.
Hence,
number of favorable outcomes = 48 * 4! * 4
sample space = 52^5
Therefore, required probability = 9/742586

bobbym · 2011-02-22 15:33:41

Hi gAr;

I do not think that gets all the possibilities:

Since you are returning the aces it could occur with 5 aces. For instance on n =4 you could have 2 aces of spades 1 Ace of clubs and 1 Ace of diamonds. Next card is the needed Ace of hearts.

Try a simulation that exactly models the problem.

gAr · 2011-02-22 15:55:27

Hi bobbym,

Did you derive any formula for that? I don't have much idea about simulations.

A second thought, since aces can repeat, required probability = 24*(51+50+49+48)/52^5 = 297/23762752.

bobbym · 2011-02-22 17:46:21

I am using a absorbing markov chain to solve it. I am thinking that it is

I got the idea from what you did! This matches the markov chains answer for n = 5. Only problem with this formula is theoretically it is possible for n = 100 or a 10000. Do we now count 52 + 51 + 50 + 49 + 48 + ...+ into negative numbers?

gAr · 2011-02-22 22:37:29

Hmmm, but I feel the problem is simple enough to use basic counting priciple!

For n = 5, ( There are 5 spaces, 5th space is the 4th distinct ace )
|*|*|*| |*| ⇒ Here, *'s are distinct Aces filled in 4! ways, the fourth space can be filled in 51 ways.
|*|*| |*|*| ⇒ 4! * 50
|*| |*|*|*| ⇒ 4! * 49
| |*|*|*|*| ⇒ 4! * 48

How did you get an extra 52?! The last ace cannot be repeated, isn't it?

bobbym · 2011-02-22 22:47:25

I was just playing with your formula and seeing what it would take to make what I think is the right answer.

I am not sure now what the right answer is. Since we replace the card and only draw one card at a time it seems like there is always 52 to choose from. What I am saying is that it is like you draw a card and mark down on a sheet of paper whether it is a new ace or not. The only fixed position is that the last card must definitely be the final ace, that is what ends the game.

You might be right, at this point I do not know.

gAr · 2011-02-22 23:02:14

Ok. For my answer to match with yours, the 5th draw wouldn't be a distinct ace. The game would end in the 4th draw.

Anyway, please get back to this when you believe you have the answer.

bobbym · 2011-02-22 23:09:20

Am I overcounting because in some, the game would end in 4 instead of 5. There are possibilities where it is not a distinct ace but it still ends in 5. For instance AS AS AC AD AH

Anyway, thanks for all your hard work. I enjoyed working with you on it so far.

gAr · 2011-02-22 23:26:43

Yeah, overcounting may be a reason. Only the last ace needs to be distinct.
Thanks to you, I got the idea from your reasoning.
And it's not hard work, it's a pleasure, nice way to forget the cruel world !
I too enjoyed working with you.

bobbym · 2011-02-23 17:00:16

Hi gAr;

Problem solved! The Markov chain I built counts the number of successes up to n. So when I got

It is the sum of P(n=4) + P(n=5)

So your answer of

is correct. If you will now state it as a formula for any n. I will give you a congratulations for a well done job!

gAr · 2011-02-23 17:31:50

Oh, cool!
For P(n=6), I get this:

It gets more tedious for higher value. There must be a pattern, I'll try.

bobbym · 2011-02-23 17:43:02

Hi;

That is correct for P(n=6).

gAr · 2011-02-23 18:08:59

Great!
I'm thinking of using partitions, so may not be able to come up with a formula for now.
Whew, what a problem!

bobbym · 2011-02-23 18:17:38

gAr · 2011-02-23 18:30:48

Awesome! I wanted to use recurrence, but couldn't derive it. Your post is indeed helpful.

bobbym · 2011-02-23 21:43:00

Hi gAr;

I have a total solution to the problem now. But it is an analytical solution and not a combinatoric one. There may not be a simple formula that involves a few binomials.

gAr · 2011-02-23 22:29:07

Hi bobbym,

Is it a solution other than the ones you posted? Analytical solution may be advanced for me, but would love to see it!
That recurrence and generating function was great! But I still cannot derive it.

bobbym · 2011-02-23 22:42:09

Hi;

It is not exactly different but it is what created the recurrence,GF and the formula. It does all the combinatoric reasoning and probability automatically. It is called a Markov chain. Originally I was working with a 44 x 44 matrix that represented the Markov chain. Thanks to a trick I was able to reduce it down to a 6 x 6. This is what it looks like with labels.

When we strip away the first row and first column which are just labels we get.

This is called an absorbing Markov chain.

If you raise M up to the nth power you will see the correct probability appear in row 1, column 5.

Math Is Fun Forum

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