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The Pigeonhole Principle

Started by Dyksexlic!, Dec 26, 11:22 AM 2010

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Dyksexlic!

Disclaimer - "The following thread contains mathematical ideas which you may not agree with. If this is likely to cause you anxiety, please refrain from reading any further. If you continue, you implicitly acknowledge that Dyksexlic is not responsible for any harm caused to your cerebral cortex."


How to prove ANYTHING with the Pigeonhole Principle:

There are FOUR steps involved.

1) Decide what the 'pigeons' are. They will be the things that youââ,¬â,,¢d like several of to have some 'special property'.

2) Set up the 'pigeonholes'. You want to do this so that when you get two 'pigeons' in the same 'pigeonhole', they have the property you want. To use the Pigeonhole Principle, it is necessary to set things up so that there are fewer 'pigeonholes' than 'pigeons'.

(Sometimes, the way to do this relies on some 'astute observation'. )

3) Give a rule for assigning the 'pigeons' to the 'pigeonholes'. It is important to note that the conclusion of the Pigeonhole Principle holds for any assignment of 'pigeons' to 'pigeonholes', so it holds for any assignment you describe. Pick the rule so that when enough 'pigeons' occupy the same 'pigeonhole', that collection has the property you want.

4) Apply the Pigeonhole Principle to your system and get the desired conclusion.


:thumbsup:
And The Truth Shall Set You Free. .

ScoobyDoo

Well, that's all well and good if you have pigeons but do you apply this concept to roulette?

Please give us some specific examples of how to use it with roulette. It would be most helpful, since I feel confident that many of us reading your post, have no idea what on earth you are referring to. It sounds interesting though.

Scooby Doo

Dyksexlic!

Quote from: ScoobyDoo on Dec 26, 01:41 PM 2010
Well, that's all well and good if you have pigeons but do you apply this concept to roulette?

The 'pigeons' are merely a mathematical metaphor.

Consider the principle a 'conceptual visualisation tool'.

A 100% Mathematically guaranteed roulette bet is possible with the correct application.

Quote from: ScoobyDoo on Dec 26, 01:41 PM 2010
Please give us some specific examples of how to use it with roulette. It would be most helpful, since I feel confident that many of us reading your post, have no idea what on earth you are referring to. It sounds interesting though.

Scooby Doo

This thread is ONLY concerned with the Pigeonhole Principle itself.

For 'winning bet selections', kindly refer to Norman Bates... [Sarcastic smile  :twisted:]!

However, consider the following example..

QUESTION -
Prove that in a streak of 10 LOW roulette spins {1, 2, . . . , 18}, the selection
includes integers a and b such that a|b (that is, a divides b ââ,¬â€œ there exists an integer k such
that ak = b).


ANSWER -
Let the 'pigeons' be the 10 spins selected.

Define nine 'pigeonholes' corresponding to the odd spins 1, 3, 5, 7, 9, 11, 13, 15, and 17.

Place each spin selected into the pigeonhole coresponding to its largest odd divisor (which must be one of 1, 3, 5, . . . , 17).

Notice that if x gets placed in the pigeonhole corresponding to the odd spin m, then x = 2km for some spin k ≥ 0.

Since 10 spins are selected and placed in nine pigeonholes, some pigeonhole contains two spins a and b, where a < b.

Suppose this pigeonhole corresponds to the odd spin t. Then, a = 2rt and b = 2st, where are < s, so that a2s−r = b.

Since s−r is a positive spin, it follows that a|b.

Welcome to the Magical Land of Mathematics..!

8)

And The Truth Shall Set You Free. .

Fripper

You lost me in the question but it was a nice read tho.

Maybe if I had read it in my own language it may have been easier to understand  ::)

Thanks for sharing the pigeon principle  :thumbsup:
All i'm doing is living my life.

Dyksexlic!

Quote from: Fripper on Dec 26, 03:45 PM 2010
You lost me in the question but it was a nice read tho.

Maybe if I had read it in my own language it may have been easier to understand  ::)

Thanks for sharing the pigeon principle  :thumbsup:

You're welcome.    :thumbsup:

Mathematics sounds the same in ANY language.

The Pigeonhole Principle is a roulette players secret weapon.

In general, it may not be so clear how to apply the Principle.

Sometimes we need to cleverly construct the 'pigeons' and the 'holes'.

If we do this correctly, the proof should be slick.

Otherwise, the problem may seem forbiddingly difficult.

When stuck, do not give up so easily!

You learn and improve the most when you are stuck.

Keep thinking of possible approaches, perhaps for a few hours,

And you might be rewarded with an elegant solution.

This is the ONLY way to learn mathematical problem solving.

Cheers.
And The Truth Shall Set You Free. .

Ka2

Quote from: Dyksexlic! on Dec 26, 03:38 PM 2010
The 'pigeons' are merely a mathematical metaphor.

Consider the principle a 'conceptual visualisation tool'.

A 100% Mathematically guaranteed roulette bet is possible with the correct application.

This thread is ONLY concerned with the Pigeonhole Principle itself.

For 'winning bet selections', kindly refer to Norman Bates... [Sarcastic smile  :twisted:]!

However, consider the following example..

QUESTION -
Prove that in a streak of 10 LOW roulette spins {1, 2, . . . , 18}, the selection
includes integers a and b such that a|b (that is, a divides b ââ,¬â€œ there exists an integer k such
that ak = b).


ANSWER -
Let the 'pigeons' be the 10 spins selected.

Define nine 'pigeonholes' corresponding to the odd spins 1, 3, 5, 7, 9, 11, 13, 15, and 17.

Place each spin selected into the pigeonhole coresponding to its largest odd divisor (which must be one of 1, 3, 5, . . . , 17).

Notice that if x gets placed in the pigeonhole corresponding to the odd spin m, then x = 2km for some spin k ≥ 0.

Since 10 spins are selected and placed in nine pigeonholes, some pigeonhole contains two spins a and b, where a < b.

Suppose this pigeonhole corresponds to the odd spin t. Then, a = 2rt and b = 2st, where are < s, so that a2s−r = b.

Since s−r is a positive spin, it follows that a|b.

Welcome to the Magical Land of Mathematics..!

8)



Sorry but this doesnt prove anything. It's obvious that in 10 low numbers from 1 to 18, there would be at LEAST (pigeonwise) 1 number that is even or odd. You only need 1 even number to make ak = b count.

NoBody

Hi all,

I notice that the first post from regarding the pigeonhole principle was deleted.

Does anyone has any latest tips, hints or post from Dysexlic?

Do PM me, I am interested.

Thank you so much! ^. ^

Regards,

NoBody

VLS

Hi There NoBody, we did experience database corruption but it has been addressed.

You should see all posts recovered now.

Anything else, just shout.

Cheers!
Vic
🡆 ROULETTEIDEAS․COM, home of the RIBOT FREE software bot, with GIFTED modules for the community! ✔️

Ka2

Quote from: NoBody on May 03, 09:31 PM 2011
Hi all,

I notice that the first post from regarding the pigeonhole principle was deleted.

Does anyone has any latest tips, hints or post from Dysexlic?

Do PM me, I am interested.

Thank you so much! ^. ^

Regards,

NoBody


My advice is. Forget about this Idiot. His system does not work and can not work for several simple reasons. I was blinded by the pigeonhole theory at the beginning because it was mathematically sound. I can still kick myself in the head for putting so much time in investigating his ideas.

Luckily I'm much wiser now. And gave up on roulette a long time ago. I would suggest all of you to do the same, but I'm affraid that advice is just for deaf man's ears  :)

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