Chess with Maths

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Mr_Mathematician
MARattigan wrote:
Mr_Mathematician wrote:
MARattigan wrote:

@Mr_Mathematician

English is an imprecise language and almost anything that is said depends on context. I think the great majority of readers would take the phrase "a queen of each colour" to mean a black queen and a white queen in this context.

I can't understand "Also the reason for doubling and answer without changing id is we are placing two queens one by one outside, it's not a chess tournament ." in any way, but doubling the figure 2576 in @Optimissed's calculation which has already accounted for the same positions that would be arrived at by starting instead with the black queen is incorrect.

Try it with a black and white knight on a chessboard reduced to two adjacent squares. If you come up with the answer 4 using the same reasoning, you need to keep working on it.

I may but you are seeing the final result as outcome but in questions there are no. Of ways

Let we fix black position first there are ways for white

And we again fix white position for black there are ways for black.

Ok let me clarify you more suppose there are three cities A B C we have to go to the another city D which have 3 ways going via A , B , C

Now a person would be 3 possibilities whether he would like to go with A , B or C cities finally he reaches on city D . So as we are observer, with context to your thoughts he would have must followed only one way since his final position is same No

The two opposite colours are taken to be equally likely even we would consider both of them.

W B

BW

are different, even in making most probable move by computers they also prefer them as separate

Probablity of Occurrence of each event is ½ here

This comes back to the point I made about context.

If the question, "how many ways are there of choosing 2 things from 6", or similar questions are asked then people expect the answer to be based on the final outcomes.

Of course you can say

I can take two with my left hand, that's 1 way,

I can take the first one with my left hand and the second with my right, that's 2 ways.

I can sniff the first one up against my left nostril then shake the rest up until one jumps out, that's 3 ways.

I can let the dog run through them and choose two that have moved furthest (with my right nostril), that's 4 ways.

I can glue on labels 1,2,...6 and stick pins in an old fashioned telephone directory until I've got two matches ...

And the answer may well come out differently.

But in the context people will generally assume that a count of final outcomes is what is required. If your question intends something else, it should make that explicit.

That is another case choosing 2 balls together , choosing 2 balls one by one answer will come up different

I agree that in these cases final outcomes but in chess board placing black after white or placing white after black is equally probable but you have to count both case

Ok let's go to probality suppose my question was find the probability of the same problem if white plays first answer would be 1/2 here in numerator 1 is the outcomes of only white but in the denominator there is the case of white + black both 

suppose now according you only we have to count one type of case either of black or white so your answer would be 1 ( probality) but this contradicts that fact that if I play with black then also we get same probality which is not possible ( an event cannot have probality greater than one here we are getting two by adding)

MARattigan
Mr_Mathematician wrote:
MARattigan wrote:
Mr_Mathematician wrote:
MARattigan wrote:

@Mr_Mathematician

English is an imprecise language and almost anything that is said depends on context. I think the great majority of readers would take the phrase "a queen of each colour" to mean a black queen and a white queen in this context.

I can't understand "Also the reason for doubling and answer without changing id is we are placing two queens one by one outside, it's not a chess tournament ." in any way, but doubling the figure 2576 in @Optimissed's calculation which has already accounted for the same positions that would be arrived at by starting instead with the black queen is incorrect.

Try it with a black and white knight on a chessboard reduced to two adjacent squares. If you come up with the answer 4 using the same reasoning, you need to keep working on it.

I may but you are seeing the final result as outcome but in questions there are no. Of ways

Let we fix black position first there are ways for white

And we again fix white position for black there are ways for black.

Ok let me clarify you more suppose there are three cities A B C we have to go to the another city D which have 3 ways going via A , B , C

Now a person would be 3 possibilities whether he would like to go with A , B or C cities finally he reaches on city D . So as we are observer, with context to your thoughts he would have must followed only one way since his final position is same No

The two opposite colours are taken to be equally likely even we would consider both of them.

W B

BW

are different, even in making most probable move by computers they also prefer them as separate

Probablity of Occurrence of each event is ½ here

This comes back to the point I made about context.

If the question, "how many ways are there of choosing 2 things from 6", or similar questions are asked then people expect the answer to be based on the final outcomes.

Of course you can say

I can take two with my left hand, that's 1 way,

I can take the first one with my left hand and the second with my right, that's 2 ways.

I can sniff the first one up against my left nostril then shake the rest up until one jumps out, that's 3 ways.

I can let the dog run through them and choose two that have moved furthest (with my right nostril), that's 4 ways.

I can glue on labels 1,2,...6 and stick pins in an old fashioned telephone directory until I've got two matches ...

And the answer may well come out differently.

But in the context people will generally assume that a count of final outcomes is what is required. If your question intends something else, it should make that explicit.

That is another case choosing 2 balls together , choosing 2 balls one by one answer will come up different

But not if you're counting final outcomes, which would be the normal interpretation.

I agree that in these cases final outcomes but in chess board placing black after white or placing white after black is equally probable but you have to count both case

Similarly. If you're counting final outcomes it doesn't matter which way round you place the pieces. Do you understand where the flaw was in @Optimissed's post? If not try the problem with two knights on a 2x1 board and you should see.

Ok let's go to probality suppose my question was find the probability of the same problem if white plays first answer would be 1/2 here in numerator 1 is the outcomes of only white but in the denominator there is the case of white + black both 

How you define the probability of a problem? And you removed the White to move specification - is it meant to be a purely combinitorial problem or are you envisaging playing on? (As I mentioned previously, not actually legal because the positions are dead.) 

suppose now according you only we have to count one type of case either of black or white so your answer would be 1 ( probality) but this contradicts that fact that if I play with black then also we get same probality which is not possible ( an event cannot have probality greater than one here we are getting two by adding)

You appear to be confusing the order in which the pieces are added to the board to make the position with side to move or some order of moves played from the position. You will have to clarify what you're saying if it is to make sense to me. If you want to include side to move as an attribute of what is to be counted then naturally the number of cases doubles, but if so I think you need to make that explicit in your question. Similarly if you wanted to include the ply count under the 75 move rule you need to multiply by 150, but again you would need to make that explicit.

Mr_Mathematician
MARattigan wrote:
Mr_Mathematician wrote:
MARattigan wrote:
Mr_Mathematician wrote:
MARattigan wrote:

@Mr_Mathematician

English is an imprecise language and almost anything that is said depends on context. I think the great majority of readers would take the phrase "a queen of each colour" to mean a black queen and a white queen in this context.

I can't understand "Also the reason for doubling and answer without changing id is we are placing two queens one by one outside, it's not a chess tournament ." in any way, but doubling the figure 2576 in @Optimissed's calculation which has already accounted for the same positions that would be arrived at by starting instead with the black queen is incorrect.

Try it with a black and white knight on a chessboard reduced to two adjacent squares. If you come up with the answer 4 using the same reasoning, you need to keep working on it.

I may but you are seeing the final result as outcome but in questions there are no. Of ways

Let we fix black position first there are ways for white

And we again fix white position for black there are ways for black.

Ok let me clarify you more suppose there are three cities A B C we have to go to the another city D which have 3 ways going via A , B , C

Now a person would be 3 possibilities whether he would like to go with A , B or C cities finally he reaches on city D . So as we are observer, with context to your thoughts he would have must followed only one way since his final position is same No

The two opposite colours are taken to be equally likely even we would consider both of them.

W B

BW

are different, even in making most probable move by computers they also prefer them as separate

Probablity of Occurrence of each event is ½ here

This comes back to the point I made about context.

If the question, "how many ways are there of choosing 2 things from 6", or similar questions are asked then people expect the answer to be based on the final outcomes.

Of course you can say

I can take two with my left hand, that's 1 way,

I can take the first one with my left hand and the second with my right, that's 2 ways.

I can sniff the first one up against my left nostril then shake the rest up until one jumps out, that's 3 ways.

I can let the dog run through them and choose two that have moved furthest (with my right nostril), that's 4 ways.

I can glue on labels 1,2,...6 and stick pins in an old fashioned telephone directory until I've got two matches ...

And the answer may well come out differently.

But in the context people will generally assume that a count of final outcomes is what is required. If your question intends something else, it should make that explicit.

That is another case choosing 2 balls together , choosing 2 balls one by one answer will come up different

But not if you're counting final outcomes, which would be the normal interpretation.

I agree that in these cases final outcomes but in chess board placing black after white or placing white after black is equally probable but you have to count both case

Similarly. If you're counting final outcomes it doesn't matter which way round you place the pieces. Do you understand where the flaw was in @Optimissed's post? If not try the problem with two knights on a 2x1 board and you should see.

Ok let's go to probality suppose my question was find the probability of the same problem if white plays first answer would be 1/2 here in numerator 1 is the outcomes of only white but in the denominator there is the case of white + black both 

How you define the probability of a problem? And you removed the White to move specification - is it meant to be a purely combinitorial problem or are you envisaging playing on? (As I mentioned previously, not actually legal because the positions are dead.) 

suppose now according you only we have to count one type of case either of black or white so your answer would be 1 ( probality) but this contradicts that fact that if I play with black then also we get same probality which is not possible ( an event cannot have probality greater than one here we are getting two by adding)

You appear to be confusing the order in which the pieces are added to the board to make the position with side to move or some order of moves played from the position. You will have to clarify what you're saying if it is to make sense to me. If you want to include side to move as an attribute of what is to be counted then naturally the number of cases doubles, but if so I think you need to make that explicit in your question. Similarly if you wanted to include the ply count under the 75 move rule you need to multiply by 150, but again you would need to make that explicit.

By this you proving concept of probability wrong 75 move rule is another thing that's not take place in random events

Mr_Mathematician

You must need to be hold good concepts on probability, permutation and combination then you will slowly realizes why to count 2 cases

How the variation can be more .

Elroch
MARattigan wrote:
Elroch wrote:
Mr_Mathematician wrote:

🎯Brain teaser '‼️

  • In how many way two queens shall be placed into 8×8 chessboard such that they never get cancel by each other ( there are only two queens on board place).Cancel means possibilities of capture.

Can you answer this question with logic

If it was rooks, I'd answer it with 'logic' - a rook always takes away 15 squares, so the answer is 64 * (64-15) / 2. The 2 is because the rooks are not distinguished.

For queens it's more a matter of counting. Classify squares based on the number of diagonal moves available. Then add up:

N_squares * (64 - 15 - N_diagonal_moves) / 2

where N_squares is the number of squares with a specific number of diagonal moves.

[In truth both are a combination of logic and counting, it's just that the first is narrower, with only one type of square].

To quibble, a rook takes away 15 squares if you assume the rooks cannot occupy the same square. That would be illegal, of course, but then it's illegal anyway.

It's not a quibble, it's the interpretation of the question. In the question "In how many ways can two rooks be placed on a chessboard so that they do not attack each other?" virtually everyone would understand the implicit assumptions that:

  1. each rook is placed on a single square - not overlapping them
  2. the two squares are distinct

You are suggesting the consideration of another question, which was neither intended nor understood by almost everyone who reads it.

MARattigan

Yes, I agree with that, but doesn't that mean it is indeed a quibble?

MARattigan
Mr_Mathematician wrote:

You must need to be hold good concepts on probability, permutation and combination then you will slowly realizes why to count 2 cases

How the variation can be more .

I have done my time in the kitchen with all three.

Suppose we are placing a white knight and a black knight on a 2x1 chessboard.

We can place the white knight as below

N .

. N

In each case there is 1 place to put the black knight giving

N n

n N

so you can count the number of cases as 1.1+1.1=2.

Now you want to double that, you say, because there are 2 knights.

That gives 4 cases.

SHOW ME THEM!

Mr_Mathematician
MARattigan wrote:
Mr_Mathematician wrote:

You must need to be hold good concepts on probability, permutation and combination then you will slowly realizes why to count 2 cases

How the variation can be more .

I have done my time in the kitchen with all three.

Suppose we are placing a white knight and a black knight on a 2x1 chessboard.

We can place the white knight as below

N .

. N

In each case there is 1 place to put the black knight giving

N n

n N

so you can count the number of cases as 1.1+1.1=2.

Now you want to double that, you say, because there are 2 knights.

That gives 4 cases.

SHOW ME THEM!

Case 1 :When white plays first in 2×1 after black

Fixing white on first ( a1 )

W□>> we have one ways to fill b1 ( let ) --1

Fixing white on b1

□ W >> we have left black place a1 ---2

Now Case 2:- When black plays first then after white

Fixing black on a1

B□ >> we have one ways to fill for white b1

Fixing black on b2

□B>> We have one ways to fill for white on a1

Note that cases got repeated but here sequence matters which we are playing first

MARattigan
Mr_Mathematician wrote:
MARattigan wrote:
Mr_Mathematician wrote:

You must need to be hold good concepts on probability, permutation and combination then you will slowly realizes why to count 2 cases

How the variation can be more .

I have done my time in the kitchen with all three.

Suppose we are placing a white knight and a black knight on a 2x1 chessboard.

We can place the white knight as below

N .

. N

In each case there is 1 place to put the black knight giving

N n

n N

so you can count the number of cases as 1.1+1.1=2.

Now you want to double that, you say, because there are 2 knights.

That gives 4 cases.

SHOW ME THEM!

Case 1 :When white plays first in 2×1 after black

Fixing white on first ( a1 )

W□>> we have one ways to fill b1 ( let ) --1

Fixing white on b1

□ W >> we have left black place a1 ---2

Now Case 2:- When black plays first then after white

Fixing black on a1

B□ >> we have one ways to fill for white b1

Fixing black on b2

□B>> We have one ways to fill for white on a1

Note that cases got repeated but here sequence matters which we are playing first

There is no connection between the order the pieces are assumed on the board in a method of counting the diagrams and any side to play. There is no mention of side to play in your current enunciation and in the original White to play was specified, so the assumption of separate cases depending on side to play is invalid on any sensible interpretation. You might as well double your cases by counting them as separate depending on whether it's sunny or raining or multiplying by 150 to account for the ply count (actually make that infinity because the positions aren't legal so no reason why the ply count should be).

The salient point is that the final outcomes (diagrams) in your Case 1 and Case 2 are the same, so if you count both you're double counting. A straightforward mathematical error.

You say here , "If the game has 2 queens you have to count for the both first and second that's the reason answer get doubled ." So which is it to be? Are you doubling because there are two queens, or are you doubling because you've decided to have two separate cases for WTP and BTP? If the latter then you need to add it to the question, because it's either been specifically ruled out or not mentioned so far.

Elroch
MARattigan wrote:

Yes, I agree with that, but doesn't that mean it is indeed a quibble?

If you wish. 🤷

TopazScorpion
magipi wrote:

Just out of curiosity, what does "never get cancel by each other" mean in the puzzle? Everyone seems to understand it, but I don't.

This means only one queen can go to a square, and not the other. If both queens are able to go to a square, it doesn't count.

MARattigan

Both queens can go to e4. Does that mean it doesn't count?

TopazScorpion
MARattigan wrote:

Both queens can go to e4. Does that mean it doesn't count?

COrrect.

MARattigan

So then no diagrams count? That should make the answer easy.

TopazScorpion
MARattigan wrote:

So then no diagrams count?

Hm?

MARattigan

Can you put two queens anywhere on the board such that there is no square they can't both reach?

TopazScorpion
MARattigan wrote:

Can you put two queens anywhere on the board such that there is no square they can't both reach?

I don't think so, sadly.
sad.png

MARattigan

@Optimissed https://dictionary.cambridge.org/

Mr_Mathematician
MARattigan wrote:
Mr_Mathematician wrote:
MARattigan wrote:
Mr_Mathematician wrote:

You must need to be hold good concepts on probability, permutation and combination then you will slowly realizes why to count 2 cases

How the variation can be more .

I have done my time in the kitchen with all three.

Suppose we are placing a white knight and a black knight on a 2x1 chessboard.

We can place the white knight as below

N .

. N

In each case there is 1 place to put the black knight giving

N n

n N

so you can count the number of cases as 1.1+1.1=2.

Now you want to double that, you say, because there are 2 knights.

That gives 4 cases.

SHOW ME THEM!

Case 1 :When white plays first in 2×1 after black

Fixing white on first ( a1 )

W□>> we have one ways to fill b1 ( let ) --1

Fixing white on b1

□ W >> we have left black place a1 ---2

Now Case 2:- When black plays first then after white

Fixing black on a1

B□ >> we have one ways to fill for white b1

Fixing black on b2

□B>> We have one ways to fill for white on a1

Note that cases got repeated but here sequence matters which we are playing first

There is no connection between the order the pieces are assumed on the board in a method of counting the diagrams and any side to play. There is no mention of side to play in your current enunciation and in the original White to play was specified, so the assumption of separate cases depending on side to play is invalid on any sensible interpretation. You might as well double your cases by counting them as separate depending on whether it's sunny or raining or multiplying by 150 to account for the ply count (actually make that infinity because the positions aren't legal so no reason why the ply count should be).

The salient point is that the final outcomes (diagrams) in your Case 1 and Case 2 are the same, so if you count both you're double counting. A straightforward mathematical error.

You say here , "If the game has 2 queens you have to count for the both first and second that's the reason answer get doubled ." So which is it to be? Are you doubling because there are two queens, or are you doubling because you've decided to have two separate cases for WTP and BTP? If the latter then you need to add it to the question, because it's either been specifically ruled out or not mentioned so far.

That's not mathematical error you should prove it if is so

https://youtu.be/Km024eldY1A?feature=shared

Just you see sequence matters in chess board whether to put black aur white queens first .There is total possible cases not only yours case ( which is assumed to be default case only by your mind ) total cases would definitely counted finally be multiplying 2 just because of sequence.

Yes if originally white as to play first then only Cases without multiplying it but now there is both possibilities of black as well as white .

MARattigan
Mr_Mathematician wrote:
MARattigan wrote:
Mr_Mathematician wrote:
MARattigan wrote:
Mr_Mathematician wrote:

You must need to be hold good concepts on probability, permutation and combination then you will slowly realizes why to count 2 cases

How the variation can be more .

I have done my time in the kitchen with all three.

Suppose we are placing a white knight and a black knight on a 2x1 chessboard.

We can place the white knight as below

N .

. N

In each case there is 1 place to put the black knight giving

N n

n N

so you can count the number of cases as 1.1+1.1=2.

Now you want to double that, you say, because there are 2 knights.

That gives 4 cases.

SHOW ME THEM!

Case 1 :When white plays first in 2×1 after black

Fixing white on first ( a1 )

W□>> we have one ways to fill b1 ( let ) --1

Fixing white on b1

□ W >> we have left black place a1 ---2

Now Case 2:- When black plays first then after white

Fixing black on a1

B□ >> we have one ways to fill for white b1

Fixing black on b2

□B>> We have one ways to fill for white on a1

Note that cases got repeated but here sequence matters which we are playing first

There is no connection between the order the pieces are assumed on the board in a method of counting the diagrams and any side to play. There is no mention of side to play in your current enunciation and in the original White to play was specified, so the assumption of separate cases depending on side to play is invalid on any sensible interpretation. You might as well double your cases by counting them as separate depending on whether it's sunny or raining or multiplying by 150 to account for the ply count (actually make that infinity because the positions aren't legal so no reason why the ply count should be).

The salient point is that the final outcomes (diagrams) in your Case 1 and Case 2 are the same, so if you count both you're double counting. A straightforward mathematical error.

You say here , "If the game has 2 queens you have to count for the both first and second that's the reason answer get doubled ." So which is it to be? Are you doubling because there are two queens, or are you doubling because you've decided to have two separate cases for WTP and BTP? If the latter then you need to add it to the question, because it's either been specifically ruled out or not mentioned so far.

That's not mathematical error you should prove it if is so

That's what I just did.

https://youtu.be/Km024eldY1A?feature=shared

Just you see sequence matters in chess board whether to put black aur white queens first .There is total possible cases not only yours case ( which is assumed to be default case only by your mind )

And, as I read it, the minds of almost all of the other contributors, which is not surprising given the wording of your question:

In how many ways two queens shall be placed into 8×8 chessboard such that they never get cancel by each other ( there are only two queens on board place).Cancel means possibilities of capture.

Who would ever read that to mean in how many ways can the queens be placed into the board and a side to move be chosen?

And how would choosing a side to move be described by, "If the game has 2 queens you have to count for the both first and second that's the reason answer get doubled"?

total cases would definitely counted finally be multiplying 2 just because of sequence.

Yes if originally white as to play first then only Cases without multiplying it but now there is both possibilities of black as well as white .

In the original, the comment that White was to play appeared to be irrelevant to the question. I assumed that was why you deleted it.