@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
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)