The king moves to either the c file or the g file. If the king moves to the c file, the rook closest to the a file, goes to the d file. If the king moves to the g file, the rook closest to the h file goes to the f file.
Chess960 Explained!
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i think its a great game that is excellent for teaching and practicing on ur tatics. I use it all the time. Once i show the lil kids how to set up the board and move the pieces. going to google that tranactional chess though want to see what thats about
I recently started playing chess 960 with a friend on chess.com. We finished 3 games now, and the strange thing is that all of these 3 games were with the SAME starting position, and the 4th game we just started as well. The frequency of this happening by chance is 1: (960x960x960x960), so I guess there must be a different reason.
can someone explain this to me?
Chris
I think this happends when you do "rematch" rather than creating a new one from scratch.
Thanx bishop...I will check this with my friend (it was he who did the challenges)
Actually, the probability of four games in a row with the same position (with random draws) is 1/(960x960x960).
you are absolutely right ironic :-)
can you also explain why there are 960 different possible starting positions? I still find it hard to believe.
chris212121 wrote:
you are absolutely right ironic :-)
can you also explain why there are 960 different possible starting positions? I still find it hard to believe.
There are four ways to place a bishop on a light square and four ways to place a bishop on a dark square. Once the bishops are placed there are six ways to place the queen, then five ways to place the first knight, and four ways to place the second knight. Once all those are placed, there is only one way to place the king and the rooks (since the king has to be between the rooks). Keep in mind that we have double counted the knights: we have counted placing the first knight on the a file and the second on the g file, and we counted placing the first on the g file and the second on the a file. Both situation are the same.
4 x 4 x 6 x 5 x 4 = 1,920 / 2 = 960
wow...thanx for the brilliant lecture !!
even I understood the logic in it
Thank you Eric for the information. I will try a few games and see if i like the variation.
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Thankzzzzzz
The first of the 4 games is already a given thing, Paul. We want to know the probability of the THREE next games being the same in order to know the probability of four consecutive identical games.
paul211 wrote:
You simply are wrong my friend.
He really isn't. The probability calculation would go as follows:
the probability of any given setup is 1/960.
the probability of any given setup four times is 1/960^4.
the probability that the same (initially unknown) setup is repeated four times is: 960*1/960^4 = 1/960^3.
We have to multiply our probability of getting a given setup four times by the number of ways that we could have obtained the observed result. I.e. think of the coefficients in a binomial distribution for me.
Alternatively consider the following example. What is the probability of rolling the same number on two fair six sided dice?
For any particular number the probability of rolling it twice is 1/36. However we don't care which number it is, so we multiply 1/36 by six to account for the six ways that we could achieve the result. Thus we obtain 1/6.
If you throw a dice, and you get let's say a 2, the probability of getting a 2 again on the second throw is 1/6 , not 1/36....agree Paul ??
The probability of getting 2 times 2 is 1/36, but the probability of getting the same number twice is 1/6, not too difficult to understand is it?
I am sorry Paul, but you are wrong...the same way as I was wrong in the first place and was corrected by ironic_begar.
The situation is perfectly comparable to the example of the dice I gave.
I started a game of 960 chess..in position A. Started game 2 and got position A again (probability one on 960). Started game 3 and again got position A (probability 1 on 969x960). Started game 4 and AGAIN position A (probability 1 on 960x960x960) if all this was based on pure chance (which it wasn't)
I throw a dice and get a 2. Throw it again and get another 2 (probability 1 on 6) I throw it a third time and get another 2 (probability 1 on 6x6). I throw it a fourth time, and get yet another 2 (probability 1 on 6x6x6)
Agree?
My original question was ' What is the probability of getting four games od chess 960 in a row, all starting with the same position'
The answer is 1 on 960x960x960
lobachevskii has it right. There are 960 ways to get the same position four times in a row, one way for each position. Each of those 960 ways has a probability of 1/960^4. The 960 ways goes into the numerator and divides out, giving you 1/960^3.
Another way to look at it (the one I usually use) is that the first probability is 1. You will get a position, some position. The probability that the second position matches the first is 1/960, and that's the same for the probabilities that the third matches the second and the fourth matches the third. Multiply them together and you get 1/960^3.
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The only change in rules is that castling is funky - when you castle, the king/rook end up in the same position as when they castle in regular chess (king on g-file and rook on f-file for kingside, king on c-file and rook on d-file for queenside).
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(For more info, read here: http://en.wikipedia.org/wiki/Chess960 )
Obviously this is not so clear case for chess960 castling. So, please someone competent enough for this particular realisation of chess960 to put a simple table with exact positions of king and rook before and after a castling. It shall not be so hard to do it. And will solve all the troubles.