How can I calculate which opponent I get?in tournament

How can I calculate which opponent I get?in tournament?

Live here uses a pretty standard swiss pairing algorithm. For the first round, players are listed in rating order, top half of the field, plays the bottom half.

With 10 players

1 plays 6

2 plays 7

3 plays 8 .....

After each round, the same process is done, but it is done by score groups. So, after round 1 all players with 1 point are their own group, all players with 0, etc, and the same method from above is used. There are methods to handle odd number point groups, so some players will get paired with someone in a different score group. You also have to worry about color allocations and making sure two players don't face each other more than once.

https://en.wikipedia.org/wiki/Swiss-system_tournament

thank you!

how about sometimes the Black and white doesnt match? Or there are also times a player already played black black or white white..etc.. and their next round must be opposite color.

Suppose 10 players, and there was no draw.  

Top half has 1.0, lower half has 0.0

I think Player 5 will play player 6.  Will player 1 always play player3 even if it may cause player3 to play the same color again in round2?

Often, the highest rated player in the the 0 group gets moved up to the 1.0 group for pairing and the top half plays bottom half in that grouping. So player 1 would pair against player 4.

The official rules of swiss tells us when pairing isnt possible the next group combines. Each player can have max. 2 times the same color in following games.

That is not totally true.

We had a weird case in the 2011 continental class championship, expert section (under 2200) in October 2011.  It was a 7 round event, and in the first 2 rounds, White won in over 80 to 90 percent of the games, which means many upsets, especially in Round 1.

I do not recall the exact number in the section, but let's say it was 42.  I know I played the 3-seed in Round 1 and I was Black, so in that hypothetical situation, 1 got White, and you'd then have 1 vs 22, 23 vs 2, 3 vs 24 (me), 25 vs 4, 5 vs 26, 27 vs 6, etc, all the way to 21 vs 42.

If the higher seed had won each game, 22 thru 42 would have lost, and I, 24 would face 34 (1 thru 20 face, 21 (Bottom winner) faces 22 (top loser), and 23 thru 42 face).

But because of all the upsets, I was BARELY in the top half of those that lost round 1, and had to face the bottom seed in round 2.  Because so many of those with 1 point were due Black and so many of those with 0 were due White, there were a lot of people with WW or BB the first 2 rounds.

With most the higher rated players getting Black in the 1-group (higher rated got due color), that group corrected itself, and those with WW mostly had 1, which is the largest score group after 2 rounds.

But the 0's had the opposite problem.  Higher rated got due color in cases where both were due White, and they, including myself, won.

So the vast majority with 2 had WB, those with WW or BW mostly had 1, and a lot of 0's had BB (There were a few 1.5 or 0.5 in the mix).  So the 2's were not a problem, most got WBW or WBB first 3 rounds.  A lot of the 1's got WWB or BWW, but the 0's still had 1 more than half with BB, so one had to get BBB, and that was my round 2 opponent.

There were a couple of zeroes with 1 White going into the 3rd round, but still too many with BB, that the guy I faced round 2 got Black again, and here was his schedule that tournament:

Rd 1 - Black

Rd 2 - Black (vs me)

Rd 3 - Black

Rd 4 - Full Point BYE

Rd 5 - White

Rd 6 - White

Rd 7 - White (Equalization takes priority over alternation)

So yes, while rare, you CAN get 3 in a row of 1 color in a tournament!

I seem to recall for me it was BWWBWBW.