user33433: Who receives the second round full point bye, if there are an odd number of people scheduled to play in round two?

The lowest rated player who has not received a bye.
The lowest rated player in the lowest score group who has not received a bye and is not due a bye in a later round.
The lowest rated player in the lowest score group whether he/she has received a bye or not.
Any player you expect to lose every round.


user4418: The process depends if the tournament is an all-play-all or a Swiss. For an all-play-all all the players play each other once with a different player having the bye in each round.
For a Swiss it is more complicated. There are a number of FIDE documents which have relevant information on this.
First, for pairing purposes, the players are ranked according to estimated playing strength. This is defined in C.04.2 General handling rules for Swiss Tournaments:

2 Before the first round the players are ranked in order of, respectively
a. Strength (rating)
b. FIDE-title (GM-IM- WGM-FM-WIM-CM-WFM-WCM-no title)
c. alphabetically (unless it has been previously stated that this criterion has been replaced by another one)

Next, C.04.1 Basic rules for Swiss Systems has general rules which include this for byes:

A player who has already received a pairing-allocated bye, or has already scored a (forfeit) win due to an opponent not appearing in time, shall not receive the pairing-allocated bye.

Finally, C.04.3 FIDE (Dutch) System details the actual FIDE Swiss pairing rules. These rules basically state that the player who gets a bye is the player in the bottom score group who is left without an opponent.
This makes it more difficult than the four possibilities you present.


The lowest rated player who has not received a bye.


No. This would basically mean the lowest rated player receiving the bye every round and effectively being excluded from the tournament.


The lowest rated player in the lowest score group who has not received a bye and is not due a bye in a later round


No. If the player has requested a bye in a later round that does not exclude them from receiving the bye. Also it is possible to be the lowest rated player in the lowest score group and be allocated an opponent by the pairing algorithm.
Example: 3 players in the lowest score group, A, B and C in rating order. A and B are expected to have white according to the pairing rules for colour and C is expected to have black. The pairing will be A vs C and B, who is the second lowest rated player, will have the bye.


The lowest rated player in the lowest score group whether he/she has received a bye or not


No. No player may be allocated a pairing bye (i.e. not a requested bye) twice.

Any player you expect to lose every round

Obviously not.

33433: <p>Who receives the second round full point bye, if there are an odd number of people scheduled to play in round two?</p>
<ol>
<li>The lowest rated player who has not received a bye.</li>
<li>The lowest rated player in the lowest score group who has not received a bye and is not due a bye in a later round.</li>
<li>The lowest rated player in the lowest score group whether he/she has received a bye or not.</li>
<li>Any player you expect to lose every round.</li>
</ol>
 4418: <p>The process depends if the tournament is an all-play-all or a Swiss. For an all-play-all all the players play each other once with a different player having the bye in each round.</p>
<p>For a Swiss it is more complicated. There are a number of FIDE documents which have relevant information on this.</p>
<p>First, for pairing purposes, the players are ranked according to estimated playing strength. This is defined in <a href="https://handbook.fide.com/chapter/C0402" rel="nofollow noreferrer">C.04.2 General handling rules for Swiss Tournaments</a>:</p>
<blockquote>
<p>2 Before the first round the players are ranked in order of, respectively</p>
<p>a. Strength (rating)<br />
b. FIDE-title (GM-IM- WGM-FM-WIM-CM-WFM-WCM-no title)<br />
c. alphabetically (unless it has been previously stated that this criterion has been replaced by another one)</p>
</blockquote>
<p>Next, <a href="https://handbook.fide.com/chapter/C0401" rel="nofollow noreferrer">C.04.1 Basic rules for Swiss Systems</a> has general rules which include this for byes:</p>
<blockquote>
<p>A player who has already received a pairing-allocated bye, or has already scored a (forfeit) win due to an opponent not appearing in time, shall not receive the pairing-allocated bye.</p>
</blockquote>
<p>Finally, <a href="https://handbook.fide.com/chapter/C0403" rel="nofollow noreferrer">C.04.3 FIDE (Dutch) System</a> details the actual FIDE Swiss pairing rules. These rules basically state that the player who gets a bye is the player in the bottom score group who is left without an opponent.</p>
<p>This makes it more difficult than the four possibilities you present.</p>
<blockquote>
<ol>
<li>The lowest rated player who has not received a bye.</li>
</ol>
</blockquote>
<p>No. This would basically mean the lowest rated player receiving the bye every round and effectively being excluded from the tournament.</p>
<blockquote>
<ol start="2">
<li>The lowest rated player in the lowest score group who has not received a bye and is not due a bye in a later round</li>
</ol>
</blockquote>
<p>No. If the player has requested a bye in a later round that does not exclude them from receiving the bye. Also it is possible to be the lowest rated player in the lowest score group and be allocated an opponent by the pairing algorithm.</p>
<p>Example: 3 players in the lowest score group, A, B and C in rating order. A and B are expected to have white according to the pairing rules for colour and C is expected to have black. The pairing will be A vs C and B, who is the second lowest rated player, will have the bye.</p>
<blockquote>
<ol start="3">
<li>The lowest rated player in the lowest score group whether he/she has received a bye or not</li>
</ol>
</blockquote>
<p>No. No player may be allocated a pairing bye (i.e. not a requested bye) twice.</p>
<blockquote>
<p>Any player you expect to lose every round</p>
</blockquote>
<p>Obviously not.</p>
