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Cutthroat Pickleball (Posted on 2024-08-21) Difficulty: 3 of 5
Pickleball is played on a court similar to a tennis court: two sides of a net, each side divided into a left and a right side, as seen by the players on that side, so the two right sides are diagonally opposite each other.

The most popular form is doubles, where each team of two plays against the other. In this form, when the serve switches from one team to the other, the player on the right serves first. This happens many times before the game is ended by one team scoring 11 points (by at least 2).

In the less popular singles form, a player serves from the right if his score is even and from the left when his score is odd.

While one side is serving, in each form, each time a score is made (by the server, the only player who can), the service is shifted from left to right or vice versa.

A form has been devised to allow three people to play--each player being given the opportunity to be the winner. It's called Cutthroat Pickleball, or Australian Cutthroat Pickleball. At any given time one player is on the "singles side" of the court and plays only one half of that side, both in serving and defending. That player stays on that side until he or she fails to win a point (only the server can score a point, but if that fails, the service passes to the next player).

The passage of service to the next player occurs by having all three players rotate clockwise: the left defensive player goes to the serving side of the net; the right side player goes to the left and the former server goes to the right side of the defensive side of the net.

Unaddressed in online descriptions of the game is which side (left or right) the person becoming the server in the rotation, should start on as a rotation is taking place. Is it the left, as in doubles play, or dependent on the parity of his score, as in singles play? Why does it matter? It matters as, if the service always starts from the right, her service will always be to her predecessor in the rotation, as service is always cross-court to the diagonally opposite player. This is the only service that's guaranteed to take place, and this other player may be stronger or weaker than the remaining player of the three.

So the puzzle is this:

Alice, Becky, and Carol are playing. Becky will be the starting server and will start from the right side (uncontroversial, as she's beginning to serve and also her score is zero, an even number).

As it happens, Alice will score 60% of the time if serving to Becky and 70% of the time when serving to Carol. Becky will score 40% of the time when serving to Alice, but 60% of the time when serving to Carol. Carol will score 30% of the time when serving to Alice and 40% of the time when serving to Becky.

Remember that a given player continues to serve so long as she continues to score points, and shifts between left and right each time she does score.

The question is: What is the percent of times does Becky win under these four scenarios:

  • doubles rule -- new server starts on right side
    • Order: Becky, Alice, Carol, ...
    • Order: Becky, Carol, Alice, ...
  • singles rule -- new server starts on side based on score parity
    • Order: Becky, Alice, Carol, ...
    • Order: Becky, Carol, Alice, ...

Which rule results in less dependence on the order of service?

  Submitted by Charlie    
No Rating
Solution: (Hide)
                     winner
              Alice  Becky  Carol
doubles rule
order B,A,C    86.8  11.8    1.4

doubles rule
order B,C,A    72.2  26.7    1.1

singles rule
order B,A,C    83.2  15.6    1.2

singles rule
order B,C,A    79.8  18.8    1.5

The singles rule shows less dependence on the order of serving.

Stats are from simulation:

clearvars

  rots=[2 1 3;2 3 1];
for type=0:1
  fprintf('type %d\\n',type)
for o=1:2
  rot=rots(o,:);

% rot=[2 1 3];
adv=[.1 0 -.1];
st0=0;
wins=[0,0,0];


for trial=1:1000000
  scores=[0 0 0];
  sscores=sort(scores);
  rotplace=1;
  while sscores(3) <11 || sscores(3)-sscores(2)<2
    server=rot(rotplace);
    side= mod(scores(server),2);
    if type==0
      side=0;
    end
    while true
      rp1= rotplace+1+side;
      if rp1>3
        rp1=rp1-3;
      end
      rcvr=rot(rp1);
      probscore=.5+adv(server)-adv(rcvr);
      if rand<probscore
        scores(server)=scores(server)+1;
        side=1-side;
        sscores=sort(scores);
        if max(sscores(3))>=11 && sscores(3)-sscores(2)>=2
          wins(server)=wins(server)+1;
          break
        end
      else
        break
      end
    end
    rotplace=rotplace+1;
    if rotplace>3
      rotplace=rotplace-3;
    end
  end
end
disp(wins)
fprintf('%6.1f %6.1f %6.1f \\n',wins*100/trial)
end
end

Comments: ( You must be logged in to post comments.)
  Subject Author Date
No SubjectObecam2024-09-17 05:01:37
moto x3m moto x3m2024-08-28 22:07:40
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