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Mar 04 2014

Asian Tour Event Four: Now Underway

Davison

A short post today to remind readers that the final event of the 2013/14 Asian Tour is now underway, with four main tour places, as well as eight Players Championship Grand Finals spots up for grabs this week in Dongguan, China.

The even began earlier today, notable pro scalps being Daniel Wells, Cao Xinlong and Li Yan, while Mark Davis was the highest ranked player to progress today, earning himself another 360 points in the race to a Crucible spot.

You can view my thoughts on the event following the publication of the draw last week here, while the scores and Asian Tour Order of Merit can be found at snookerorg.

Meanwhile, the seventh and final regular group of the Championship League came to a conclusion today, Ryan Day defeating Andrew Higginson 3-0 to join Martin Gould, Shaun Murphy, Stephen Maguire, Judd Trump, Joe Perry and Ricky Walden in the winner’s group, which starts tomorrow.

  • Mark Ruesch gen. Klaas

    Not for first time, I am deeply suspicious of the draw
    for the AT events.
    It is always surprising, how the European players are
    grouped, so that they tend to throw each other out of the tournament.
    Since I know a little bit of stochastics, I calculated
    the odds for AT4 and I found the chances for the draw as it is, a meagre 0.4 %!
    Is the draw public and under legal supervision?
    If not the probability is in my opinion low enough to
    allege that somebody is tampering with the draw to favor the local players.

    • arne b

      I do not understand what your number says.
      As you put it, the probability for this exact draw to happen is 0.4% – not surprisingly because the prob. for any specific draw would be this low, so there is nothing wrong about this.
      To test your allegation, I think you need to calculate the probability for this kind of draw (i.e. Europeans grouped together as they are) to happen and compare it to all AT draws so far (Large numbers are necessary for statistics to work properly). Of course, first you need to define what “grouped together” means. In this draw, half of the Europeans (12 of 24) are in the second quarter, the rest are divided 2/5/5. So are these 12 a grouping? Or is the number actually quite likely to occur? (I don’t know, but I can imagine it being a not too rare number.)

      I think the mathematics needed to support your allegation is a lot more than you have written here, although your thesis itself is of course interesting and could be important, if it is true.

      • Mark Ruesch gen. Klaas

        You are right, I need to phrase more exactly. There
        are 23 European players (or 24?) in a field of 128 (I counted the byes as players for simplicity, although I know they are not placed randomly).

        Under these circumstances, the probability of four (or
        more) European-European-pairs in the first round is 0.4 %. So 0.4 % is not the probability of the actual draw, this would somewhere in the 10 e -100 region, but for this “type” of draw.

        I did not examine the later rounds, but they look
        suspicious to me, too. However, because of the influnece of the outcome of the first round, the mathematics get real tricky.

        • mikets

          I believe your calculations are wrong. I suspect you haven’t taking into account the seeding. The draw was divided into 61 seeded players and 61 non-seeded players (plus the three top seeds who got byes) so each player had 61 possible opponents rather than over 120. There were 16 seeded Europeans and 7 non-seeded; the chances that 4 or more non-seeded Europeans are drawn against seeded Europeans are, according to my calculations, about 12%, so you would expect it to happen about 1 time in 8.

  • mark

    Hi matt just wondering could you help me. Im a big angles fan and a win tom will leave him leavel with guodong in the last place f grand finals. What happens if finish tied.
    Ta