3-move drawing program?

[Chexhero]

Personally, if both sides of a ballot are going to be played by each player, I don't see why the 3 move deck and openings being completely random is not sufficient. Even if you draw the same opening twice, a different opponent still has many ways to play the line differently against you. Even the dreaded double cross has many variations to it. But, I am ok with using the software. Alex, Ingo, and Eric all support it, who are much smarter than I am.

[Ingo_Zachos]

Dear checker friends,

thx for your input. All are very welcome.
I will try to answer some of them.
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There appears to be some rather woolly thinking going on.
As mentioned above the 3 move ballots are included in the deck on the basis that they are sound for a Draw (given best play) not on whether they are strong or weak, any imbalance in that regard being countered by the fact of having to play both sides of the opening.

But it now seems to be proposed that if someone sitting at the next table draws the Single Corner, then a computer program should ensure that one is spared the horrors of the Octopus or Skullcracker.
....

1) Liam, you made a terrible thought mistake. The distroubtion of opening difficulty in a match or single encounter is always equal, as your weak side is the opponent`s strong side, as long as you play both sides of the opening.

BUT: In a tournament with 9 rounds it is very unlikely that all players had the same openings, despite the drawing has been random all the time. The distribution is now random, but not equal.

That means, for instance two players in a team competion can have two very unequal distributions of opening difficluty.
Imagine, a USA - GB &Ireland match is played, and Mr. McCosker and Mr. Devlin both play 10 rounds.
Mr. McCosker gets the Dyke (1.11-15 22-17, 2. 15-19) in all ten rounds, which is as likley as any other distribution.
Mr. Devlin gets 10 different openings, each one is one of the new 12 openings.

Both score 4 wins and 16 draws.

Do you think that the opening difficulty for both players was equal, even though the draw was random?

You made a though mistake like thinking that 1 is odd and 3 is also odd, thus their sum must be odd. But in fact it is even.

Your thought mistake is that a random draw over many rounds does only in two cases always produce an equal distribution: the two players case (match) and a tournament with an indefinate number of rounds.

In all other cases it is likely to produce an imbalance, in contrast to what you claim.
Random draw = very likely imbalanced opening distribution is correct.

Since most tournemnts are swiss system or round robin, with much more then teo players and less then indefinate round numbers, you will not very likely get an equal opening difficutly.
But: a few rules in the drawing each round may produce a distribution that is closer to an equal distribution then just blindly following a random walk.
Like: do not allow the same opening twice in the same event for one player. In this case, Mr. McCosker would have playes the Dyke only once.
or: do require a maximum number of new ballots, say one third of the rounds. In this case, Mr. Devlin would only get 3 new openings.

2) I m saying that after a Single Corner it should be more likely to get a Skullcracker or an Octopus and impossible to get another Single Corner.

Not less likely as you seemed to understand.
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Well and what happened if unexperienced referee select a wrong option ... and pick openings ... who will be blamed ?

And what happened if there will be an error in program ... who will be responsible ... or we allow "opps" ?

If a card deck with 144 openings is used, as the referee mistook them ... who will be blamed
And what happens if there is a mistrint on the cards ... willl you allow "oops" ?

Note: practical mistakes always happen, no matter which system you apply.

You can only check which deck it is before, count the number of the cards (always do that in any card game!), and check for misprints.

A software can be checked and tested as well.
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Question: is the goal here for every board to play the same opening each round, for every board to play an opening of comparable difficulty, or for every board to play a randomly-chosen opening?

No, the goal is to secure an equal or at least fair distribution of openings for all players over the tournement. Still the draw is random, but say after round one the opening that was played in round one by both players is not in the pot anymore, so its 154 or 155 openings which can be randomly chosen, and if the number of say, new openings is over 1/3 of all rounds, then the new openings are not in the pot anymore, but still enough left for a random draw.

The Italians use the same opening each round for all pairings, thus also equalizing opening difficulty, but if you insist on a different opening each pairing each round, it does not mean we could not reduce the chance of an highly unequal distribution of openings. We don`t have to tolerate the influence of chance in our game. We are not playing a card game after all.
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It is a simple beginner level coding project.

Everyone see it on their level.

But we went too far. For the time now WCDF should put announcement and offer: $50, $500 or $5,000 and go from there. Someone (usually business analyst) should write specification for coder. And we also need QA team (WCDF members or volunteers) for testing.

Ingo, WCDF is ready to spend $50 on this ? If yes, Eric got a paid job.

Alex

lol. To save money ... take him by the word.

Seriously, Alex, this is not an idea of the WCDF, but by personal idea.
Had it when a player in the German Open got the same openings in the first two rounds.

And if it can be done in an Excel spreadsheet I will first try it myself and save even more money.
But I might take his offer sometimes later this year, in case I can`t do it with Excel.

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Greetinx from a cloudy and grey Dortmund day,

Ingo Zachos

[Alex_Moiseyev]

We don`t have to tolerate the influence of chance in our game. We are not playing a card game after all.

Scratching my head and laughing