[bd71]

The possibility of "randomizing" penalty doubles -- say a double of a tentatively bid game when your hand is weak -- is briefly mentioned in an article ("Strategic Doubles") in the latest issue of Bridge World (September 2011).

[iviehoff]

Maybe OP can get away with this, or not far off, by removing reference to the randomisation. After all, plenty of people plenty of the time choose not to make a 3-level pre-empt with a hand that meets the formal minimal requirements they have defined on their card, which is probably something like 0+ points and 6+ cards. Or consider for example weak 2s, a lot of people put down 5+ card suit, but in practice there are plenty of players who would rarely use the bid with only 5 cards. Having a mixed meaning bid with strong meanings as well as the pre-empt, well that's just like the Multi-2D.

[barmar]

The spirit of full disclosure is that the opponents are entitled to any information you have about partner's hand. Since you don't know what random decision he's made, you don't know any more than the opponents.

What I find interesting about this particular example is that the disclusure isn't really needed when the 2♥ bid is made. When he makes that bid, you can definitively state that he has either an intermediate heart or weak spade hand, not much different from other multi bids. It's the PASSES that are really notable, since he might hold a weak spade hand that could have opened 2♥.

I'm not sure how you would describe that in any useful way, though. "He either has a normal non-opening hand, or he has 5-8 HCP 5+♠; 80% of the time when he has the latter hand he passes." Now, for someone to figure out the likelihood that he holds the weak spade hand, they need to know the frequency that non-opening hands fit that description, and then reduce that by 20%.

Another aspect of this is what it means when he passes, and then bids 1♠, either as a response or overcall. There's 20% less chance of him holding 5-8 and 5+ cards than there would be for most other players. So he's a little more likely to either have 9+ HCP or only 4 cards in the suit. So do you alert all these spade bids so you can disclose the small difference in probabilities?

[semeai]

barmar, on 2011-September-01, 10:13, said:

What I find interesting about this particular example is that the disclusure isn't really needed when the 2♥ bid is made. When he makes that bid, you can definitively state that he has either an intermediate heart or weak spade hand, not much different from other multi bids. It's the PASSES that are really notable, since he might hold a weak spade hand that could have opened 2♥.

The 2♥ bid is affected as well. It changes the probability that he has a heart hand or a spade hand. Without the 20% it would be much more likely it was the spade hand, but with that it's probably close to even.

Using Bayes' Theorem:

prob(spade hand | 2♥) = prob(2♥ | spade hand) * prob(spade hand) / prob(2♥) = .2 * prob(spade hand) / prob(2♥)
prob(heart hand | 2♥) = prob(2♥ | heart hand) * prob(heart hand) / prob(2♥) = prob(heart hand) / prob(2♥)

So the ratio prob(spade hand | 2♥) : prob(heart hand | 2♥) is equal to .2 * prob(spade hand) : prob(heart hand) instead of the same expression without the .2 if the 20% provision weren't there.

Here spade hand = 5-8, 5+ spades, heart hand = 11-15, 6+ hearts.

Added: According to BBO's "deal source" widget, prob(spade hand) = 5.06%, and prob(heart hand) = 1.80 %. So without the 20% provision, it's more than twice as likely it's the spade hand. With it, it's almost twice as likely it's the heart hand.

[barmar]

So who's responsibility is it to perform these calculations? Is it reasonable to describe the hand as in the OP, and expect opponents to calculate the probabilities themselves at the table? Or should the pair playing this system figure this out, so that they can provide more useful disclosure?

[semeai]

barmar, on 2011-September-01, 10:53, said:

So who's responsibility is it to perform these calculations? Is it reasonable to describe the hand as in the OP, and expect opponents to calculate the probabilities themselves at the table? Or should the pair playing this system figure this out, so that they can provide more useful disclosure?

I have no laws expertise, but I think OP (plus randomization method) is enough disclosure, and it's up to the opponents to work out that this makes it less likely it's the spade hand.

[hrothgar]

gombo121, on 2011-August-31, 23:38, said:

Concerning practical realization of the RNG, there is a very simple method - shuffle your hand before you look at it and then interpret red cards as zeros and black cards as ones - you get 13 bits of randomness, which should be enough for any practical purpose!

Unfortunately, these bits are strongly correlated with the shape of your hand...

A better scheme might be to measure "transitions"

Shuffle your cards.
Look at card 1

If card 1 is higher in rank than card 2, code a 1
If card 2 is lower in rank than card 2, code a 0
If you have a tie, ignore this bit

Move to card 3

If card 3 is higher in rank than card 4, code a 1
If card 3 is lower in rank than card 3, code a 0
If you have a tie, ignore this bit

Move to card 5...

You'll end up with less than six bits of randomness, but I think this will eliminate a lot of issues with covariance...

[gombo121]

semeai, on 2011-September-01, 10:43, said:

The 2♥ bid is affected as well. It changes the probability that he has a heart hand or a spade hand. Without the 20% it would be much more likely it was the spade hand, but with that it's probably close to even.

That's exactly the point of the method - to introduce a weak option, but don't let it to be too frequent.

2hrothgar: Thank you for very enlightening piece.

[gombo121]

hrothgar, on 2011-September-01, 11:44, said:

Unfortunately, these bits are strongly correlated with the shape of your hand...

Valid point. Then we can use "even" cards as zeros, "odd" as ones and ignore aces - that should not to correlate to any game-related factor.

[hrothgar]

gombo121, on 2011-September-01, 12:01, said:

Valid point. Then we can use "even" cards as zeros, "odd" as ones and ignore aces - that should not to correlate to any game-related factor.

>> Rank = 1:13;
>> Rank = Rank'

Rank =

     1
     2
     3
     4
     5
     6
     7
     8
     9
    10
    11
    12
    13

>> sum(Rank(1:2:end))

ans =

    49

>> sum(Rank(2:2:end))

ans =

    42

Trust me, transitions is the way to go
If you need additional bits you can build in whether the suit is higher or lower in rank...
(more chances of a tie here, but you'll still get a few more bits)

[gombo121]

mgoetze, on 2011-September-01, 09:44, said:

Or he might open a 3532 15-count sometimes 1NT and sometimes 1♥. Does it really matter if it is random, semirandom or completely deterministic which of these hands he chooses which opening bid with?

Well, it may.
Like, you counted opponent's hand to be 3532 and know him to hold AK in hearts in K in both spades in diamonds;you need to place him with either ♥Q or ♣Q. If the opponent never opens 1NT with weak doubleton, you have a clear-cut case for ♣Q; if he choose randomly, it is probably for ♥Q; if it is semirandom, you'd really like to have an idea about factors affecting his choice.

True, it is not often when it would matter, but it may.

[gombo121]

hrothgar, on 2011-September-01, 12:10, said:

ans = 42

I like the answer, but I didn't get the point. I suggested interpreting sequence like 6 5 Q A K 10 2 J as 0101001 (ace is ignored so that number of zeros and ones are equal).

And, yes, transitions certainly work.

[hrothgar]

gombo121, on 2011-September-01, 12:23, said:

I like the answer, but I didn't get the point. I suggested interpreting sequence like 6 5 Q A K 10 2 J as 0101001 (ace is ignored so that number of zeros and ones are equal).

And, yes, transitions certainly work.

I am (obtusely) noting that there is a correlation between hand strength and bits...

[barmar]

hrothgar, on 2011-September-01, 12:55, said:

I am (obtusely) noting that there is a correlation between hand strength and bits...

Is there? Take a hand and convert it to bits using the odd/even rule . Replace one of the cards with the next higher card, and one of the bits will flip. But the same thing happens if you replace that card with the next lower card. The bits for QJT QJT QJT QJT9 are the same as for 432 432 432 5432.

Ignoring aces seems to mess this up, since replacing a King with an Ace causes you to lose bits. I suggest that Aces be assigned bits based on the color or shape of the suit (not rank, as that would then correlate with whether your hand is major or minor-oriented).

[hrothgar]

barmar, on 2011-September-02, 00:35, said:

Is there? Take a hand and convert it to bits using the odd/even rule . Replace one of the cards with the next higher card, and one of the bits will flip. But the same thing happens if you replace that card with the next lower card. The bits for QJT QJT QJT QJT9 are the same as for 432 432 432 5432.

Which hand is stronger?

♠ QT86
♥ QT86
♦ QT86
♣ QT

or

♠ KJ97
♥ KJ97
♦ KJ9
♣ KJ

On average, the "even" cards are weaker than the odd cards.

[Free]

gombo121, on 2011-August-31, 14:06, said:

While discussing building a system with a friend I arrived to a curious question: suppose I'd like to introduce an two-way opening 2♥, which promise
1) 11-15, 6+♥
2) 5-8, 5+♠, 20%(!)
but with a twist - second variant is used not each time when a suitable hand presents itself, but in only, say, 20% cases chosen completely at random.

Is there any general regulations against such agreements (besides it is being obviously brown sticker)? I feel that probably there are some, but cannot come up with any concrete example.

I haven't read everything in the thread, but I want to point out that this opening is NOT brown sticker. There's only 1 weak version in there, and the suit is known in that case.

[JanM]

Free, on 2011-September-03, 06:35, said:

I haven't read everything in the thread, but I want to point out that this opening is NOT brown sticker. There's only 1 weak version in there, and the suit is known in that case.

I think you are misreading the definition of Brown Sticker bids. The rule that a bid is not BS if its only weak meaning promises a specific suit is:

Quote

The bid always shows at least four cards in a
known suit if it is weak. If the bid does not show a known four
card suit it must show a hand a king or more over average
strength.

10-15 does not promise a king or more over average strength.

[ggwhiz]

Kudos to whomever arranged to have such calls known as "BS" bids.

This one is more like "Probable Ballistic".

[Vampyr]

It does seem like any attempt at randomisation will be highly influenced by the shape and/or strength of the hand. But that is OK, because since the method of randomisation must be disclosed to the opponents anyway, this will promote fuller disclosure.

[hrothgar]

Vampyr, on 2011-September-03, 10:57, said:

It does seem like any attempt at randomisation will be highly influenced by the shape and/or strength of the hand. But that is OK, because since the method of randomisation must be disclosed to the opponents anyway, this will promote fuller disclosure.

Wrong

The high-low / low-high transition scheme that I described in this thread is uncorrelated with shape or strength