In principle, it's a simple problem. Let's assume that we have five hearts and therefore, after counting partner's guaranteed doubleton, there are six hearts to be distributed among the three unseen hands. Naturally, I used a method based on the principle of Vacant Places and with the starting assumption that opener has only 5 vacant places (remember, he has two cards guaranteed in each suit) and that the unseen hands have 13 vacant places each. Still, because there are six hearts to "deal out", it quickly led to quite a bit of complexity (if you were to examine every case, there are 3^5 = 243 different ways to deal those cards). Since there were eight probabilities in total he was seeking (he was also interested in the situation where we have a six-card major), I decided that a little Java programming was in order.
The results are a little surprising perhaps: the most likely holding opposite is three, with a probability of 44% or so (almost 50%!), regardless of our own length. Another way of looking at it is that only roughly one third of the time will I be disappointed and find a doubleton in his hand opposite my own longish suit. I also added the figures for a four-card suit in responder's hand.
The following table shows the specific probabilities, always assuming of course that opener is "allowed" to open 1NT with a five card suit (but no longer), but is not allowed to open with any suit shorter than a doubleton:
| Partner's length: | 2 | 3 | 4 | 5 | |
|---|---|---|---|---|---|
| My length: | |||||
| 4 | 25.14% | 44.00% | 25.14% | 5.71% | |
| 5 | 31.34% | 44.77% | 20.35% | 3.54% | |
| 6 | 38.74% | 44.03% | 15.31% | 1.91% |
So, now I'm well set up for doing any vacant-places calculations of any complexity. Comments welcome.
Interesting!
ReplyDeleteI think that these results are a little miss leading to the (p)-1NT-(p)-? and 1NT-(p)-? as the pass information by the opponents isn't included in the analysis. My quick thoughts lead me to think the knowledge of the pass, only decreases the likelihood of partner having 3 or more card support. The more cards your partner has in your suit, the more cards the opponents have in other suits, thus increasing the likelihood they will make a non-pass call.
The results are also more realistic when discussing a strong NTs vs. weak as more passes are expected by the opponents when partner has 15+ HCP.
I often wonder when its right to bid a 5 card suit over a weak (10-12/10-13/12-14). Again the more points in your NT range, it seems more likely partner will have a fit. This said, even going with the odds of the fit, doesn't necessarily say you want to announce the fit, especially over weak NT, as now the opponents might be encourage to bid.
Overall, I applaud the work, but am cautious to make any decisions based on it.
Brian
Interesting points. Enemy bidding was not part of the terms of reference for the problem as posed (although I do agree that it's very relevant). However, I think that you've over-simplified the issue of interference over 1NT a little (good opponents are perhaps more apt to try to disrupt a strong notrump with a distributional hand -- most people like to have some real values before interfering over a weak notrump).
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