You open with 1♥ or 1♠ (5 cards Major). And your partner responded with 1NT. Holding 5332, what would you bid then? Is it safe to pass 1NT response? We will dig the answer in great detail to prepare you for the more sophisticated problem down the line.
The hand of 1NT responder
Point-wise, there are two different approaches to play the Standard system:
- The 2-over-1 Game Forcing: meaning a bid in a new suit in level-2 shows length in that suit with 12+HCP. Hence, will not stop bid until reach a game contract. So, assuming your minimum point to respond is 6 HCP, the 1NT bid will have 6-11 HCP range.
- The Standard 2-over-1: where a bid in a new suit in level-2 shows length in that suit but only with 10+HCP. If the responder repeats that suit, the opener may pass (i.e a “non-forcing” bid). So, in this case, with 1NT bid, the responder would have 6-9 HCP range.
How often it happens…
Since 1NT response includes any distribution at all, the frequency of it happening is determined by the point range described above. And, the chances of your partner have 6-11 HCP (or 6-11 HCP) is also determined by how many points you have:
- If your hand has 12-HCP, the odd of the responder have 6-11 HCP is 55.8%1Please read “Companion hand HCP Tables” by Richard Pavlicek for further information. So, using the same table, for 6-9 HCP, the odd is only 36.8%
- When your hand is a 14-HCP one, the chance of 6-11HCp is reduced to 57.8%. For 6-9 HCP, the odd becomes 39.7%
- Therefore, for 12-14 HCP, we can just average the two numbers above to read:
- For 6-11 HCP, the odd is 56.8%
- For 6-9 HCP, the odd is 38.3%
Your hand situation
As described in the beginning, this problem only happens when you have a 5-3-3-2 distribution. And the chance of you has this distribution every time you open a 1 Major2See Spectrum of Level-1 Opening is 7.76%
If you play Strong NT (i.e your 1NT opening is 15-17 HCP range), then the frequency of having exactly a 5-3-3-2 with 12-14 HCP is 7.76 x 20.6% = 1.6% (~ 1 every 63 boards)
Now, If you play a weak NT system (1NT = 12-14 HCP), then your 5-3-3-2 hand here is in the range of 15-17 HCP. Since the responder can have at least 9-HCP (or 11-HCP for 2-o-1 GF), you cannot pass 1NT response, i.e: it is not safe to pass 1NT respond. The reason is obvious, with 15 – 11 HCP (or even 15-9 split), a game is still possible.
Hand distribution aspect
First of all, you need to make sure that your system doesn’t allow the responder to bid 1NT when the responder has a fit in the Major. This means, with 1NT bid, responder only can have a maximum of 2 cards in that Major suit.
We can calculate the odds of this happening by combining the odds of 5-0 fit(2.5%), 5-1 fit (13.9%) and 5-2 fit (29.2%)3 Please read “Bridge Partnership (Two Hands) Probability Analysis” in particular in the section “Given you have X cards in a suit what is the probability partner holds Y cards in the same suit?”. So, the total is 45.6%
Therefore at this point, we can tell the frequency of 1♠-1NT is:
- For no-fit, 6-11 HCP 2-o-1 GF), the odd is 56.8% x45.6% = 25.9% (approximately 1 in every 4)
- For no-fit, 6-9 HCP (standard), the odd is 38.3%x45.6% = 16.5% (approximately 1 in every 6)
For 1♥-1NT, we need to account the odds of NOT having 4 or more card in Spade. If you have 2 cards Spade, the probability your partner has 4 or more Spade is 54%4 Please read “Bridge Partnership (Two Hands) Probability Analysis” in particular in the section “Holding X cards in a suit what is the probability of a fit of N or more cards in that suit?”. If you have 3-cards Spade the odds of at least 7-cards fit is: 44.%. Wanting to use just one figure, we can just average them to 49%. Therefore, the odd that your partner NOT bid 1♠ over your 1♥, i.e directly bid 1NT is 51%
So, for 1♥-1NT, the frequency is:
- For no-fit, 6-11 HCP 2-o-1 GF), the odd is 25.9% x 51% = 13% (approximately 1 in every 8)
- For no-fit, 6-9 HCP (standard), the odd is 16.5% x 51% = 8.4% (approximately 1 in every 12)
Checking The Specific Hand Distribution
Now that we know the frequency when 1M-1NT sequence happens, we need to know what kind of distributions that is possible for the responder. What are the odds and whether or not any particular distribution is suitable for playing 1NT contract? See the table below:
|%||out of||OK for|
|OK for 1NT|
- The first 3 column is a rewrite of the original 39-hands distribution statistics.
- Each distribution has the total number of combination for a specific distribution, shown in the fifth column. For example:
- There are 12 specific distributions for 5-3-3-2. They are: 5332, 5323, 5233, 3532, 3523, 2533, 3352, 2353, 3253, 2335, 3235 and 3325
- For 4-4-4-1 distribution, there are 4 specific distributions as follows: 4441, 4414, 4144 and 1444.
- Then among those specific distributions, there are only some distributions that qualify for 1M-1NT bid. For example:
- For 5-3-3-2 distribution, the qualified distributions are those with a doubleton in the opening Major. In the case of 1♠ opening, the doubleton has to be ♠. In other words, Spade can only be placed in 1 (one) of the ‘spot’ of distribution, i.e: the ‘2’ from 5-3-3-2. Spade cannot be in the ‘3’ or in the ‘5’ as 1NT supposes to indicate a misfit in 1♠.
- For 5-5-2-1 distribution, the opening Major suit can be placed in as the doubleton or the singleton. (Can fill two ‘spots’)
- For 7-2-2-2 distribution, the opening Major suit is good in three spots.
- So, the percentage number in column #4 is indicating how many ‘spot’ can be:
- For 3 spots, it is 75%
- When it is good for 2 spots, it is 50%
- For a single spot, it becomes 25%
- And, if the distribution is illegible for 1M-1NT then it is 0%. For example 4-3-3-3. The 1NT response cannot be a 4-3-3-3 distribution because it means it has at least a 3-cards support in the opening Major suit.
- Then on 6th column, a ‘v’ is the mark that the distribution is considered suitable for 1NT contract. Let me elaborate:
- Any long 1-suiter distribution (say 7-cards or longer) can be better played in that long suit rather than 1NT. (The responder’s hand is the weaker hand, so will have less entry to access the long suit)
- Any 2-suiter can also be better played in one of the suits rather than NT.
- Any distribution with at least doubleton (no void or singleton) can be played in 2 of the Major
- The smallest fit for a suit in 1NT play should be at least a 3-2 fit.
- The last column is the actual probability after considering the eligible distribution only.
Distributions that are decent for 1NT contract
So, from the table above, there are only 8- distributions that can be suitable for 1NT contract:
The total probability of all the above is 14.98% out of 29.87% (all distribution that is eligible for 1M-1NT). Therefore, the 1NT contract only is acceptable for 14.98%/29.87% = 50.15% ==> ~50%. The other half would have another better contract to choose.
The case of 3rd Hand Opening
Imagine you are on the 3rd hand (or 4th hand) position. So, in this case, your partner actually had a chance to bid first but unable to. Then, when you open 1♥ or 1♠, your partner responds with 1NT. This means:
- Your partner has no long 1-suiter hand (otherwise, your partner would open with a Weak Two or pre-emptive bid in level-3 or higher.
- Depending on your opening bid list, your partner also did not have any of the 2-suiters hand that can be opened.
Therefore, what does your partner have? Well, it is likely your partner has one of the 8-hand distributions that are decent for 1NT contract (see above). So, this can be concluded that after a passed hand, it is safe to pass 1NT respond of 1 Major opening.
So, is it safe to pass 1NT response after 1♥ or 1♠?
As a conclusion, the answer to the questions Is:
- If you play Weak NT: No. You may miss a game if you are passing 1NT response.
- If you play Strong NT:
- No, if your partner has never passed. Because about 50% of the time, there would be a better contract to have
- Only if your partner has passed, then yes, it is safe to pass the 1NT respond.