Bridge hand - what are the chances of making 7NT with these hands?

The Game Room
1

My partner and I bid to 7NT with:

S KQx
H KQxx
D xx
C AQxx

S Ax
H ATx
D AKxx
C KJxx

The lead was a club and there are 12 top tricks (4 clubs, 2 diamonds, 3 hearts and 3 spades.) A 13th trick will have to come from the hearts or the diamonds.

The contract makes if:

  • the JH is singleton or doubleton
  • the hearts break 3-3
  • one defender has 4+ hearts (including the J) and 5+ diamonds (since they can be squeezed)

Obviously there’s always a chance of a mistaken discard (the play started with 4 rounds of clubs, then 2 hearts, then 3 spades…), but that’s not mathematically calculable.

On the day the JH was singleton. :sunglasses:

But what is the % chance of making the contract? (I know that a 3-3 break is 35%…)

2

I am not good enough at bridge to check your logic, but assuming that each of those bullet’d possibilities is sufficient to make 7NT alone, then the chance that at least one of them occurs is the inverse of the chance that the opposite of each occurs together, or P(7NT) = (1 - P(!JHsd)*P(!H3-3)*P(!3rd thing)).

Let’s try the second one first, since it’s simplest. There are 6 hearts that can be in two possible places. That’s 2^6 total possibilities. The number of those that have exactly 3 in each hand is n choose k, or n! / (k!(n-k)). For n=6 and k=3, nCk = 20.

So there are 20 ways out of 64 possible ways to have exactly 3 hearts in each hand, meaning that the middle term, P(!H-3-3) is the inverse of that, or 44/64.

I leave the other two terms as an exercise to someone with more time on their hands. And probably also to correct whatever errors I made above.

3

Thanks for that!

4

One thing to bear in mind is that they can’t occur together. The three options are mutually exclusive. So if you can work out each you can just add them together.

5

Something doesn’t add up here, and I think it has to do with isolating the suits.
20 out of 64 is .3125 or 31%, which is not the 35% glee mentions in the OP.

You get 35 … actually 35.5% if you calculate the number of different hands you can get out of the 26 unknown cards, 10,400,600 and the number that have 3 hearts 20*184756 and divide the latter by the first.

6

Ah, yes, I did miss that. The other cards do make a difference in possible distribution.

7

Does it also make with a 4-2, 5-1, or 6-0 split if the J is in the East and you finesse to the 10?

8

Good question!

I should have stated that I wanted to make 7NT with no risk - i.e. taking no finesses (unless they are marked.)

So apart from the chances I gave, I can add a remote possibility.

Take the player with ATx in hearts. If the opponent on their right holds all 6 missing hearts, then when the KH is led, the finesse of the T on the second round is marked.

From t’Internet, a 6-0 split is about a 2% chance - so for the specific player to have all 6 is a 1% chance.

9

The first part of the calculation is simple. The second very much isn’t.

The chance of a 3-3 Heart break is 35.5%
The chance of a 4-2 Heart break is 48.5%; 1/3 of the time the Jack will be on the short side of the split, so the chance of a doubleton Jack is 16.2%
The chance of a 5-1 Heart break is 14.5%; 1/6 of the time the Jack will be the singleton, so the chance of a singleton Jack is 2.4%
The chance of a 6-0 Heart break is 1.5%, 1/2 the time it will be on the right side to mark the finesse, so the chance of a marked finesse is 0.75%
These are all exclusive, so we can add them up. The chance of the Hearts lying favourably is 54.9% - which is better odds than the second-round Heart finesse, so you were right not to take it :slight_smile:

Then it gets messy. The odds of the 7 missing Diamonds not splitting 4-3 is 37.8%, but the odds of a player holding 5 or more Diamonds given that he already holds 4 or more Hearts to the Jack isn’t simply half of this (the player with the long Hearts has only 9 or fewer non-Hearts, while his partner has 11 or more). I don’t know how to calculate this one exactly.

10

Thanks for that!

11

So is it possible to play 3 rounds of hearts, then if the jack hasn’t dropped, go for the squeeze? I don’t know how that works, sorry for the basic question. This being a concrete example of the play may help. But I thought a squeeze usually involved losing the lead, which is obviously not an option here. I’m probably confusing it with an endplay.

12

No problem - and you’ve practically got it. :sunglasses:

I play the hand as follows:

  • win the opening lead and cash the 4 club tricks (1)
  • next win the KH (2)
  • now play the AH (3)
  • play 3 rounds of spades, discarding a diamond and cash the QH (4)

Now my two hands are:

S void
H x
D xx
C void

S void
H void
D AKx
C void

If one player started with 4+ hearts and 5+ diamonds, then they have to discard from:

S void
H J
D Qxx
C void

(It doesn’t matter which diamonds they actually hold - the point is that if they started with 5, then their partner started with just 2.)
So if the defender throws the JH, I have a master H (plus AK diamonds);
if the defender throws a diamond, then I cash the AK diamonds and my last diamond is a winner!

(1) the opposition may make a bad discard, but in any case, this helps set up any possible squeeze
(2) if the opponent on their right holds all 6 missing hearts, I now know to lead a heart and finesse the TH
(3) if the JH is doubleton, the TH has become my 13th trick
(4) this is to leave everyone with just 3 cards

You mention losing the lead - and this does happen with many squeezes (but not in a grand slam!)
If there is a potential squeeze, but there is a suit the key defender can just discard (as his partner guards it), the Declarer can play one round of that suit. This losing a trick you were going to lose anyway is known as ‘rectifying the count’, since at the end the key defender has no safe discard.

13

Excellent post, many thanks. I now understand it a lot better (and I particularly enjoyed the numbered footnotes, showing how you ‘gain’ another way to win/a few more percentage points at each step). I very much doubt I would be able to pull it off at the table, or even spot the opportunity, but then I’m a very casual and inexperienced player.

14

I think the easiest way to calculate the last one (at least in a way where I’d be confident in the result) is to look at the hand that doesn’t have the 4+ hearts and 5+ diamonds.

That hand has 2 or less hearts, not including the Jack, 2 or less diamonds, and a corresponding number of clubs and spades, but the distribution of those doesn’t matter.

That’s nine different variations of x and y, where x is the number of hearts and y is the number of diamonds. Each variation would have this many combinations:

5 choose x * 7 choose y * 13 choose (13 - (x + y))
5 is the number of remaining hearts, not counting the J. 7 The number of remaining clubs. And 13 the remaining clubs and spades.
x and y both range between 0 and 2.

That’s more calculator punching than I care to do though. :slight_smile:

15

The problem was annoying me, so I sat down and tried to slog it through. Warning - maths incoming

If a defender has exactly 4 Hearts, then the chance of him drawing exactly 5 Diamonds in his remaining 9 cards (out of the 7 Diamonds and 13 black cards for him to draw from) is:

(7/20 x 6/19 x 5/18 x 4/17 x 3/16 x 13/15 x 12/14 x 11/13 x 10/12) x (9! / (5! x 4!))) where the second term is the number of distinctly different orders in which he can draw diamonds and black cards

Cancelling some terms, this evaluates to 277,200 * 126 / 390,700,800, which is about 8.9%

But the defender might draw 6 Diamonds, the chance of which is (7/20 x 6/19 x 5/18 x 4/17 x 3/16 x 2/15 x 13/14 x 12/13 x 11/12) x (9! / (6! x 3!))), which works out to 1.2%
Or he might draw 7 Diamonds (0.046%), for a total chance of 10.15%

But the defender might have 5 Hearts, in which case he has only 8 draws to get 5 Diamonds (total chance about 5.25% for 5, 6 or 7 Diamonds), or even 6 Hearts, (total chance about 2.2%)

So:
48.5% of the time the Hearts break 4-2, 2/3 of those the Jack is with the long Hearts and in 10.15% of those cases the hand with 4 Hearts has 5 or more Diamonds, so the total added chance is 3.3%
14.5% of the time the Hearts break 5-1, 5/6 of those have the Jack with the long Hearts and in 5.25% of those cases the hand with 5 Hearts has 5 or more Diamonds, so the total added chance is 0.6%
1.5% of the time the Hearts break 6-0, half of those the finesse isn’t marked, and in 2.2% of those cases the hand with 6 Hearts has 5 or more Diamonds, so the total added chance is 0.01%

So finally summing the exclusive percentages, the chance of the Hearts lying favourably is 54.9%, and the chance of the red-suit squeeze working given that the Hearts are not favourable adds another 3.9%, giving a total probability of 58.8% of making the slam with best play on both sides.

From which we can conclude that 7NT was the wrong contract, since a 58.8% chance of success in the grand slam. is not worth gambling your dead-certain 6NT. :wink:

We can also conclude that slam bidding can never be an exact science, since if you change the Heart 10 to the Jack the grand slam is 100%, while if you change it to the 9 it goes down under 50%.

Congratulations on your grand slam!

16

Brilliant post. At the level at which I play, I’m sure the percentage of incorrect discards by the defence increases the probability of the slam making to over 100%. I’m also amused to note that, again at my level, in a lot of situations players are going to make the grand slam by doing the squeeze by accident (i.e. cashing all their winners and then playing out the last few cards and hoping for the best!).

17

Dead_Cat, glad you are enjoying it!

Merrick, thanks for the maths. :sunglasses:
You are certainly correct to say that in competition, you should have an excellent chance of making a Grand Slam before you risk bidding it, because:

  • if you make it, you get a top score
  • if you go one down, you lost to not only to the 6NT bidders, but also to the wimpy 3NT contracts.

Perhaps a fair guideline is having a 2 in 3 chance of making…

Incidentally I should adjust the % chances once more - if the player on lead leads a heart into the ATx, then 7NT makes as well.

There are at least 3 contracts worth considering on the original two hands (repeated for convenience.)

S KQx
H KQxx
D xx
C AQxx

S Ax
H ATx
D AKxx
C KJxx

In order of reward, they are:

  • 7NT is around 60% (using Merrick’s excellent maths, plus the small chance of a heart lead.)
  • 7clubs is 100% if the clubs are 3-2 (and no immediate ruff); it’s around 60% if the clubs are 4-1 and fails if the clubs are 5-0
  • 6NT is 100%.

Bridge is a fascinating game. :heart_eyes:

18

It depends a bit on the scoring system - on raw aggregate points, bidding 7NT not-vulnerable gains 500 points if it works but loses 1040 if it fails, while vulnerable you stand to gain 750 or lose 1540, which fits the 2/3 rule almost exactly.
However if you translate to IMPs, you stand to gain 11 or lose 14 non-vulnerable and gain 13 or lose 17 vulnerable - which sets the break-even odds around 56%, so maybe you were right to bid the 59% grand after all! (This surprised me - I’d always used the “you need a better than 2/3 chance to bid 7” heuristic and never done the maths!).

And if you’re playing matchpoints, then it depends on what you think everyone else is doing. If you can trust everyone at the table to be in either 6NT or 7NT, then statistically you should bid 7 if you think it’s better than a 50% chance, since that gives you an odds-on chance of the top score rather than the bottom one. In practice, of course there will always be someone in 3NT or 6 of a suit, or who drops a trick in the play, so 6NT+1 is unlikely to be an absolute bottom, but 7NT-1 may well be.

19

Yes indeed.

For example, take a club bridge night with 11 tables playing duplicate.
Based on my experience, the other 10 tables might well get these results:

  • 1 pair in 7NT
  • 7 pairs in 6NT
  • 1 pair in 5C
  • 1 pair in 3NT

So if my partner and I reach 7NT, we will either make the contract (scoring 19/20) or go 1 down (scoring 1/20.)
Assuming the odds of 7NT success are about 60%, we would expect to get 59%.

If we bid 6NT (always makes), we will get either 11/20 (when 7NT makes) or 13/20 (when 7NT fails.)
Assuming the odds of 7NT success are about 60%, we would still expect to get 59%. :open_mouth:

I hope I’ve got my calculations right - it seems extraordinary that the expected result is identical in both contracts!

20

@glee how did your bidding get you to 7NT?