It’s just odds…the answer

This hand is from the 2002 Interstate. The question is, how to make 3NT after the D10 opening lead from South. The commentary is mine from a report on the tournament.

	NORTH Q875 AKJ7 74 1086
WEST AKJ1096 Q84 J3 42		EAST 2 953 K62 AKJ953
	SOUTH 43 1062 AQ10985 Q7

Here’s a tip that will gain you buckets in the long run: NEVER play 6 card suits in 2NT. Those suits make either lots of tricks or none. At our table versus SA in the second round robin we defended 2♠ and shot it 2 tricks for +200. But our teammates with the EW cards played 2NT and went 5 down for –500 and 7 IMPs out. More sensible is to bid the EW cards to 3NT, like Horton-Markey in the Open for SA. It is true that they went 5 down, but at least they were getting a decent payout if they made, unlike the pusillanimous 2NT contract. Of course, better still is to make 3NT as Brad Wein did when sitting East. He cashed clubs from the top, intending next to finesse the spade, but when the ♣Q dropped that wasn’t necessary.

After finding out that Wein for Victoria and Horton for SA took opposite views, I decided to ask the mathematicians what they thought. I asked Robert van Riel and Bruce Neill how East should play 3NT on the lead of the ♦10, with the jack holding in dummy. Should spades be cashed before a club finesse, or clubs before a spade finesse? And does it change the odds if South happens to take a bid?

Bruce emailed me back and said:

‘Before I could do a calculation, I would need to think about what assumptions to make!! But, my gut feel is that you should cash the AK of clubs then finesse spades. Especially so if LHO is known to have 6 diamonds. That makes it very unlikely s/he has 4 rag spades, which is the main case where cashing AK of spades then finessing clubs wins.’

Meanwhile Robert’s first impressions came to the opposite conclusion!

‘I don’t think South’s 2♦ interpolation makes much difference, except that it more definitely places ♦AQ. There’s enough high cards left to make it impossible to place ♠Q and ♣Q. The heart honours are probably split; give South (say) ♥A and ♦AQ109xx and she really doesn’t need much more to bid 2♦ at equal vul. Equally South might have something like ♠Qxx-♥AKJ-♦AQ109x-♣Qx.

Most likely they can beat me if I lose the lead (certainly if I lose it to North), so I’ll need 6 tricks in one of the black suits. I think the normal line would be to cash ♣A, ♠AK (and the remaining spades if ♠Q falls) and then finesse ♣J.’

All very educational. One might have thought that at the table a mathematician would ‘simply’ work a hand like this out. But it turns out to be pretty complicated stuff. Indeed, what we are looking at in this article is nothing more than some of the calculations and discussions which took part.

When one looks at the maths provided by van Riel it makes one realise that at the table intuition, not analysis is so often the way to go:

Van Riel:
‘The probability of spades providing 6 tricks by cashing ♠AK (i.e. ♠Q or ♠Qx in either hand) is
(2/12 x 0.1453) + (10/30 x 0.4845) = 0.0242 + 0.1615 = 0.1857 (18.57%) and the probability of spades providing 6 tricks by means of the finesse (i.e. ♠Q, ♠Qx or ♠Qxx with South) is
(1/12 x 0.1453) + (5/30 x 0.4845) + (10/20 x 0.3553) = 0.0121 + 0.0807 + 0.1777 = 0.2705 (27.05%)

(1) Clubs 5-0 (3.91%, 2 combinations) I’ll take the spade finesse: 0.0391 x 0.2705 = 0.0106 (1.06%)

(2) Clubs 4-1 (28.26%, 10 combinations) (a) ♣Q singleton with North. I’ll cash ♣K to make sure and take the spade finesse: (1/10 x 0.2826) x 0.2705 = 0.0076 (0.76%) (b) ♣10 singleton

with South. I’ll cash ♣K in case it’s from ♣Q10 and take the spade finesse: (1/10 x 0.2826) x 0.2705 = 0.0076 (0.76%)
(c) ♣Q singleton with South. I’ll belt out ♠AK and if ♠Q doesn’t fall finesse ♣9 (I can’t lose):
(1/10 x 0.2826) = 0.0283 (2.83%)

(d) In the remaining 7 combinations, I’ll fail unless ♠Q drops under ♠AK: (7/10 x 0.2826) x 0.1857 = 0.0367 (3.67%)

(3) Clubs 3-2 (67.83%, 20 combinations) I’ll start by assuming South plays like a tiger: holding ♣Q10, she’ll play ♣Q, luring me into line 2(c); and from ♣10x, she’ll play ♣10, luring me into line 2(b), failing in each case unless ♠Q falls. This means I’ll succeed, regardless of what happens in spades, when North holds ♣Q10 or ♣Qx (4 combinations) and when North holds ♣Q10x (3 combinations). In the remaining 13 combinations, I’ll need ♠Q to fall.

(a) ♣Q10, ♣Qx or ♣Q10x with North: (7/20 x 0.6783) = 0.2374 (23.74%)

(b) ♠Q falling under ♠AK otherwise: (13/20 x 0.6783) x 0.1857 = 0.0819 (8.19%)

Now I’ll assume that South isn’t aware of these deceptive plays. Now I’ll succeed whenever North holds ♣Q and also when South holds ♣Q10. The rest of the time, I’ll need ♠Q to fall.

(c) ♣Q with North or ♣Q10 with South: (11/20 x 0.6783) = 0.3731 (37.31%)

(d) ♠Q falling under ♠AK otherwise: (9/20 x 0.6783) x 0.1857 = 0.0567 (5.67%)

Totting it all up: if I make a favourable deduction about South’s defensive prowess, my chances are 52.06% (which is close enough to what I calculated yesterday); if I make an unfavourable deduction, my chances are 41.01%. The practical odds must be somewhere between these; even if South finds a false
card, I’m allowed to second guess her. This all makes the alternative line, with a success rate of 46.84%, quite attractive, since at least it spares you any agonising decisions to do with false cards.’

Bruce did an equally vast amount of maths and took a look at van Riel’s.

Bruce Neill:
‘…you might like to comment on why the best play changes if LHO is known to have long diamonds. Basically it’s because the main benefit of cashing spade A-K is the chance of dropping Q doubleton offside. That chance is much less when LHO has long D’s PLUS enough hearts that RHO didn’t overcall. In contrast cashing club A-K has a real chance of winning.’

Faced with this, and a spreadsheet Bruce had constructed van Riel went on to say:

Van Riel:
‘My calculations take no account of South’s diamond length.

I don’t entirely agree with Bruce when he says that “the main benefit of cashing spade A-K is the chance of dropping Q doubleton offside”. The main benefit of my line is that, looking at the club suit in isolation, the club finesse is the most likely way to muster 6 tricks. Of the 20 possible 3-2 breaks (67.83%), North holds ♣Qx or ♣Qxx 10 times, whereas South holds ♣Qx
only 6 times.

I agree that diamond length with South reduces the likelihood of a short ♠Q with North. On the other hand, it increases the probability of short clubs with South. When clubs break 3-2 then, North is more likely to hold the 3, which means that she’s more likely to hold ♣Q, so that the club finesse line is more likely to work. It’s hard to quantify this, doing the sums by hand.

I haven’t checked Bruce’s figures, but they imply to me that when South has 6 diamonds, cashing ♣AK is only marginally better than taking the club finesse (4.006% against 3.795% of the relevant distributions). I’d expect this to go the other way if South had only 5 diamonds.

Whatever, the lines are fairly close, but I’d say the club finesse line was superior if South could be trusted to play honest cards.’

It all reminds me of crossing the road. A vast amount of calculation takes place, but all we are aware of is an intuitive answer. ‘Cross now’ ‘Don’t cross now’. Perhaps Horton got punished for a slightly better line, maybe Brad got lucky. But they were both near the money. And unlike road crossing, nothing that bad can happen to you if you get it wrong.

I remember once going down in a critical contract by taking a line which was maybe 1% worse than the other. At the scoreup my teammates all looked at me a bit like I was a cane toad. On the one hand they all wanted to squash me. On the other, nobody wanted to get that close…

What is the solution to this dilemma? (1) is to put a lot of work into learning the odds. (2) is to get teammates who aren’t that good at them. Take my advice – try (2) first!

See you tomorrow.