Board 5
Our side vulnerable
Our side vulnerable
| ♠ A K 8 5 2 ♥ 8 ♦ A J ♣ K 9 8 7 5 |
Two passes to me. I know some people open one club with this pattern, and I used to be one of them. But I never had much success with that approach. (I can recall one deal where it did work spectacularly well. At rubber bridge, I opened one club with six-six in the black suits, LHO overcalled one spade, and partner made a negative double. Even with that result, however, I still think I've been a net loser opening one club.)
I open one spade, partner responds one trump, I bid two clubs, and partner corrects to two spades. I'm glad I'm not playing forcing notrumps, since I have only one more club than I've promised rather than two more. If partner weren't a passed hand, my advantage over the Eastern Science Fiction players would be even more pronounced, since my partner's upper limit in high cards would be lower. I would probably pass anyway, but I pass a bit more comfortably in my style.
LHO passes as well and leads the three of diamonds:
| NORTH
♠ 7 3 ♥ Q 9 7 5 3 ♦ K 7 2 ♣ A 10 3 |
||
| SOUTH
♠ A K 8 5 2 ♥ 8 ♦ A J ♣ K 9 8 7 5 |
| West | North | East | South |
| Pass | Pass | 1 ♠ | |
| Pass | 1 NT | Pass | 2 ♣ |
| Pass | 2 ♠ | (All pass) |
East plays the queen, and I take the ace. Frequently, when one is concerned about a bad break in a side suit, it is right to start the side suit before touching trumps. If the side suit splits badly, you may be able to compress your trump losers and your side suit losers. With only two trumps in dummy, however, this doesn't look like the right approach. If someone ruffs the second club and plays a trump, I may be sorry I started clubs early. I cash the ace and king of spades. West plays ten-jack; East plays four-six.
A priori, the odds are about four to three that trumps are four-two rather than three-three. Has West's carding changed that? West could have begun with (A) jack-ten doubleton, (B) queen-jack-ten, (C) jack-ten-nine, or (D) queen-jack-ten-nine. Let's assume that, initially, each of these four cases is equally likely. (Yes, (B) and (C) are slightly more likely, but not by enough to make much of a difference in our analysis.) The principle of restricted choice dictates that the true relative frequency of each case is the a priori frequency divided by the number of ways West might play his cards. In theory, West could play his cards in any order with any of these holdings. Certain orders, however, wouldn't make sense. For example, queen-ten, queen-nine or, to a lesser extent, jack-nine would be a poor choice, since declarer would be disinclined to play West for a doubleton after such a sequence. So let's assume that West would always choose touching cards if he played his cards in descending order. That means he has two ways to card from (A), five ways from (B) or (C), and nine ways from (D).
The number of cases where the suit can split three-three, then, is one fifth (for B) plus one fifth (for C): a total of .40. The number of cases where the suit can split four-two is one half (for A) plus one ninth (for D): a total of .61. Thus the odds of a four-two versus a three-three break are roughly three to two. Of course, I didn't go through this calculation at the table. I already know that the odds of a four-two break go up when restricted choice comes into play. And on this particular deal, I don't care by how much. I included this calculation in the post only because I've never seen anyone discuss the touching-card assumption before or how it affects the calculation.
I cash the diamond jack--four--seven--eight, and play a club to dummy's ace. West plays the deuce; East, the four. I play the king of diamonds--ten--heart eight--five, then lead the ten of clubs. East plays the six. I doubt West has a stiff club, but it probably won't hurt to let this ride. Even if West wins and gives his partner a club ruff, it will probably be with a natural trump trick. I play low. West wins with the jack and plays the nine of diamonds, on which East pitches the queen of clubs.
I ruff, and I'm down to:
| NORTH
♠ -- ♥ Q 9 7 5 ♦ -- ♣ 3 |
||
| SOUTH
♠ 8 5 ♥ -- ♦ -- ♣ K 9 8 |
West was either 3-3-5-2 or 2-4-5-2. If I play a trump now, I'll make five in the former case or go down one in the latter. If I play clubs, I guarantee making four. Since this looks like a normal contract and since declarer can be held to three on a heart lead however trumps split, I suspect making four is above average. So I see no reason not to settle for that result. I play the king of clubs. Making four:
| NORTH
♠ 7 3 ♥ Q 9 7 5 3 ♦ K 7 2 ♣ A 10 3 |
||
| WEST
♠ J 10 ♥ K 6 4 2 ♦ 9 6 5 4 3 ♣ J 2 |
EAST
♠ Q 9 6 4 ♥ A J 10 ♦ Q 10 8 ♣ Q 6 4 | |
| SOUTH
♠ A K 8 5 2 ♥ 8 ♦ A J ♣ K 9 8 7 5 |
Because of the favorable position in the black suits, we can make five clubs. But I can't see getting there even if I bid three clubs over two spades. Plus 170 turns out to be worth nine out of twelve matchpoints. Three other pairs were plus 170, two were plus 140, and one reached four spades and went down.
Me: +170 (9 MP)
Total: 39 MP (65%)
Current rank: 1st
A 3-3 break is 35.5% (I use 36% at the table). There are 20 different combinations West may have, so each is 1.8% likely. So if W has three touching cards (QJ10 or J109) he will presumably play J then 10 1/5 of the time, by your reasoning. This makes the likelihood of seeing J, then 10 from a 3-card holding 0.72% (approximately).
ReplyDeleteA 2-4 break is 24.2% (I use 24% at the table), and there are 15 different doubletons W may have. From J10, he will play J then 10 1/2 the time. This is 1/30th of the time, or about 0.8%.
Similarly, the probability that W was dealt QJ109 is also 24/15, or 1.6%. Using your reasoning, he will play J then 10 0.18% of the time, so you will see J then 10 a total of 0.98%. This makes the odds of a 4-2 split where you see the J then 10 from a perfectly random LHO 0.98/(0.98+0.72) or about 59%.
So your calculation is good enough to make the right decision, but I question one of your assumptions. From QJ109, it is probably correct to play 10-9 or 9-10 randomly, so declarer needs to consider Q109 as well as J109. If you accept that, then when you see J then 10, the odds become 0.8/(0.8+0.72) or about 10:9 that the suit is 4-2.
The validity of my statement is shown by considering that 10 then 9 will be from either Q109 or J109, and the probability of a 3-3 split will be doubled, or 1.44%. The odds for declarer will then be to play for a 3-3 break.
This is all too tedious to work out at the table while defending, of course. But a quick general rule for which card or cards to randomize is to figure out what holding might matter to declarer. From J109, if it is possible for him to take a 3rd round finesse into your hand, then you could make a trick if partner has neither the 8 nor the Q (but, oddly enough, not if he has either), so you should randomize from 3, but not J-9. If it is not possible for him to finesse into your hand on the 3rd round, you are toast unless partner has the Q, so you should randomize as Phil suggests. But from QJ109 you should randomize between the bottom two cards only, to take advantage of the above.
Unless somebody sees something wrong with my math....
My way of counting cases rather than adding up probabilities does ignore the fact that a specific three-three break (1.8%) and a specific four-two break (1.6%) are not equally likely. But the difference is so small that it usually doesn't matter, and counting cases makes the computations considerably easier. If my analysis shows a tie between two plays, then I go back consider which play catered to more three-three breaks.
ReplyDeleteI don't quite get your point about QJ109. I don't see what's wrong with playing J-10, 10-J, or 9-J. Although I do see that any sequence that includes the queen is probably wrong (a point I missed in the write-up), since each such sequence is consistent with only one possible three-card holding while other sequences are consistent with two. At least I think that's what I see. It is Memorial Day after all, and I've had too many strawberry daiquiris to think about his clearly. I'll give it another shot tomorrow.
In the light of day, all I can say for sure is that defender's correct play from QJ109 is considerably more complicated than I realized. I don't think this fact has any bearing on my calculation in the post, since all that matters for that calculation is how people actually DO card, not how they SHOULD card. But I am curious how one actually should card from QJ109. When I have some spare time, I'll work it out and post again.
ReplyDeleteI think the point I was trying to make was, with the N-S spade layout AK?xx opposite xx, a W with QJ109 should try to card keeping as many of the two-card and three-card holdings included as possible.
ReplyDeletePlaying the Q and 9 in descending order is hopeless for concealment as you point out. Playing 9 then Q leaves only QJ109, QJ9, or Q9 possible.
Q and 10 lead to similar analysis, you must play them up the line, and the relevant holdings are Q10, QJ10, or QJ109.
Q and J can be played in either order, but still allow only three possibilities - QJ, QJ10, or QJ109.
J and 10 can again be played in any order, they allow for J10, QJ10, J109, or QJ109, for initial holdings instead of 3.
J and 9 again should be played in ascending order, they allow for J9, J109, QJ9, or QJ109, again four initial holdings.
10 and 9 can be played in either order, and allow for 109, Q109, J109, or QJ109, again 4 combinations.
So there are really only 5 rational ways to play from QJ109 - play 9-10, 9-J, 10-9, 10-J, or J-10. When you choose these options, it will be "percentage" for declarer to play for a 3-3 break, since you have two remaining "live" original 3-card holdings and one each 4-card and doubleton. As your calculation notes, doubletons are weighted 1/2, tripleton sequences are weighted 1/6, and non-sequential sequences are weighted 1/2, and the 4-card holding is weighted 1/5 (rather than your 1/9).
Using those weights, a 4-2 break is (1/2+1/5) of 1.6%, or 1.12%. The 3-3 breaks are weighted differently. 9 and 10 in either order will be 1/2 of Q109 and 1/6 of J109, for a total of 2/3 of 1.8%, or 1.2% - percentage will be to play for the drop. 9, then J is 1/2 of QJ9, and 1/6 of J109, again 1.2%. But J and 10 in either order is 1/6 of each of QJ10 or J109, or only 1/3 of 1.8% - percentage will be to play for the 4-2 break.
In summary - when you hold a 3-card sequence and want to convince declarer to play for a 4-2 break, you want to randomize the middle card with the card next to the one you hope partner holds. When you hold a 4-card sequence, and want declarer to play for a 3-3 break, randomize the two lowest cards.
Getting me to put Excel away is usually a good way to get me to make a point in a way normal people might be able to understand.
Not quite. See the next post.
ReplyDelete