02-27-2013, 02:57 PM

richardpaulhall posted this hand over in the bidding discussion:

For this discussion, let's say we got this hand in 1st seat, bid 65, and took it right there. Let's consider this strictly as a play problem. At trick 1, is a no-brainer. After that, what you would desperately like to do is get partner in while any club/diamond aces might still be worth tricks. So, we'll also assume partner does NOT have the other 3 diamond aces, and therefore doesn't signal. We'll also assume neither LHO nor parnter melded aces...but we'll consider the implications if RHO melded aces.

Options:

a)

b)

c)

d) spades from the top

There's no reason to ever lead a club, based on your hand.

Quick comment about the defense, for the first 3 cases. From their perspective...a 65 bid, cash only 1 side ace...either a great trump suit or a big sandbag with a 2-suiter (say, 9-7 with an second suit or so). In either case, the defense has to worry that we have short suits. Ergo: they may have to cash in the side suits ASAP.

a) can work in one of two ways. First, it suggests you're looking for partner to ruff diamonds; perhaps you have a hand with

where you can strip trumps when you've got the secondary diamonds ready to go. This MIGHT induce LHO to duck the diamond, which greatly increases the chance you'll get partner in. Second, without any trickery, it hits partner when LHO has no , and partner has at least 1.

What are the odds of this? I'll use the notation X=Y=Z to denote who has the missing aces, *in clockwise order* from you. So 1=2=0 would mean LHO has one, and partner the other 2.

Each ace can go fundamentally equally to each player, so there are 3^n (n == number of aces missing) possible X=Y=Z triples, each equally likely (not strictly mathematically true, but close enough for our purposes).

So with 3 aces missing, there are 27 aces. Our 'success' cases require LHO to have no aces, and partner to have 1 or 2 aces. (If he has all 3, we'll assume he signals.) This is 2 of the 27 aces...or roughly a 7% chance. UGH.

b) suggests hearts is our second suit...it's actually telling a better story. The whole hand might be this:

, is playing aggressively, but it is more logical than the story I offered for case a). My 's are much more valuable if I retain my .

The chance that LHO has no , and partner has at least 1...there are 4 such cases, out of the (now) 81 possible. That's 1 in 20...5%. BAD. So really, you'd be hoping to mislead the defenders and induce a mistake.

c) . The straightforward lead trying to reach partner. There are 9 cases...you reach partner when the trump aces are 0=1=1 or 0=2=0. 2 cases, 22%. Not great, but way better than the other 2.

Now we can consider the impact if RHO melds aces. In case a) there are now 2 missing diamond aces, and partner *probably* doesn't have both (no signal). We're down to 9 cases, with 0=1=1 being the successful one. The diamond play rises to 11%. The heart play similarly rises to 3 cases out of 27, or 11%. The ...with only 1 missing ace, that rises to 33%. Pretty nice improvement!

d) Spades from the top is a strategy to reduce your trump losers to 1. If you can, you now have 12 tricks, so your estimate is pulling 30. Give partner 1 heart ace, and you're at 31. If the trumps are 3-3-2 (the - separator means 'in any order', while the = means 'in that order)...you never lose more than 1 trump trick.

So...what's better? There are 8 cards missing, which means 3^8 or 6561 cases.

(and wins) loses when partner has A or Ax, and RHO the other A. Partner has singleton ace, RHO has ace...there are 6 cards distributed between the opponents. This can happen 64 ways. Partner has Ax, RHO A...5 missing cards, 32 cases. Therefore, is WRONG, 96 times.

loses (and wins) when *LHO* has A or Ax, and partner has at least 1 more card in the suit with the required A. If LHO has the bare A, partner must have Ax, and there are 5 cards between partner and RHO...32 cases. If LHO has Ax, partner must have Axx, so there are only 3 cards be be spread around...8 cases. Therefore loses only in 40 cases. Oh, and LHO can have AA, with partner having Txx or longer. That's another 8 cases. (Yes, partner might play the T on your first ace...but with the holding you have, the ONLY non-point missing, is 1 J. He can play a K to give the point.)

Still, *in the cases that matter*... is effective twice as often as , when the goal is to reach partner, and that really needs to be your first priority. at trick 2 may make you lose 2 trump tricks, when would hold the loss to 1...but getting partner in, before either defender, is worth more than 2 tricks in clubs and diamonds. Partner will have, on average, 1 diamond ace, and 1 1/3 club aces.

For this discussion, let's say we got this hand in 1st seat, bid 65, and took it right there. Let's consider this strictly as a play problem. At trick 1, is a no-brainer. After that, what you would desperately like to do is get partner in while any club/diamond aces might still be worth tricks. So, we'll also assume partner does NOT have the other 3 diamond aces, and therefore doesn't signal. We'll also assume neither LHO nor parnter melded aces...but we'll consider the implications if RHO melded aces.

Options:

a)

b)

c)

d) spades from the top

There's no reason to ever lead a club, based on your hand.

Quick comment about the defense, for the first 3 cases. From their perspective...a 65 bid, cash only 1 side ace...either a great trump suit or a big sandbag with a 2-suiter (say, 9-7 with an second suit or so). In either case, the defense has to worry that we have short suits. Ergo: they may have to cash in the side suits ASAP.

a) can work in one of two ways. First, it suggests you're looking for partner to ruff diamonds; perhaps you have a hand with

where you can strip trumps when you've got the secondary diamonds ready to go. This MIGHT induce LHO to duck the diamond, which greatly increases the chance you'll get partner in. Second, without any trickery, it hits partner when LHO has no , and partner has at least 1.

What are the odds of this? I'll use the notation X=Y=Z to denote who has the missing aces, *in clockwise order* from you. So 1=2=0 would mean LHO has one, and partner the other 2.

Each ace can go fundamentally equally to each player, so there are 3^n (n == number of aces missing) possible X=Y=Z triples, each equally likely (not strictly mathematically true, but close enough for our purposes).

So with 3 aces missing, there are 27 aces. Our 'success' cases require LHO to have no aces, and partner to have 1 or 2 aces. (If he has all 3, we'll assume he signals.) This is 2 of the 27 aces...or roughly a 7% chance. UGH.

b) suggests hearts is our second suit...it's actually telling a better story. The whole hand might be this:

, is playing aggressively, but it is more logical than the story I offered for case a). My 's are much more valuable if I retain my .

The chance that LHO has no , and partner has at least 1...there are 4 such cases, out of the (now) 81 possible. That's 1 in 20...5%. BAD. So really, you'd be hoping to mislead the defenders and induce a mistake.

c) . The straightforward lead trying to reach partner. There are 9 cases...you reach partner when the trump aces are 0=1=1 or 0=2=0. 2 cases, 22%. Not great, but way better than the other 2.

Now we can consider the impact if RHO melds aces. In case a) there are now 2 missing diamond aces, and partner *probably* doesn't have both (no signal). We're down to 9 cases, with 0=1=1 being the successful one. The diamond play rises to 11%. The heart play similarly rises to 3 cases out of 27, or 11%. The ...with only 1 missing ace, that rises to 33%. Pretty nice improvement!

d) Spades from the top is a strategy to reduce your trump losers to 1. If you can, you now have 12 tricks, so your estimate is pulling 30. Give partner 1 heart ace, and you're at 31. If the trumps are 3-3-2 (the - separator means 'in any order', while the = means 'in that order)...you never lose more than 1 trump trick.

So...what's better? There are 8 cards missing, which means 3^8 or 6561 cases.

(and wins) loses when partner has A or Ax, and RHO the other A. Partner has singleton ace, RHO has ace...there are 6 cards distributed between the opponents. This can happen 64 ways. Partner has Ax, RHO A...5 missing cards, 32 cases. Therefore, is WRONG, 96 times.

loses (and wins) when *LHO* has A or Ax, and partner has at least 1 more card in the suit with the required A. If LHO has the bare A, partner must have Ax, and there are 5 cards between partner and RHO...32 cases. If LHO has Ax, partner must have Axx, so there are only 3 cards be be spread around...8 cases. Therefore loses only in 40 cases. Oh, and LHO can have AA, with partner having Txx or longer. That's another 8 cases. (Yes, partner might play the T on your first ace...but with the holding you have, the ONLY non-point missing, is 1 J. He can play a K to give the point.)

Still, *in the cases that matter*... is effective twice as often as , when the goal is to reach partner, and that really needs to be your first priority. at trick 2 may make you lose 2 trump tricks, when would hold the loss to 1...but getting partner in, before either defender, is worth more than 2 tricks in clubs and diamonds. Partner will have, on average, 1 diamond ace, and 1 1/3 club aces.