Help
kenrexford, on Feb 8 2007, 12:14 PM, said:
Kelsey is probably figuring in the reality that, in many layouts, LHO cannot play small (Kx/Qx).
I will have to think on this claim a while. When LHO follows to the second round in this suit, you care correct Hx as well as small singleton is removed. That leaves nine possible holdings for LHO (where x is any small card and H is the honor not lost at the first trick).
xx
xxx
xxxx
xxxxx
Hxx
Hxxx
Hxxxx
Hxxxxx
Since there is only five "x's", there is only one possible way LHO can hold xxxxx (all the smalls and none of the two bigs), so that 1 combination. Likewise, there is only two way LHO can hold Hxxxxx (his partner has to hold the singleton K or Q == the other honor. Using such there is ten ways he could hold Hxxxx (his parter holds the either of the H and any one of the five little "x's").
If you think of it this way, there is a grand total (after EAST wins the first trick and west follows LOW to the second) of 78 possible combinations (numbers shown below -- with those where the "winning" play of finessee again is in bold).
10 xx
10 xxx
5 xxxx
1 xxxxx
20 Hxx
20 Hxxx
10 Hxxxx
2 Hxxxxx
If you total the number of winning cases (where the finessee) works, you find 52 of them. 52/78 = 0.66667, the expected value for this restricted choice play -- even taking into account WEST can not have Kx or Qx nor x.
So to answer the question in this thread, assume you are vul, lets see how you do...
1. Play the ACE on second round. You drop the second honor offside one time out of 78, and you go down 1 the other 77 times, for an average of -91.03 per play.
2. Hook the second time. You will make 52 times out of 78 and go down two 26 times out of 78 for an average result of +333.33 points per board.
Which looks better to you? I know what you are thinking, imps scale is not linear, so what would be the expectet result for both plays if our opponents were in the same contract. Fair enough. Assume your opponents are in 3NT and take the second finessee how do you fair?
On those hands where you play ACE on second round. If RHO has KQ doubleton, you win 12 imps, and on the other 25 boards where the second honor is offside, you win 3 imps. However, on the 52 boards where the honor is onside, you go down one (-100) while they make so you lose 12 imps on each. The net average loss per board is -6.88 imps for this line of play.
If you choose the same play as your opponent (finesse), you of course average 0 imps. Now imagine your opponent rested in 1NT, making either +90 or +150 how do the various plays do? On the 52 hands where hook wins, they are +150, on the others they are +90. If you play the ACE on second round you are +600 once, and win 10 imps on that one, but you are -100 on all the others, losing 5 imps on all the other 77 boards, for essentially a certain 5 imp/board loss.
On the other hand, if you finesse on the second round, you win 10 imps on 52 boards (520 imps) and lose 7 imps on 26 boards (-182 imps). 520-182 = net +338 imps, divided by 78 = average gain of 4.33 imps.
So, in each case, risking down two is much, much better than playing for the fairly sure down two.. don't you think?