At 5:22 and 16:50 -- there's only 2 ways (not 3) W can have Tx doubleton (one or the other of the two spot cards). Low toward the 9 followed by running the jack works for T-KQxx (1 x 4:1) Tx-KQx (2 x 3:2) That's 9.6%. What suit breaks am I missing that takes 3 tricks? Sorry if being dense.
Oh wow! Thanks for pointing this out! I completely miscounted the number of Tx. So SuitPlay assumes low to the 9 picks up 2 Tx and 1 KQ for a total of 10%. Ultimately the conclusion is the same. Since declarer will pick up KT, QT, and both Tx combinations in practice, that's 4 4-2 combinations and gives the 13.6% odds of low to the 9. So this is still the better play for 3 tricks. But purely from the viewpoint of presuming defenders will always make declarer's life as hard as possible, SuitPlay's analysis is correct.
Firstly low the 9 and then running the Jack doesn't work for T-KQxx. First trick is x-T-J-H. I eventually cover the 9 with the 2nd H and remain with 76 vs 8x Secondly there are actually 4 T- holdings with west KT QT Tx Ty I think best plan is, if low to the 9 loses to an H run jack -> Tx, Ty if an honour appears, run jack -> KT, QT That leaves us with 2 interesting points. 1) what is best if the T appears 2) what would West play from any of these 4 holdings? if west completely randomized we could do no better than picking up 2 from 4, In practice, as pointed out in the video, it is very hard to play the T from HT, especially if J9 are in dummy. Also playing the T from Tx is very counterintuitive. In practice the T should not appear from any of the four T-holdings (or should it?), making low to the 9 picking up all 4 ten+1 holdings in west more often than it should. We don't pick up KQ tight in west, since we should run jack on the 2nd round, as restricted choice makes it twice as likely that the H was from HT than KQ tight. Also low to the 9 has some potential for a defensive error, while banging out the A has not. For instance West really shouldn't play Q from Qxx, but it happens. I'm probably missing some interseting points. I like the problem very much.
@@janiswehner4071 Thanks for this comment. 🥰. I watched the video and played with suit play and knew that I was missing something important to my understanding of the problem.
Awesome video. Suit combinations is the weakest part of my game, and I frankly don't know how to study them beyond pure memorization. So I have often used SuitPlay to "check" my intuition, when I come across a hand I (mis)played, and wasn't sure what the best line was for X tricks. When I see what SuitPlay recommends, it just tells me the answer with zero understanding as to why, other than to list out all the possible suit combos that exist. This feels like a poor study method, since it's rooted in memorization. How do you go calculating combinatorics? Its not a branch of math I've seen much. For example, you said "5 choose 3 is 10 ways". Is there an intuitive way I can calculate this on the fly?
The formula is {n choose k} = n! / (k! * [n-k]!) and you can see a lot of these numbers cancel. So for the 5 choose 3 example I do the simplified calculation: Remove 3! from the top and bottom so now its 5*4 / (5-3)! = 5*4 / 2! = 20/2 = 10 It's useful to note that {5 choose 3} = {5 choose (5-3)} so to use the above method you would want to pick the larger number to cancel the most terms. It's less effective for numbers like {13 choose 8} where doing the math in your head is harder.
At 5:22 and 16:50 -- there's only 2 ways (not 3) W can have Tx doubleton (one or the other of the two spot cards).
Low toward the 9 followed by running the jack works for
T-KQxx (1 x 4:1)
Tx-KQx (2 x 3:2)
That's 9.6%. What suit breaks am I missing that takes 3 tricks? Sorry if being dense.
Oh wow! Thanks for pointing this out! I completely miscounted the number of Tx. So SuitPlay assumes low to the 9 picks up 2 Tx and 1 KQ for a total of 10%.
Ultimately the conclusion is the same. Since declarer will pick up KT, QT, and both Tx combinations in practice, that's 4 4-2 combinations and gives the 13.6% odds of low to the 9. So this is still the better play for 3 tricks.
But purely from the viewpoint of presuming defenders will always make declarer's life as hard as possible, SuitPlay's analysis is correct.
Firstly low the 9 and then running the Jack doesn't work for T-KQxx. First trick is x-T-J-H. I eventually cover the 9 with the 2nd H and remain with 76 vs 8x
Secondly there are actually 4 T- holdings with west
KT
QT
Tx
Ty
I think best plan is,
if low to the 9 loses to an H run jack -> Tx, Ty
if an honour appears, run jack -> KT, QT
That leaves us with 2 interesting points.
1) what is best if the T appears
2) what would West play from any of these 4 holdings?
if west completely randomized we could do no better than picking up 2 from 4, In practice, as pointed out in the video, it is very hard to play the T from HT, especially if J9 are in dummy. Also playing the T from Tx is very counterintuitive. In practice the T should not appear from any of the four T-holdings (or should it?), making low to the 9 picking up all 4 ten+1 holdings in west more often than it should.
We don't pick up KQ tight in west, since we should run jack on the 2nd round, as restricted choice makes it twice as likely that the H was from HT than KQ tight.
Also low to the 9 has some potential for a defensive error, while banging out the A has not. For instance West really shouldn't play Q from Qxx, but it happens.
I'm probably missing some interseting points. I like the problem very much.
@@janiswehner4071 Thanks for this comment. 🥰. I watched the video and played with suit play and knew that I was missing something important to my understanding of the problem.
your youtube recommendations look surprisingly similar to mine.
Awesome video. Suit combinations is the weakest part of my game, and I frankly don't know how to study them beyond pure memorization. So I have often used SuitPlay to "check" my intuition, when I come across a hand I (mis)played, and wasn't sure what the best line was for X tricks. When I see what SuitPlay recommends, it just tells me the answer with zero understanding as to why, other than to list out all the possible suit combos that exist. This feels like a poor study method, since it's rooted in memorization.
How do you go calculating combinatorics? Its not a branch of math I've seen much. For example, you said "5 choose 3 is 10 ways". Is there an intuitive way I can calculate this on the fly?
The formula is {n choose k} = n! / (k! * [n-k]!) and you can see a lot of these numbers cancel.
So for the 5 choose 3 example I do the simplified calculation:
Remove 3! from the top and bottom so now its 5*4 / (5-3)! = 5*4 / 2! = 20/2 = 10
It's useful to note that {5 choose 3} = {5 choose (5-3)} so to use the above method you would want to pick the larger number to cancel the most terms. It's less effective for numbers like {13 choose 8} where doing the math in your head is harder.
Many thanks. Excellent
There are 2 other 10x combination: K10 and Q10. W plays Q or K, and you play for KQ stiff. This is the problem.