mishovnbg Posted March 4, 2004 Report Share Posted March 4, 2004 ------------------------------------------------------------------Hi all! Both me and and one of the best bulgarian players go down in this 3NT... At my table I notice that dealer(W) have a short thinking before to pass. BiddingW---N---E---SP---1♣*-P---3NT**, all pass * Precision, (15)16+** 14-15 bal Lead ♠Q [hv=d=w&v=n&n=skxhaxxda108xxcaxx&s=saxhqjxxdq9xckq8x]133|200|Scoring: IMP[/hv] ------How do you play it and why? ----------------------------------------------------------------Misho Quote Link to comment Share on other sites More sharing options...
Gerben47 Posted March 4, 2004 Report Share Posted March 4, 2004 The break in tempo doesn't help much because you only have a one-way double finesse in ♦. But first things first, ♠K, ♣AKQ. If they're 3-3 concede a heart. If not: let the ♦Q run, they knock out ♠A and let the ♦9 run. Doesn't work? Too bad, next board. What on earth was west thinking about if he has no diamond honor and just 6 cards in ♠? Quote Link to comment Share on other sites More sharing options...
the hog Posted March 4, 2004 Report Share Posted March 4, 2004 Gerben beat me to it. I play the same line. Quote Link to comment Share on other sites More sharing options...
bridgeboy Posted March 4, 2004 Report Share Posted March 4, 2004 I would play differently. From the slight pause of West, I think it is more likely he has 6 S and 4 H, considering to preempt with 4 card major. Hence I would cash Ace D play small D towards the Q. I think the differences in handling the D suit are very small, all near 75%.(maybe 4-0 breaks matter) :huh: Quote Link to comment Share on other sites More sharing options...
Free Posted March 4, 2004 Report Share Posted March 4, 2004 I play ♠K and low ♥ to the Q in trick 2. If ♥K is in front, ♥ break 3-3 or ♣ break 3-3 I'll make 9 tricks. If none of the suits break even, and ♥K was behind my ♥Q, I'll play on the ♦s, but I don't know how exactly (because I haven't got enough info on how the suits break). Perhaps ♦K is a stiff so you won't need to finesse... Quote Link to comment Share on other sites More sharing options...
Gerben47 Posted March 4, 2004 Report Share Posted March 4, 2004 Ah, I didn't say yet what I think west was thinking about. He was thinking about opening a weak two on a 5-card suit:QJT9x x Kxx Jxxx Or he might have both minor suit Jacks and didn't like all his quacks enough to open.QJ9xxx x Jx Jxxx Quote Link to comment Share on other sites More sharing options...
inquiry Posted March 4, 2004 Report Share Posted March 4, 2004 [hv=d=w&v=n&n=skxhaxxda108xxcaxx&s=saxhqjxxdq9xckq8x]133|200|Scoring: IMP[/hv]T1. Win in dummy.T2. Club King in handT3. Club ACE in dummy... did JT, J9, or T9 of clubs fall from WEST (?)T4. Heart ACE (if they two club honors fell or not). assume no king falls T5. Club from dummy, now..... if two club honors fell, hook the club 8 otherwise play to club Q At this point, you make if ♥K was singleton (not a great chance, but after taking clubs you will not get a chance to play the ace of ♥ and take the ♦ hook(s) later. If west plays two of the missing high clubs, the odds on taking the ♣ hook are so very much better than double hook in ♦.. West having specifically the JT9 of clubs tripleton are very low indeed (1.7%). Of course if EAST shows out of ♣ on third round, you don't hook, and if he plays the missing third honor you win and take your fourth club and set up a second heart trick. If all this goes wrong... ♣ not 3-3 and no luck with the JT9, and ♥K not fall, now I revert to the the double finessee in diamonds. I guess I will start with the ♦Q.... BTW, this hand shows why the 9 or 8 in your "long suits" can be of extra values. If my ♣ suit was KQ32... the chances to pick up the suit with 4-2 split goes away totally. as you can not benifit from JT, T9, or J9 doubleton behind you occur four card suit. Ben Quote Link to comment Share on other sites More sharing options...
luis Posted March 4, 2004 Report Share Posted March 4, 2004 Ok let's seeTake the 1st trick in hand.Club K and club to the A. If west dropped 2 club honors cash the dA if the dJ or dK fall continue diamonds and claim. If no diamond honor falls play heart ace and heart to the Q. Then take the spade in dummy and hook clubs if the hK didn't appear singleton or doubleton in east. If west didn't drop 2 club honors then cross to hand with a third club, if clubs are 3-3 duck a heart for the 9th trick. If clubs are not 3-3 then diamond to the Ace, if no diamond honor appears play for hearts 3-3 playing heart ace and heart to the Q. If a diamond honor appears just play on diamonds for 9 tricks. I win when a Clubs are 3-3 and west doesn't have JT9b West has 2 club honors doubletonc Any player has a singleton diamond honor or KJ doubleton d Somebody has the singleton hKe East has the doubleton hKf Hearts are 3-3 and west doesn't have JT9 of clubs Quote Link to comment Share on other sites More sharing options...
mishovnbg Posted March 5, 2004 Author Report Share Posted March 5, 2004 -----------------------------------------------------------------Hi all! ------ The player at other table played like Luis. ♥K onside or all chances in ♣ or ♥ 3-3 or K♦ fall. This summary is percentage play, better than direct double finesse in ♦. ------ At my table, as I wrote, W had a slight delay before pass. Like majority of players in bulgaria, opps play multi+dutch - they have weak 2 and major-minor 2 suiters. So only 2 hands are not covered - both minors/majors. From lead and my cards is clear - he can have only major 2 suiter, probably 6♠-4♥. To be 4 cards ♥ a suit for him, need to be headed by honour, so I have no chances with ♥ suit and probably with any 3-3 break. Stll 3-3 in ♣ is possible, if W is singleton/void ♦. Even my guess about W hand is not right, he had certainly some distribution, which means to play for ♦ length in E. I played A ♦ and another DI to hand and go down, because W had KJx in ♦. Yes, he had a distribution, good 6 cards in ♠, but his problem was not 4 cards in ♥, but far too much defense to open weak 2 in not vul, someting about I even can't suppose with our 29hcp :) . Full hand:[hv=d=w&v=n&n=skxhaxxda108xxcaxx&w=sqj10xxxhkxdkjxc9x&e=sxxxhxxxxdxxcj10xx&s=saxhqjxxdq9xckq8x]399|300|Scoring: IMP[/hv] ------Instead of comment play of world class players, I will give you an oppinion of Boian - one of the best among them. His line: Take ♠ lead with K, ♣ to K, ♦Q, if W don't cover with ♦K - take A♦, ♥A and another ♥... Note I will not describe variations of falling honours. --------------------------------------------------------------Misho Quote Link to comment Share on other sites More sharing options...
inquiry Posted March 5, 2004 Report Share Posted March 5, 2004 I win when a Clubs are 3-3 and west doesn't have JT9b West has 2 club honors doubletonc Anyone ♦K, ♦J or ♦KJ d Somebody has the singleton ♥Ke East has the ♥Kxf Hearts are 3-3 and west doesn't have JT9 of clubsJustice was done when the mathematically superiour line works... so ron, gerben and I would have made, luis, bridgeboy and free would have gone down. I rank my line as 83% chance, and luis's line at 66% Here is a comparision if someone would like to confirm my math (please, cause I make a lot of mistakes). First luis and I start off the same testing ♣ and for singleton ♥K. First, if the Guy has JT9 of clubs…. behind you…you are down, so I subtracted this 1.766% chance out at the bottom from the last percentage number Luis's line from above odds.... cumulative oddsa ♣s are 3-3 and west doesn't have JT9 33.752 33.752b West has 2 ♣ honors doubleton 4.84 36.96c Anyone has ♦K, ♦J or ♦KJ 11.304 44.087d Somebody has the singleton ♥K 2.422 45.44e East has the doubleton hK 8.075 49.84f ♥3-3 and west doesn't have JT9 of ♣s 35.528 65.899 my line from above (essential ron's and gerben's too)a ♣s are 3-3 and west doesn't have JT9 33.752 33.752b West has 2 ♣ honors doubleton 4.845 36.961c Somebody has the singleton hK 2.422 38.488d WEST has K or J of ♦s. 76 83.47 How is it that Luis's line which caters to 6 distributional conditions is worse mathematically than the my line that handles only 4 (if my math is right)? My a, b and c are the same as his a, b and d, but my condition d is a 76% chance, while his cef, add up to only 54.9% and they are not really additive, you have to take the ones out that worked from subsequent calculations... This is accounts for the difference I think. No one I know can work out the odds at the table, but if I got the odds right here, luis;s line is inferior to simply hooking twice in ♦'s (which would be 76%). However, while we can't work out the odds at the table, we should have a good feel for double hook (76) verses 3-3 or Kx on side. Double hook is better odds. But, if you want to take the hesitation before bidding into account, it is often right to throw the math tables out...so I am not sure how much that influenced peoples various line. I choose to ignore it here.Ben Quote Link to comment Share on other sites More sharing options...
Trpltrbl Posted March 7, 2004 Report Share Posted March 7, 2004 Ben, Are you, when playing online or in real, also sitting there with a calculator and calculating all your chances ? Because in that case you need to score about an 90+% on each brd to even out with all the time penalties hehehehe. Mike :( Quote Link to comment Share on other sites More sharing options...
Trpltrbl Posted March 7, 2004 Report Share Posted March 7, 2004 And on the hand's play I will go to my hand with a ♣ and play Q of ♦. I am sure they will not make it easy and cover with the K, now I can play either the ♥ King to be onside or to be 3-3 or the ♣ to be 3-3 or have 2 of J109 of ♣ have shown up, and have west only have 2 ♣. Against me suits never break 3-3 so I am going after the ♦'s. Mike :( Quote Link to comment Share on other sites More sharing options...
mishovnbg Posted March 7, 2004 Author Report Share Posted March 7, 2004 -------------------------------------------------------------Hi Ben! -------Your math about counting percentage is wrong my friend. 3 general mistakes, imho: 1. The percentage of stiif honours must be calculate by permutations not by variations. This mean for example that stiff ♥K from missing 6 cards is 14.53%/(1*2*3*4*5)=0.12% 2. Remaining percentage in paticular suit is changing after any turn. For example, after you take ♥A distribution 6-0 as well as 5-1 with stiff ♥K is no longer possible, so percentage of other distributions becomes more likely. I never have interest to know exactly raise of percents, but probaly is enough accurate to count them raising with proportion of percent of already impossible distributions. This mean after taking ♥A original percentage of 3-3 in ♥ = 35.53 will change to 35.53 + 1.49%*35.53/98.51% = 36.07% 3. The cumulative percentage must be count based on remaining part of 100% success after try of some percentage, using changed percent for remaing cards. For example counting percentage for play in ♥ suit by rounds (without counting changing of percents because ♠Q lead and follow all in suit :D ): - Take ♥A, ♥K didn't drop -> 0.1% unsuccess ->99.9% remain - Low to ♥QJx, E follow, W win his ♥K -> 50%*99.9%=49.95% unsuccess -> 50.4% remain. - Take ♥Q for 3-3 in ♥ -> (35.53%+(1.49%+14.53%)*35.53%/83.98%)*50.4%/100=21.32% of unsuccess -> 29.08% remain....... This mean 70.92% without counting yet percentage for stiff ♦K (1.18%), 3-3♣ without J109(34.28%) or stiff 2 of ♣J109(3.02%) without above corrections(1,2), but from remaining chance of success(3) of 29.08% mean additional 11.19% of success for total of 82.11% Counting your own percent I will leave for you friend as homework :(. ------I did above hard and unnecessary work imho only for you Ben and if I right, I ask you to promisse me to argue using percents only using real mathematics, instead of juggle with percents :) . ---------------------------------------------------------Misho Quote Link to comment Share on other sites More sharing options...
inquiry Posted March 7, 2004 Report Share Posted March 7, 2004 Ben, Are you, when playing online or in real, also sitting there with a calculator and calculating all your chances ? Because in that case you need to score about an 90+% on each brd to even out with all the time penalties hehehehe. Mike :( No, playing on line or in real world, I don't try to calculate the odds. I try to use what I think is the superior line..... I thought I made this clear when I said "No one I know can work out the odds at the table" Quote Link to comment Share on other sites More sharing options...
inquiry Posted March 7, 2004 Report Share Posted March 7, 2004 And on the hand's play I will go to my hand with a ♣ and play Q of ♦. I am sure they will not make it easy and cover with the K, now I can play either the ♥ King to be onside or to be 3-3 or the ♣ to be 3-3 or have 2 of J109 of ♣ have shown up, and have west only have 2 ♣. Against me suits never break 3-3 so I am going after the ♦'s. Mike :D Good idea, :( you should definately always play for not 3-3 suit breaks…. Hope that works for you. looooool Quote Link to comment Share on other sites More sharing options...
the hog Posted March 7, 2004 Report Share Posted March 7, 2004 Ben, you are older than Rado! What upsets me is that I am older than you - but just a little bit. :) Quote Link to comment Share on other sites More sharing options...
inquiry Posted March 8, 2004 Report Share Posted March 8, 2004 Hi all, unless you are interested in long (and probably wrong) math calculations using combination math and the like, feel free to skip this one. But anyone who is a math genius, read on and help misho and I correct our misconceptions Your math about counting percentage is wrong my friend. 3 general mistakes, imho: 1. The percentage of stiif honours must be calculate by permutations not by variations. This mean for example that stiff ♥K from missing 6 cards is 14.53%/(1*2*3*4*5)=0.12% First, I used tables and a spreadsheet to approximate the numbers. But first. First, my friend, you calculation of stiff King is very flawed. You don't divide the percenttage of 5-1 split by 5 factorial (5!)... This is combinational math, but I can show it simplier with a table of the possible 5-1 splits, assuming missing K65432 One opp....other oppK6543 2K6542 3K6532 4K6432 5K5432 665432 K So there is only six combinations, this works out to be 14.52%/6, no by 120, and gives the odds of a singleton King at 2.42%. If you don't want to work these odds out yourself, Theodore Triandaphyllopoulos has a great website with a bridge odds calculator where you can do this quickly, and without the need for a spreadsheet (and keeps from introducing horrible errors)... seeOdds calculator 2. Remaining percentage in paticular suit is changing after any turn. For example, after you take ♥A distribution 6-0 as well as 5-1 with stiff ♥K is no longer possible, so percentage of other distributions becomes more likely. I never have interest to know exactly raise of percents, but probaly is enough accurate to count them raising with proportion of percent of already impossible distributions. This mean after taking ♥A original percentage of 3-3 in ♥ = 35.53 will change to 35.53 + 1.49%*35.53/98.51% = 36.07% Yes, I understand that the odds change as cards are played. The odds calculator above even allows for you to enter cards played known in opponents hand for figuring out the odds. Let's take your example and see what the calculator gives us. 1) Both opponents played a spade, and 2) both played a small ♥. So what is the odds that the ♥ are 3-3? Well, they have gone up to 41.35%. And how has the odds of a 3-3 ♣ split been affected by the ♥ and ♠ played by both opponents? It has changed from 35.53% to 36.48%. For practical purposes at the table, I would tend to ignore these small changes. The calcuations, however are not that hard with the help of a computer. First, you have to know the general equation. When you can see 26 cards (dummy exposed), the number of hands the opponents can have is Combination(26:13) which comes out to be 10,400,600 hands. If you wanted to calculate the odds for a stiff ♥ king here, where x is the number of cards you want them to hold in the suit ♥ and y is the total number of ♥ in the opponents hand. Here x=1 (singleton), and y=6 (they have six ♥. This is the generall equation.... xCy * (26-y)C(13-x) This reduces to 1C6 * 20C12 = 6 * 125,970 = 755,820 If you divide 755,820/10,400,600 = 0.072670807, or 7.267%. This is the odds of one player having a singleton, so you have to double it for anyone having a singleton which is the 14.53% you quoted earlier. Now, if you want to take into account both played a ♠, before, the 26 in the equation becomes 24, and the 13 becomes 12, and the total number of hands is now 24!/12! = 2,704,156 1C6 * 18C11 = 6 * 31,824 = 190,944 When you divide this by the remaining number of combinations you get...190,944/2,704,156 = 0.070611. Just the odds the odds calculator above shows for 5-1 split in one hand when 1 card known in both opponents hand. You double this because either can have a singleton. So after the ♠ lead, the odds of dropping a singleton king isn't really 2.42% (14.53/6), but rather 2.35 (14.122/6), just as the table shows. Ok. Now we are getting some where. My line, chance of dropping the singleton K is 2.35%. That doesn't happen, so we move on to the chance, 3-3 Clubs, we know about the 1♠ and 1♥ in their hands, so we cash two clubs ending in dummy. If both follow, we also know about two clubs... So there are 4 known cards, with 2 clubs remaining out. The equation drops to... 1C2 * (18-2)C(9-1) = 2 * 12,870 = 25,750 At this point, with 4 cards played, the total number of combinations is 18!/9! = 48,620, so the chance of 3-3 ♣ split is 25,750/48,620 = 52.94. This should look fairly familiar, this is the odds of dropping a singleton king missing two cards, which is what this reduced too. Now, we didn't factor in the odds of JT9 tripleton behind South....If West dropped the T9, playing for the drop 3-3, works if East has stiff J, but loses if EAST has Jx. When you lead a club from dummy, if EAST shows out, (23.5% - or half of the 47.06% when not 3-3), you win and know what to do. But when EAST plays low, the fact that the J has not shown up affects the odds. There were 20 3-3 splits, only one of them will the JT9 be together. There were 15 4-2 splits, with JT, T9, or J9 being 3 of them. I am not sure how to calculate the odds precisely here as to whether to hook or not, but by gut feeling is if EAST plays low, 1/20th chance of 3-3 with JT9 together is a pretty good bet, but on the other hand, mathematically this might be wrong, as the calcuated here still suggest playing for the drop is the right line. So I probably screwed up somewhere... If you calculate the odds of all three top clubs behind you before touching clubs, in 3-3 split, it is astronomically small....only 1.8%, while chance of high doubleton is 4.9%. But ok, so to keep this simple, we just play three rounds of clubs, and find the bad news of no split. Now we have a choice of cash ♦ACE or play hook ♦... Here the roads split. Cash ♦A works with singleton K or J, or doubleton KJ... So what are odds of that? We have 5 known cards from EW. The odds of singleton honor in ♦ is... 1C5 * 16-5C(8-1) = 5 x 330 = 1650. The total number of combinations is now 16!/8! = 12,870, so 1650/12,870 = 12.82%. Once again, either opponent can be singleton, so double this, this comes to 25.6% (note agreement with "odds page". However, not all singletons are useful, just singleton K or J, which is only 2/5 of all singletons, so multiple this 25% by 2/5 to get 10.3% chance of success. However, you also score with doubleton KJ, I will not calculate the odds for this, (use the odds page), you will find it is ... 7.18% (either opponent KJ doubleton). Alright, now after all this, luis's line falls back on ♥ play, needing Kx originally on right, or the 3-3. This is 51.17% chance (use odds page with 6 known cards). So let's total this up.. drop singleton ♥K = 2.35, find ♣ 3-3 after both follow twice (if someone shows out, you might choose a different line), is 52.9%, this 52.9 is actually additive to the 2.35 because we took that chance into account in getting the 52.9, so we are up to 55.25, Ok, so no luck so far, the chance of dropping a stiff K or J, or a J from KJ at this point is 10.3% + 7.18% = 17.48%. This is no longer additive... this is so this is (100-55.25)*17.48%+55.25=63.26%Now the odds of finding the ♥K like you want it is 50.17% of this difference, which is (100-63.26)*50.17%+63.26 = 81.69, higher than I thought, but I ignored doubleton KJ when I first calculated luis's odds. Now, let's take the double hook after finding out the ♣ are 4-2. With five known cards in each hand (actually, you will know 6 in one of them, five in the other, I will ignore this here), the odds of success on the double hook is up to 76.67% (look it up on odds page). So after trying ♥ and ♣ I was at the same 55.25 as in the luis line, so that this 76.67 and multiple by the remaining, or (100-55.25)*76.67% + 55.25 = 89.56%. Ok, I short-circuited two assumptions in these calculations. I ignored ♣ doubleton honor hand hook, and tripleton honor, and I assumed ♣ where no worse than 4-2. I did this for both lines, so the biases will be there for both. The real trick to figure out is playing ♦A then on ♥ better than hooking twice in ♦. I think the odds still favor the double ♦ hook. Is this what you wanted misho? Can someone else clear up the muddy thinking on the assumptions, and make sure the additive and non-additive points are correct. Thanks... ben Quote Link to comment Share on other sites More sharing options...
flytoox Posted March 8, 2004 Report Share Posted March 8, 2004 ------------------------------------------------------------------Hi all! Both me and and one of the best bulgarian players go down in this 3NT... At my table I notice that dealer(W) have a short thinking before to pass. BiddingW---N---E---SP---1♣*-P---3NT**, all pass * Precision, (15)16+** 14-15 bal Lead ♠Q [hv=d=w&v=n&n=skxhaxxda108xxcaxx&s=saxhqjxxdq9xckq8x]133|200|Scoring: IMP[/hv] ------How do you play it and why? ----------------------------------------------------------------Misho I will take a simple way. Take s from dummy. Three rounds club stopping at hand, if c 3-3, then I finesse hk. If c 4-2, then I will play D Q, planning to finesse d twice. Quote Link to comment Share on other sites More sharing options...
inquiry Posted March 9, 2004 Report Share Posted March 9, 2004 my long post in reply to misho's comments about my math being wrong, I made a simple, but knowingly wrong assumption, that ♣ were either 3-3 or 4-2, and I ignored the chances of ♣JT9 being off-side. Now, I want to proceed with the comparison, taking all splits in club suit. The missing six clubs can be divided into WEST's hand the following ways. 0C6 x 16C11 = 1 x 4368 = 43681C6 x 16C10 = 6 x 48048 = 480482C6 x 16C9 = 15 x 11440 = 1716003C6 x 16C8 = 20 x 12870 = 2574004C6 x 16C7 = 15 x 11440 = 1716005C6 x 16C6 = 6 x 8008 = 480486C6 x 16C5 = 1 x 4368 = 4368TOTAL------------- 705,432 It is worth checking that with 22 cards left in their hands, 22!/11! = 705,432 So of these, the ones that dont work are obviously the 6-0 and 5-1 splits (total 104,832). In addition, only the 4-2 splits with WEST holding JT doubleton work, which is 1/5 of the 2C6 above... so that of the 171600 for West having 2♠, 4/5 fail. Or, 137280, as does all the hands were west has 4 clubs (another 171600). Finally, of the 20 ways WEST can hold three clubs, the proposed line of hooking club 8 if two honors fall from west, 1/20th of the 3-3 splits fail, or 12870. So the losing case total here is 425,582 losing cases... so the chance for success is here 39.5%. Note if ♣ are 3-3 with JT9 tripleton offside, there is no recovery from that. When we calculated the chance of success for dropping singleton ♥K at 2.35, now the odds in teh club suit is 39.5%. So the overall chance of these lines (they get better if both opponents follow to both clubs, of course), is ...This brings our chance of success up to 40.92% (39.5%x(100-2.35)+2.35). Now on top of this, you can compare the percentages figured out the odds of catching singleton K or J or doubleton J, and failing that, finding heart King on-sides doubleton or tripleton, to finessing twice in diamonds. The odds of finding the singleton K, J or doubleton KJ, or the odds of both hooks working are affected somewhat by how ♣s divided (EAST has 4 or WEST has 4). For sake of the argument, we will make it the best case for ♥K being onside and diamond honors being offside by giving WEST four clubs.. this split favors the play for stiff diamond, then on ♥ compared to the other possible split. Odds of WEST diamond distribution when he shows up with four clubs, are.... 0C5 9C7 1 36 361C5 9C6 5 84 4202C5 9C5 10 126 12603C5 9C4 10 126 12604C5 9C3 5 84 4205C5 9C2 1 36 36 For a total of 3,432 combinations. The 6-0, and 0-6 fail (72 chances). If the 420 5-1 splits here, 3/5 fail, or 315, of the 3-2 splits, only one out of ten works, or 1134 fail when WEST has two, and only 1 out of 10 when EAST has two (another 1134 fails). And when WEST holds four diamonds, only 2/5 times will east have singleton honor, so again, 315 of these cases fail. Adding up the failures, we get 2,970 failing cases, or a percentage of success of only 13.5%. Adding this chance to the early failure cases (60.1%*13.5%)= adds 7.9% to our chances of success when playing for diamond honor drop. 39.5+7.9 = 47.4%. This having failed, now you have to play for luck in ♥, now we turn to east hand, knowing WEST had 4 clubs, east 2, and everyone one ♦, 1♠ and 1♥. What are the chances EAST will have started with Kxx, Kx or any three hearts? There are only four hearts out now, so the equation is simplified even further (we know 12 of opponents 26 cards), and 5 of EAST's cards. 0C 10C8 1 45 451C4 10C7 4 120 4802C4 10C6 6 210 12603C4 10C5 4 252 10084C4 10C4 1 210 210Total combinations = 3003 If EAST is now void (singleton ♥ originally), you are toast. If he has only one heart (480 cases), there is only a one in four chance that it is the king, so that is 1/4 of 480 or 120 winning cases. If EAST has 2 ♥, you always win, so that is 1260 winning cases, If east has three or four hearts, you always lose. So the winning cases, are 1260+120 = 1380 or ~ 46% chance of success (actually giving WEST long clubs helped in diamonds, but hurt in hearts). So you had 47.4% chance of success before playing on ♥ (meaning 52.3 failure)... multiply 52.3xthis 46% and we can add 24% more to the chance of success of this line. 47.4+24 = 71.4%. My initial guess was this line was inferior to simply taking the ♦ hook at trick one, and I stand by that estimate (I thougth it was about 68%... but I left out doubleton KJ of diamonds). As for my line with the double hook. After clubs split, for the sake of aguement using same one I gave above with WEST long in clubs (if EAST long, better chances double hook wins). The diamond distributions for wEST would be (same as above)0C5 9C7 1 36 361C5 9C6 5 84 4202C5 9C5 10 126 12603C5 9C4 10 126 12604C5 9C3 5 84 4205C5 9C2 1 36 36 If west has five or four diamonds (last two case), double hook surely wins (456). When west has 3 diamonds, double hook wins 9 out of ten times (only KJ doubleton off side loses), for 1134 success. When west has two diamonds, the hook works in 7 of the ten combinations (KJ, K2, K3, K4, J2, J3, J4... hence losing only to KJ2, KJ3, or KJ4 offsides). That is another 882 winning combinations. Finally when WEST has only one diamond, if it is the K or J (2 out of 5) the double hook wins, for an additional, 168 winning plays. The winning distributions here add up to 76.9%. Taking the chance of success in singleton heart and combined play in clubs, the winning cases before the diamond play was 40.92 (losing 59.1). 59.1x76.9% = 45.46%, adding this the the earlier 40.92, you get chance of success of this line as 84.38%. Now Misho, using the quick and dirty tables, I calculated the percentage of making on my line as 83.47... very close to the one caluculated using horrible math. I missed luis's line by a few percentage, forgetting to factor in doubleton KJ of diamonds, so my estimate of 86.5% would not have been far off the actual value caluculated above of 71.4%, and would still show that my statement that straight forward double hook in diamonds has better chance than his line. I much rather just jump to tables and roughly estimate these things when there is a question than beat it to death with this type of math. :D Ben.... Quote Link to comment Share on other sites More sharing options...
mishovnbg Posted March 11, 2004 Author Report Share Posted March 11, 2004 ----------------------------------------------Hi Ben! ------First, thank you very much, you lose much of time friend, to explain unclear for me math :) . I was wrong, but thanks to that post I have now 2 pages of right calculations and can study them rest of my life :D . ----------------------------------------------Misho Quote Link to comment Share on other sites More sharing options...
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