bhugi Posted December 15, 2003 Report Share Posted December 15, 2003 In a team match, both side go to 6NT with opponents being silent. NorthAQ72KQ732A482 SouthK96AKJ63AKT95 Both table West led Spade J. I made the contract but not using the best line of play.Plan the percentage play. ----scroll down only after you have tried ;D WestJT8586498QJ64 East43JT95QT75273 Quote Link to comment Share on other sites More sharing options...
luis Posted December 15, 2003 Report Share Posted December 15, 2003 In a team match, both side go to 6NT with opponents being silent. NorthAQ72KQ732A482 SouthK96AKJ63AKT95 Both table West led Spade J. I made the contract but not using the best line of play.Plan the percentage play. ----scroll down only after you have tried ;D WestJT8586498QJ64 East43JT95QT75273 I'd have taken the lead with the sK and play the s9 inmediately, covered. Since the 8 doesn't drop I'd have taken a finesse in clubs flotaing the 8, take the return and cash the 3 hearts and AK of diamonds, the Q of diamonds doesn't drop then another club finesse for down 1. Quote Link to comment Share on other sites More sharing options...
flytoox Posted December 15, 2003 Report Share Posted December 15, 2003 In a team match, both side go to 6NT with opponents being silent. NorthAQ72KQ732A482 SouthK96AKJ63AKT95 Both table West led Spade J. I made the contract but not using the best line of play.Plan the percentage play. ----scroll down only after you have tried ;D WestJT8586498QJ64 East43JT95QT75273 I'd have taken the lead with the sK and play the s9 inmediately, covered. Since the 8 doesn't drop I'd have taken a finesse in clubs flotaing the 8, take the return and cash the 3 hearts and AK of diamonds, the Q of diamonds doesn't drop then another club finesse for down 1. hehe, i play this line too. Quote Link to comment Share on other sites More sharing options...
inquiry Posted December 15, 2003 Report Share Posted December 15, 2003 In a team match, both side go to 6NT with opponents being silent. NorthAQ72KQ732A482 SouthK96AKJ63AKT95 Both table West led Spade J. I made the contract but not using the best line of play.Plan the percentage play. You have 10 top tricks (3S, 2D, 2C and 3H). It is difficult to say what the best percentage play is, the double club hook looks reasonable as it can gain two tricks anytime East has (Hxx, Hxxx, Hx, Hxxx, HHx, HHxx, HHxxx, HH, or HHxxxx) all in all that is about 73.6% of the time and it gains one trick and sets up a host of squeeze possibilities when EAST holds Hxxxxx (only 1.3% chance) . Against that, the double hook is down whenever WEST holds C-QJ, C-QJx, or C-QJxx (with QJxxx or QJxxxx you will discover EAST's singleton or void in clubs before playing the first or second hook). Those are very good odds indeed. With potential threats in all four suits, there is a second potential line that has some merit. Win the spade King, cross to the heart ACE, play spade to the Queen, cash one high heart (discard a club) and duck a heart (assuming both followed to the second round, if not, hook the club instead). If were 3-4, you have 11 tricks now with the long heart, and spade/club/diamond threats. If hearts were 5-2 you have to rely on a diamond hook for your 11th trick and play for a compound squeeze against whoever holds the long heart with spade, heart, diamond and clubs all threats. This line gains one trick anytime hearts are 4-3 (62.2%) and you get the extra trick via diamond hook half the time when the suit splits 5-2 or worse (half of 37.8 or 18.9%). This comes out to be 81.1% of the time you will have 11 tricks on this line (compared with the 73.6% of time you instantly gain two tricks with the double club hook. So is 81.1% chance for 11 tricks better than 73.6 for 12 (with additonal trick in 1.2% of the time?). I think so, simply because spades can be 3-3 or the squeeze ending should be easier to find. Also, the odds for the heart duck is probably slightly low estimated, as the 7-0 and 6-1 splits have been eliminated from possibilities before you duck the heart trick. If my math is right, after both opponents have followed twice to heart leads, the odds of a 4-3 split is then about 83% instead of 62%, changing the combined odds of the 4-3 split or the Diamond Q onside to gain one trick to 91.5%. Finally, the heart duck line has one feature I like. You are not automatically down when it is wrong. Spades can alwyas be 3-3 with diamond hook on as an alternative. Whereas when both club honors are off side, the double hook is down right away. Against this, some of the endings for the squeeze may be hard to read (although who guards spades will be obvious). There is thus lots to say for the immediate double club hook. Ben Quote Link to comment Share on other sites More sharing options...
luke warm Posted December 16, 2003 Report Share Posted December 16, 2003 i don't know about the percentages, but to my thinking the only way i'm set is if west has both missing club honors.. so i cross to dummy and lead a club, finessing the 9 or 10 unless Q or J played by east.. Quote Link to comment Share on other sites More sharing options...
bhugi Posted December 16, 2003 Author Report Share Posted December 16, 2003 In a team match, both side go to 6NT with opponents being silent. NorthAQ72KQ732A482 SouthK96AKJ63AKT95 Both table West led Spade J. I made the contract but not using the best line of play.Plan the percentage play. You have 10 top tricks (3S, 2D, 2C and 3H). It is difficult to say what the best percentage play is, the double club hook looks reasonable as it can gain two tricks anytime East has (Hxx, Hxxx, Hx, Hxxx, HHx, HHxx, HHxxx, HH, or HHxxxx) all in all that is about 73.6% of the time and it gains one trick and sets up a host of squeeze possibilities when EAST holds Hxxxxx (only 1.3% chance) . Against that, the double hook is down whenever WEST holds C-QJ, C-QJx, or C-QJxx (with QJxxx or QJxxxx you will discover EAST's singleton or void in clubs before playing the first or second hook). Those are very good odds indeed. With potential threats in all four suits, there is a second potential line that has some merit. Win the spade King, cross to the heart ACE, play spade to the Queen, cash one high heart (discard a club) and duck a heart (assuming both followed to the second round, if not, hook the club instead). If were 3-4, you have 11 tricks now with the long heart, and spade/club/diamond threats. If hearts were 5-2 you have to rely on a diamond hook for your 11th trick and play for a compound squeeze against whoever holds the long heart with spade, heart, diamond and clubs all threats. This line gains one trick anytime hearts are 4-3 (62.2%) and you get the extra trick via diamond hook half the time when the suit splits 5-2 or worse (half of 37.8 or 18.9%). This comes out to be 81.1% of the time you will have 11 tricks on this line (compared with the 73.6% of time you instantly gain two tricks with the double club hook. So is 81.1% chance for 11 tricks better than 73.6 for 12 (with additonal trick in 1.2% of the time?). I think so, simply because spades can be 3-3 or the squeeze ending should be easier to find. Also, the odds for the heart duck is probably slightly low estimated, as the 7-0 and 6-1 splits have been eliminated from possibilities before you duck the heart trick. If my math is right, after both opponents have followed twice to heart leads, the odds of a 4-3 split is then about 83% instead of 62%, changing the combined odds of the 4-3 split or the Diamond Q onside to gain one trick to 91.5%. Finally, the heart duck line has one feature I like. You are not automatically down when it is wrong. Spades can alwyas be 3-3 with diamond hook on as an alternative. Whereas when both club honors are off side, the double hook is down right away. Against this, some of the endings for the squeeze may be hard to read (although who guards spades will be obvious). There is thus lots to say for the immediate double club hook. Ben Yes, I played in your ways Ben... and find a squeeze to get 12 tricksAfter the lead I immediately noticed that I only got 10 tops tricks, and is very likely to get an extra one from heart or diamond, and then I will get threat cards in S, D, C or S, H, C. Another obvious line of play is of course putting your fate in the club suit only, trying to get 2 extra club winners with the highest percentage. One way is to do club hook twice. The other way is .. cash A K of diamond and play the third ...Hoping for a 3-3 or doubleton Hx. fail only when 4-2 either side hold QJxx, and for 5-1 you will stop giving the third club. Now you return back to diamond hook and Hope for a sqeeze. Of course, when that is 4-2 and QJxx you down 2 straight forward. My math is not good. Can anyone calculate the probability? ;D Quote Link to comment Share on other sites More sharing options...
inquiry Posted December 16, 2003 Report Share Posted December 16, 2003 Another obvious line of play is of course putting your fate in the club suit only, trying to get 2 extra club winners with the highest percentage. One way is to do club hook twice. The other way is .. cash A K of diamond and play the third ... Hoping for a 3-3 or doubleton Hx. fail only when 4-2 either side hold QJxx, and for 5-1 you will stop giving the third club. Now you return back to diamond hook and Hope for a sqeeze. Of course, when that is 4-2 and QJxx you down 2 straight forward. Figuring out the odds is too tough, especially at the table. Thus, their is a whole lot to be said for the simple club play: it has good chances, and it can gain two tricks at once. However, I hate to take a play that wins or loses right off the bat when I can toy around with the hand for a little while. :-) Diamond hook gains a trick - 50% of time, let's assume a squeeze can always be found then, so that line is 50% to make. I think the immediate diamond hook is the worse line. The odds of both club honors being off-sides is 24%, but you discover the problem when EAST shows out if the suit is 6-0 or 5-1, so it is down immediately only 18.4% of the time. The other 5.6% of the time, you can recover nearly half teh time when you can change horses and hook teh diamond, but you have screwed up entries for a squeeze, so you will need EAST to have four+ diamonds to the QUEEN and four hearts (or for spades to be 3-3). Let's call this roughly 50$ of the time, so the odds of the double hook winning the slam is roughly 79.2% If you win two round of hearts. If both follow, the odds of a 4-3 split is now 83.3% so ducking a heart will gain one trick, and you will be able to find a squeeze after this with threats in all three suits unless WEST has both diamond and club stopper and east guards spades. BTW, this is why I don't return the spade NINE. If west has SPADE JT doubleton, you don't want him covering. If he has JTx spades run anyway, and if he ahas JTxx then the squeeze will have to work. Keep the nine. Anytime hearts are 4-3 and a) spades are 3-3 B) West has 4+ spadesc) East has the diamond Q (50%)d) East has five or six diamondse) East has five or six clubs f) East has QJ(x) of clubs, the duck a heart play works. In addition, duck a heart when hearts are 5-2, is not automatically down. You know have a sure heart threat against one opponent, the 5-2 split occurs roughly 17% of the time. Now you need the diamond hook (or 3-3 spades, entry conditions will not allow you to try for both) for your 11th trick. Having gotten that, you go for the squeeze.. The play will depend upon which opponent has the five hearts. On 5-2 heart split with WEST having five hearts, you a) East has the diamnod Queen and west has four spades and four+ diamondsB) East has the diamond Queen and spades are 3-3.c) anyone has QJ doubleton of clubs, or if a black suit squeeze can be obtained against one of the players after a successful diamond hook. d) others? And finally, if on the second round of hearts, someone shows out, I would actually give up on hearts, and revert to the double hook in clubs. Sadly, I can't calculate odds (well, i might be able too, be able too, but it would take way too much effort), but as the line of trying hearts before clubs has the benefit that it would have worked better on the actual layout, now rather it is the right line or not is for someone else to figure out.... the club play certainly saves energy for the next hand and if it is better (or worse), it is only by a very few percentage points. ben Quote Link to comment Share on other sites More sharing options...
mishovnbg Posted December 16, 2003 Report Share Posted December 16, 2003 Hi Bhugi! If you like to learn more about percentage play, which is far more complicated than simple calculating of static percents, try to read http://rpbridge.net/7z75.htm , author Richard Pavlicek. Misho Quote Link to comment Share on other sites More sharing options...
cherdano Posted December 17, 2003 Report Share Posted December 17, 2003 Ben, I think your assumption about the heart break are too optimistic. Chances for 4-3 are 62%, for 5-2 are 31%; so after both have followed two rounds, the chance of 4-3 split is 62/93 or roughly 68 % There is a small improvement to the double finesse line by playing 2 rounds of hearts first. If West started with 2 hearts, he is endplayed after he wins the first club round. Arend Quote Link to comment Share on other sites More sharing options...
inquiry Posted December 18, 2003 Report Share Posted December 18, 2003 Arend, You may be right... my statistical analysis is very weak. However, I hope to persuade you that you are not exaclty right either. The way I calculate the odds was not to simply discard the impossible combinations and recalculate odds again (if you do that and combinational math, there is 21 possible 2-5 splits and 34 possible 3-4 with East short and 21 and 34 with West short. That comes to 112 total, and still divides out to be rough 62.5 versus 37.5. Instead I took a different mathematical approach that who knows, may be very wrong. I assumed "I knew" 8 cards in each opponents hand that were non-hearts, since the most either opponent could have is 5 hearts (after both followed twice). Four Possible distributions are: West East1 xx xxxxx 21 0.397 8.332 xxx xxxx 34 1.19 41.673 xxxx xxx 34 1.19 41.674 xxxxx xx 21 0.397 8.33 So with 8 known "non-hearts" I calculated the percentage of heart splits. Hearts can only split at this point 5-2 or 4-3. Calculated this way, you get the 80% number. To use Pavlicek terminology, I tried to correct for the heart split in a known 8:8 space. What I think is wrong with you analysis is you don't take the remaining space into account. Let me give you a very simple example. Let's say West is known to hold 6S and East 0 (bidding or play), and you know nothing else about the hand. If your apply the combinational math you do... West might hold 7 hearts (6700, or East 7, and all in between). So you would show the odds of west having 7 hearts as being identical to the odds of east, and vise versa. The number of oppossible combinations has not changed from the a priori situation (there is still 128 possible combination, and one each with EAST holding 7 and West holding seven. So using your example math, the odds of either of them having 7 hearts would be exactly equal (1/128 or ~1%). When in fact, the possibility of the fellow with 6S also having 7 hearts is very, very small (0.0001%) while the chance of the fellow with the spade void having 7 hearts is over 2%. If you stop and think about it, it seems intuitively obvious, that you have to factor something other than just the possible combination into the math. What you seem to overlook is the combinational math you present gives you the possible combinations without calculating the The probability that this distribution actually occurs. Maybe some statisticians will eventually join the discussion and set this simple example right. Ben Quote Link to comment Share on other sites More sharing options...
cherdano Posted December 18, 2003 Report Share Posted December 18, 2003 Hmm. I am actually pretty certain my math is right there. Btw, in my opinion the situation is pretty analogous to the situation considered in the section "Looking deeper" on http://rpbridge.net/7z75.htm (thanks Misho for the link!). What is the difference to the case where, say, after one round of spades we discover that spades are 6=0? One way to say it is the following: The event that spades split this way is pretty unlikely if hearts are 7=0, a little more likely with 6=1, even more with 5=2 etc. So the relative probabilities of these cases have to be adjusted once you learn about the spades split. But with hearts splitting 4-3 or 5-2, you will always get both defenders following two rounds, and so their relative probabilities do not get adjusted. Or to say it in absolute terms: spades being 6=0 rules out many of the distributions with hearts 7=0, some less so with 6=1, etc. But two rounds of hearts leave in all possibilities with hearts 5-2 or 4-3. Quote Link to comment Share on other sites More sharing options...
inquiry Posted December 18, 2003 Report Share Posted December 18, 2003 Hi Arend, Seems like my first arguement did not convince you, so let me try another track. After win club, cash heart ace, cross to dummy, and cash one heart, both opponents follow.... we are down to hearts are 5-2 or 4-3. Here we agree. What we know is that EW both have 8 sure non-hearts, now they have, among their cards, 7 hearts and 3 remaining hearts. Let's call these "h" for hearts and "N" for non-hearts. Rather than focus on the heart suit and your combinational math. Let's focus on the non-heart suit. How can these three non-hearts be split? East can have 1, 2, or 3. West can have 1, 2, 3. The combinations are: TABLE for NON-HEART Split number of west east combinations1 0 3 1 (12.5%)2 1 2 3 (37.5%)3 2 1 3 (37.5%)4 1 0 1 (12.5%) This give a total of 8 combinations, so the odds of a 1-2 split (either way) being 6/8 or 75%. Looking at the heart suit instead, the table becomes... TABLE For HEART SPLIT number of west east combinations1 2 5 21 (19%)2 3 4 34 (31%)3 4 3 34 (31%)4 5 2 21 (19%) Now, we know that when hearts are 2-5 (case #1 in the heart table, then non-hearts here have to be 3-0. But note something funny. Using math the way you propose suggest that the chance of west having 2 hearts is 19% but the chance of him having 3 non-hearts is only 12.5%. In addition, the odds of a 4-3 heart split from this table is only 62% while the corresponding odds of a 2-1 non heart split from above is 75%. How can this be? The answer is as the remaining 10 cards are being randomized, each time a heart is dealt or a non heart is dealt it affect the distribution of the other cards. Mathematically, I think the expression is you have to combine the combinations of each distribution. So it would be something like... Number of 2-5 hearts with west having two, is not... Combin(7,2) which is the 21 possible cases, but in fact is...Combin(7,2)*Combin(3,3) (which is 21) But for the other two distributions, the math is...Combin(7,3)*Combin(3,1) = 105combin(7,4)*Combin(3,2) = 105, So the odds table becomes (1-4 same hand patterns as above) 1 21 1 21 0.0833333332 35 3 105 0.4166666673 35 3 105 0.4166666674 21 1 21 0.083333333 So for the 2H-5H split, you still have the same 42 possibilities, but for the 4-3 split counting the non-heart distributions in, you have 110 possibilities. This is where I get the greater than 80% number, actually greater than 83%. Ben Quote Link to comment Share on other sites More sharing options...
inquiry Posted December 18, 2003 Report Share Posted December 18, 2003 As an aside, here is how the math goes on that hypothetical hand where WEST is known to have 6 spades and east none, and the question is the odds of heart splits. Excel is how I do this using the Combin function (abbrevated C below, with C7,1 being how many different ways a suit of seven cards can be spit 6-1 HEART LenthWest East 7 0 C7,7 C13,0 6 1 C7,6 C13,15 2 C7.5 C13,24 3 C7.4 C13,33 4 C7,3 C13,42 5 C7,2 C13,51 6 C7,1 C13,60 7 C8,0 C13,7 The math looks like this Percentage1 - 1 1 1 0.0012 x 7 13 91 0.1173 xx 21 78 1638 2.1134 xxx 35 286 10010 12.9135 xxxx 35 715 25025 32.2826 xxxxx 21 1287 27027 34.8657 xxxxxx 7 1716 12012 15.4958 xxxxxxx 1 1716 1716 2.214 This is quite different from the table where you know nothing about the EW distribution, which would look like this.... 1 - 1 0.2612 v 7 3.3913 vv 21 15.2614 vvv 35 31.0875 vvvv 35 31.0876 vvvvv 21 15.2617 vvvvvv 7 3.3918 vvvvvvv 1 0.261 As you can see, here if you know nothing about your opponents hand, the odds are more "traditional". Someone with one of the brute force simulators can probably test the first hypothesis (with 8 known non-hearts in each hand) and see if 4-3 split is closer to 80% or 62%. Ben Quote Link to comment Share on other sites More sharing options...
cherdano Posted December 19, 2003 Report Share Posted December 19, 2003 Ben, many thanks, but I am still convinced I am right ;) FWIW, while I am not a statistician, I am a mathematician, and feel pretty confident with these kind of problems. Let's first agree to simplify the problem we are discussing a bit: Say, we have 6 hearts, AKQ432 combined in hand and dummy. In the first two rounds, hearts are played, and both follow. Defenders are clever enough to know it doesn't matter which card they throw, so they will randomly select a card to play (nothing is better). Otherwise, we would have to analyze the spots they played, which would make this discussion more complicated. I agree with your math given the spades split of course. Let me explain how our situation should IMHO be calcluated: There are 19 missing non-hearts. Each defender can have between 8 and 11 of them. So there are four cases (h for hearts, N for non-hearts). By picking West cards, you get the following numbers for each case:1: h 5=2, N 8=11: C(7,5)*C(19,8) = 1,587,2222: h 4=3, N 9=10: C(7,4)*C(19,9) = 3,233,2303: h 3=4, N 10=9: C(7,3)*C(19,10) = 3,233,2304: h 2=5, N 11=8: C(7,2)*C(19,11) = 1,587,222This gives 67% for the 4-3 split. Your match would be correct e.g. if you know they each have 8 cds in the minors, so you are just missing 7 hearts and 3 spades. Exactly knowing 8 non-hearts is not the same as knowing they have at least 8 non-hearts. (Again, this is analogous to the section "Looking deeper" in Pavlicek's article.) There is a 2nd method which yields the same numbers. We know two cards of each defender, so there are 11 missing. Let's calculate the number of holdings consistent with the exact cards they have played so far: E.g., in the first case we have to pick 3 hearts, and 8 out of the 19 possible non-hearts:1: C(3,3) * C(19,8) = 75,5822: C(3,2) * C(19,9) = 277,1343: C(3,1) * C(19.10) = 277,1344: C(3,0) * C(19,11) = 75,582But due to "restricted choice" they are not equally likely: In case 1, the likelyhood that West would play the exact cards we have seen (in this order) is 1/(5*4), for East 1/2. Case 2 we get 1/(4*3) for West, and 1/(2*3) for East. This gives the following weighted number of probabilities:1,4: 75582 / (5*4*2) = 1889.552,3: 277134 / (4*3*3*2) = 3849.1Again, this gives 67% for the 4-3 split. And one more comment: Your calculation of the "a priori"-odds of the suit breaks is a little bit incorrect, too. E.g. each single 4-3 split is a little bit more likely than each single 5-2 split, because there are more ways to distribute the remaining cards in the case of 4-3. So you have to take a product of two combine functions for this case, too. Quote Link to comment Share on other sites More sharing options...
cherdano Posted December 21, 2003 Report Share Posted December 21, 2003 Ben, one more P.S.: I only now realized why you believed I must be wrong in the first place. If you thought I calculated the a-priori odds of the heart split just by counting the number of possible heart distributions, then you were quite right in pointing out that it would be wrong to do that. (Actually, they way I calculated them was just by going to www.rpbrige.net ;)) So we just differ about the way in which we have to take the fact about "at least 8 non-hearts in each hand" in account. :o Quote Link to comment Share on other sites More sharing options...
inquiry Posted December 21, 2003 Report Share Posted December 21, 2003 FWIW, while I am not a statistician, I am a mathematician, and feel pretty confident with these kind of problems. I am not a mathematician nor a statistician, and I certainly didn’t feel confident in either or our calculations. :-) I understand, that I made some assumptions that may not have been statistically correct. All I knew was that calculating the split based upon number of combinations in one suit had to be wrong. Now that you are taking the other cards into consideration, you are on the right track, and are probably much closer to correct than I. The assumption I made is that EAST and WEST are both known to hold eight non-hearts. This is of course, true, but if one followed this logic always, the odds for a 3-0 split of a suit would only be 10%. Why, because C(3,0)*C(3,3); etc comes out to be only 20 combinations, two of which are a 3-0 split (silly math is). 1 h=3, N=0; C(3,3)*C(3,0) = 12 h=2, N=1; C(3,2)*C(3,1)=93 h=1, N=2; C(3,1)*C(3,2)=94 h=0, N=3; C(3,0)*C(3,1)=1 We know this is wrong, the odds of such a split is 22%, not 10. I made the same kind of error in the heart analysis. While it is true that we know each hand holds at least 8 hearts, you can no more discard those 8 cards in that example as you can the 10 cards above. My assumption that we had 10 “known” cards was clearly incorrect. It is clear to me know that you must know what known cards for them to count. So you have seen 2 hearts and 2 spades at this point. So while you were much closer to the right answer than I (and did so with combinational math), I now believe that maybe the match taking into account the know 2 cards in hearts and 2 cards in spades, should be, 1 h=3, N=6 C(3,3)*C(15,6)=1*5005=5,0052 h=2, N=7 (C3,2)*(C15,7)=3*6435=19,3053 h=1, N=8 (C3,1)*(C15,8)=3*6435=19,3054 h=0, N=0 (C3,0)*(C15,9)=3*5005=5005 This raises the odds from the 67% you calculated to 67.3% for the 4-3 split. But either way, we have dragged my nearly 80% prediction down to 67 and kicked your earlier 62% prediction up to 67 or 67.3. And I know I certainly learned something, not the least of which is that maybe the simple double club hook maybe best after all. :-) Although at my first guess, above, like you have the 4-3 heart split at 62% (see where I said “This line gains one trick anytime hearts are 4-3 (62.2%) and you get the extra trick via diamond hook half the time when the suit splits 5-2 or worse (half of 37.8 or 18.9%). This comes out to be 81.1% of the time you will have 11 tricks on this line (compared with the 73.6% of time you instantly gain two tricks with the double club hook. So is 81.1% chance for 11 tricks better than 73.6 for 12 (with additonal trick in 1.2% of the time?). “ So now I will have to spend a whole day trying to figure out the new odds using 67.3% chance for 4-3 hearts instead of 62.2 or 83. Ben Quote Link to comment Share on other sites More sharing options...
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