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
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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 opp
K6543 2
K6542 3
K6532 4
K6432 5
K5432 6
65432 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)... see
Odds calculator
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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
Help
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you should definately always play for not 3-3 suit breaks
. Hope that works for you. looooool
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