[ochinko]

This is my 4 cards ending, and we are in dummy:

---

First case: West has longer hearts. We play a diamond to the Ace and draw the last trump.
If West has the ♣K as well, he is squeezed. With two cards left he has bared his King if he didn't drop the heart.
If West doesn't have the ♣K we make if East has both Kings. East is squeezed when we draw the last trump. When neither the last heart, nor ♦K has appeared all we have to do is to cash ♣A (Two combined 50% chances give us 75% total).

Second case: East has longer hearts. Now if one of the defenders has both Kings he is squeezed, and when we ruff our last heart all we need to do is to cash both Aces. (50% total)

And, of course, 36% when hearts are 3:3.

All in all approximately 48% chance for making 7♠ or 92% after the spades hurdle is cleared. I play for ♠1:1 (52%).

[ralph23]

ochinko, on Aug 13 2007, 07:19 AM, said:

This is my 4 cards ending, and we are in dummy.....

All in all approximately 48% chance for making 7♠ or 92% after the spades hurdle is cleared. I play for ♠1:1 (52%).

I'm sorry, I think I misunderstood. (BTW congrats for being the only one so far who's ventured into trying to caculate the odds on this mother!)

Are you saying that, once we get to trick 3 (having won the opening ♥ lead and cashed the Ace of ♠ at trick 2, and {whew!} finding the defender's ♠ did in fact split 1-1), that we now have a 92% chance of making this?

Yes, assuming that we survive until trick 3, we will prevail 36% (approx) of the time that we get to trick 3 (which happens about 52% of the time, on a 1-1 trump split) on the heart-splitting theory.

What about the other 64% of the time, when the hearts do not split 3-3? For us to come out to 92 overall (provided we make it to trick 3), we have to prevail in (56/64 = 7/8 = 87.5%) of these non-♥-splitting situations.

{nb because 56+ 36 = 92)

Do you really think that it's as high as 87.5%, just thinking about it impressionistically and not analytically?

This post has been edited by ralph23: 2007-August-13, 11:33

[ochinko]

Hm, it seems like I got carried away.

(75% * 1/2 * 64%) + (50% * 1/2 * 64%) + 36% = 76%, not 92%

So the total is 76% * 0.52 = 39.5%.

I hope I did it right this time.

[bid_em_up]

deleted

[ralph23]

ochinko, on Aug 13 2007, 01:57 PM, said:

Hm, it seems like I got carried away.

(75% * 1/2 * 64%) + (50% * 1/2 * 64%) + 36% = 76%, not 92%

So the total is 76% * 0.52 = 39.5%.

I hope I did it right this time.

Well, just a little carried away .

I'm not sure I follow all the reasoning, but before I try, let me ask a question.

Q. We are assuming (for the sake of analysis A) that West has been dealt four ♥.

Thus, East was dealt two ♥.

With this assumption in place: Is it equally probable that West and East have the King of clubs? Or is one more likely to hold His Majesty than the other?

This post has been edited by ralph23: 2007-August-13, 13:08

[FrancesHinden]

ochinko, on Aug 13 2007, 01:19 PM, said:

Second case: East has longer hearts. Now if one of the defenders has both Kings he is squeezed, and when we ruff our last heart all we need to do is to cash both Aces. (50% total)

If East has 4+ hearts, then the chance of West having the CK is 57%, once trumps are 1-1.

So after winning the lead, drawing trumps, and finding that East has long hearts, surely it is better just to take the club finesse?

[ralph23]

bid_em_up, on Aug 13 2007, 02:01 PM, said:

Lets consider the minor suit honor possible distribution in Wests (or Easts) hand:

♣K 25% of the time
♦K 25% of the time
♦K♣K 25% of the time
Neither 25% of the time

The same goes for East's hand.

Well, that is the a priori probability, assuming you know nothing about their cards.

But you know each one held precisely one spade. (Or we were down right away).

You also know that (for purposes of the analysis anyhow) one of them holds more ♥ than the other. If each holds 3 ♥, it doesn't matter who has what in the minors....

Are you with me so far?

[bid_em_up]

deleted

[ralph23]

FrancesHinden, on Aug 13 2007, 02:10 PM, said:

ochinko, on Aug 13 2007, 01:19 PM, said:

Second case: East has longer hearts. Now if one of the defenders has both Kings he is squeezed, and when we ruff our last heart all we need to do is to cash both Aces. (50% total)

If East has 4+ hearts, then the chance of West having the CK is 57%, once trumps are 1-1.

If East was dealt precisely 4♥, and the probability of West having the King of ♣ is X% in that case (where X > 50), then is the probability of West having the King of ♣ even higher when East was dealt precisely 5♥?

Same reasoning of course for a 6-0 split, except those get special treatment because West led a ♥.....

[ralph23]

bid_em_up, on Aug 13 2007, 02:17 PM, said:

You are looking at it the wrong way.

Are you with me so far?

.......
So in the end, I think it all equals out in trying to figure out what to play for, and playing for the drop of the club K rates to be the winner.

1. No, I'm not with you at all. Sorry, you will have to be more explicit. I don't see how I am "looking at it wrong" -- please let me know...

2. In the end, you say that it "all equals out"??? hmmm..... not sure how that works mathematically.... or really what it means....

But let me ask you this, if you haven't already made up your mind for sure, and closed your mind to other possibilities ... would you consider that perhaps Line A of play might be superior if the ♥ split 4-2, and Line B might be superior if the ♥ split 5-1?

I'm not saying that's true, or not. I'm asking whether or not you would consider that to be plausible.

[han]

FrancesHinden, on Aug 13 2007, 02:10 PM, said:

If East has 4+ hearts, then the chance of West having the CK is 57%, once trumps are 1-1.

How did you get this Frances?

When I check it I also got 57%, but I thought it was a lot of work, using Pavlicek's dual calculator and then adding the percentages and even dividing them. Did you use a clever trick?

[bid_em_up]

Now I remember why I usually don't attempt to calculate percentages.

[Trumpace]

Hannie, on Aug 13 2007, 03:18 PM, said:

FrancesHinden, on Aug 13 2007, 02:10 PM, said:

If East has 4+ hearts, then the chance of West having the CK is 57%, once trumps are 1-1.

How did you get this Frances?

When I check it I also got 57%, but I thought it was a lot of work, using Pavlicek's dual calculator and then adding the percentages and even dividing them. Did you use a clever trick?

Let me venture a guess:

LHO has at 1 heart and 1 spade (11 vacant spaces). Rho has 4 hearts and 1 spade (8 vacant spaces).

So, chances that LHO holds a specific card is 11/19 = (11/(11+8))

[ochinko]

Hannie, on Aug 13 2007, 11:18 PM, said:

FrancesHinden, on Aug 13 2007, 02:10 PM, said:

If East has 4+ hearts, then the chance of West having the CK is 57%, once trumps are 1-1.

How did you get this Frances?

When I check it I also got 57%, but I thought it was a lot of work, using Pavlicek's dual calculator and then adding the percentages and even dividing them. Did you use a clever trick?

Frances is quite right. It depends on how many cards you assume known in the defendants. With 4:6 which is minimum in order to know the distribution in hearts and spades it's 56.2%. With 5:7 it gets 57.1%, etc. I have my own calculator, but there are many.

[han]

Trumpace, on Aug 13 2007, 03:47 PM, said:

Hannie, on Aug 13 2007, 03:18 PM, said:

FrancesHinden, on Aug 13 2007, 02:10 PM, said:

If East has 4+ hearts, then the chance of West having the CK is 57%, once trumps are 1-1.

How did you get this Frances?

When I check it I also got 57%, but I thought it was a lot of work, using Pavlicek's dual calculator and then adding the percentages and even dividing them. Did you use a clever trick?

Let me venture a guess:

LHO has at 1 heart and 1 spade (11 vacant spaces). Rho has 4 hearts and 1 spade (8 vacant spaces).

So, chances that LHO holds a specific card is 11/19 = (11/(11+8))

First of all, there are 6 hearts out there, not 5. Secondly, Frances said 4+, not 4.

[ochinko]

Trumpace, on Aug 13 2007, 11:47 PM, said:

Hannie, on Aug 13 2007, 03:18 PM, said:

FrancesHinden, on Aug 13 2007, 02:10 PM, said:

If East has 4+ hearts, then the chance of West having the CK is 57%, once trumps are 1-1.

How did you get this Frances?

When I check it I also got 57%, but I thought it was a lot of work, using Pavlicek's dual calculator and then adding the percentages and even dividing them. Did you use a clever trick?

Let me venture a guess:

LHO has at 1 heart and 1 spade (11 vacant spaces). Rho has 4 hearts and 1 spade (8 vacant spaces).

So, chances that LHO holds a specific card is 11/19 = (11/(11+8))

Close. You have to draw one spade in order to see that they break 1:1, but you have to draw hearts three times in order to see that they are 4:2. So now you know that the one with 2 hearts has 9 slots to put ♣K, the other one - 7.

Oops, my bad. The one with 4 hearts has 8 slots, not 7, because there's only one more card that is known, which makes it 52.9% to find the King in the guy with fewer hearts.

Edit: I am not able to calculate how much the chance for finding both Kings in the hand with fewer hearts increases, but bid_em_up definitely has a point so maybe we should always go for the squeeze as being more attractive than a finesse.

[Trumpace]

Hannie, on Aug 13 2007, 03:58 PM, said:

Trumpace, on Aug 13 2007, 03:47 PM, said:

Hannie, on Aug 13 2007, 03:18 PM, said:

FrancesHinden, on Aug 13 2007, 02:10 PM, said:

If East has 4+ hearts, then the chance of West having the CK is 57%, once trumps are 1-1.

How did you get this Frances?

When I check it I also got 57%, but I thought it was a lot of work, using Pavlicek's dual calculator and then adding the percentages and even dividing them. Did you use a clever trick?

Let me venture a guess:

LHO has at 1 heart and 1 spade (11 vacant spaces). Rho has 4 hearts and 1 spade (8 vacant spaces).

So, chances that LHO holds a specific card is 11/19 = (11/(11+8))

First of all, there are 6 hearts out there, not 5. Secondly, Frances said 4+, not 4.

I never said there were only 5 hearts.

The information we know is:

LHO has a heart. RHO has 4 or more (assumed). Spades split 1-1.

So, LHO has 11 vacant spaces and RHO has 8.

Chances that LHO holds the 6th heart is ~57% (assuming he is capable of leading singleton J etc), same as the chances that he has the CK, same as the chances that he has DK etc.

This, is assuming LHO will lead a card at random from his 13 cards i.e we assume that the lead was a heart gives us no other information except the fact that LHO has at least one heart.

[ralph23]

♣♦♥♠
Let's go s..l..o..w....e.....r......

1. Assumption (or Hypothesis) 1: West has 6♥; well, in this case we are always going down, so let's ignore this one.

2. Assumption 2: West has 5♥, and East has 1♥.

In this assumed or hypothesized case #2, West has 6 cards known to be in the majors, and East has 2 known to be in the majors.

Therefore, of the available 18 minor card slots, West has 7 of those and East has 11 of those.

Therefore, if Assumption 2 is true, then West has 7 chances out of 18 to "draw" any minor suit card, and East has 11 chances out of 18.

So, if Assumption 2 is true, then the odds of East having any particular minor card -- whether the King of ♣ or the 6 of ♦ -- is 11/18 (or 61.1%). West has the inverse of 38.9%.

Anyone disagree with this so far?

[han]

What's the inverse of 38.9%?

[ralph23]

Hannie, on Aug 13 2007, 05:14 PM, said:

What's the inverse of 38.9%?

61.1 ??

I.e. 61.1 + 38.9 = 100. They are inverses of each other wrt 100. Or maybe complement is the right term.