Useful bridge/math tidbit

[junyi_zhu]

hi!

 

i certainly agree with fred about the arithmetical side of his demonstration

 

however, the fact that lho has got an early ruff basically means that he led to get one..do we all do that with a potential trump trick?

Against Henri, it could be different. :D

[bucky]

I need a clarification here: Does LHO's play to the "decision trick" count as a shown card?

 

e.g. LHO starts with a singleton, RHO wins (he has 5) and gives LHO a ruff. LHO exits in a side suit (RHO following). Trump Ace produces x from both opps. So far, it is 1 lead+2 trumps+1 side suit for LHO vs. 5+1+1 for RHO (4 vs. 7 --> finesse)

 

When you play a trump and LHO plays x, does this x count? Is it now 5 vs. 7? Or do I use the count before play to this trick of 4 vs. 7?

You can count the vacant space only when the suit distribution is clear. Suppose for one certain suit, opps have 8 cards, you play two rounds, both follow, you can't count this suit yet, because the distribution of this suit is still unclear. vacant space count can only be applied when the suit distribution is clear(or one card missing in your key suit). So for your example, the correct way to count vacant space is:

LHO: 13-1(singleton) -3(three trumps)=9

RHO: 13-5(5 in that side suit) -1 (one trump) = 7

So it is 9:7 better to play finesse vs. dropping.

I think the answer to the original question is: the x counts, so now it is 5 vs. 7. You always count the vacant space at the decision point, which is AFTER lho followed 2nd trump and BEFORE rho has played. When the side suit breaks 1-5, the vacant space is: LHO 8, RHO 6.

[kgr]

....

You can count the vacant space only when the suit distribution is clear. ....

What is the logic about this?

(Before I read this I thought making a post to ask this & I was almost sure that you could do it always.)

Logically you would think that if LHO leads a Heart then vacant spaces are 12:13..?

I can only remember this if I understand it.

[gwnn]

I think, kgr, the argument I wrote in this thread applies to this case as well. Caveat emptor.

[bucky]

Logically you would think that if LHO leads a Heart then vacant spaces are 12:13..?

I can only remember this if I understand it.

Actually yes, IF LHO picked a totally random card to lead, and that you need to make a decision right before RHO plays. After trick 1 is completed (assuming RHO follows heart), the vacant space is back to 12:12.

[kgr]

I think, kgr, the argument I wrote in this thread applies to this case as well. Caveat emptor.

I didn't understand it. I'll reread it later

[kgr]

Logically you would think that if LHO leads a Heart then vacant spaces are 12:13..?

I can only remember this if I understand it.

Actually yes, IF LHO picked a totally random card to lead, and that you need to make a decision right before RHO plays. After trick 1 is completed (assuming RHO follows heart), the vacant space is back to 12:12.

Are you saying?:

that if LHO leads a random card then probability that RHO holds another not shown specific card is 13:12.

that if LHO is a bridger and leads a card then probability that RHO holds another not shown specific card is 13:13.

(what if LHO is a beginner and does not have a clue what he is doing??)

[junyi_zhu]

....

You can count the vacant space only when the suit distribution is clear. ....

What is the logic about this?

(Before I read this I thought making a post to ask this & I was almost sure that you could do it always.)

Logically you would think that if LHO leads a Heart then vacant spaces are 12:13..?

I can only remember this if I understand it.

Because vacant space calculation is an approximation in some situations. For example, suppose opps have 8 spades, you draw two rounds, both follow. Therefore, spades can't be 8-0 or 7-1. However, 6-2, 5-3, 4-4 are all possible, so you just don't really know the exact distribution of that suit. In this sense, you can't include this suit into your calculation. All you know is that spades distribution is rather unknown and it's impossible to be 8-0 or 7-1, which is still a very small percentage comparing with all the possible spades distributions. Therefore, It's still fairly good approximation to assume that you don't know the spade distribution at all. In the other thread talking about the probabilities, one post was very well said: what's the average number of your heart suit if you open 1H showing 5 or more? I guess it can't be 6 and should be in the middle of 5-6, because frequency of 5, 6, 7 hearts holdings are very different. In this case, it's the same logic. When they hold 8 card, even if you know the distribution can't be 7-1 or 8-0, you still don't know the exact distribution and 7-1 8-0 are very rare events, so you can't really take those out of your approximation easily. Therefore, you'd just stick to what you have and consider the distribution of spades is unknown. The real percentage can be calculated by computer simulations in a quite accurate way, which should be slightly different from the vacant space calculation. However, at the table, vacant space counting is still your best friend. Still, in many situations, vacant space counting actually gives the accurate result.

[bucky]

Are you saying?:

that if LHO leads a random card then probability that RHO holds another not shown specific card is 13:12.

that if LHO is a bridger and leads a card then probability that RHO holds another not shown specific card is 13:13.

(what if LHO is a beginner and does not have a clue what he is doing??)

After the lead, 13:13 no longer applies, there is just not this much room for hidden cards.

 

When I said about random card, I am not using information on bidding and lead agreement. You can always rule out certain shape from the bidding, and the lead will also tell you something (if the lead is a low heart, even a beginner LHO probably wouldn't have AKQx in that suit). But strictly in terms of vacant space, after the opening lead and before RHO plays to the first trick, a particular unseen card can be anywhere among 12 remaining cards in LHO hand, or 13 remaining cards in RHO hand.

 

Of course in reality it is not going to be 13:12, you always have clues in the bidding and lead. If you can rule out the possibility of RHO being void in hearts even before he plays to the first trick, you are back to 12:12. If you are certain that LHO has longer hearts than RHO, you will know that RHO is more likely to hold a specific card in a side suit.

[MBV53]

Good point/tip fred!

MBV

[kgr]

....

You can count the vacant space only when the suit distribution is clear. ....

What is the logic about this?

(Before I read this I thought making a post to ask this & I was almost sure that you could do it always.)

Logically you would think that if LHO leads a Heart then vacant spaces are 12:13..?

I can only remember this if I understand it.

Because vacant space calculation is an approximation in some situations. For example, suppose opps have 8 spades, you draw two rounds, both follow. Therefore, spades can't be 8-0 or 7-1. However, 6-2, 5-3, 4-4 are all possible, so you just don't really know the exact distribution of that suit. In this sense, you can't include this suit into your calculation. All you know is that spades distribution is rather unknown and it's impossible to be 8-0 or 7-1, which is still a very small percentage comparing with all the possible spades distributions. Therefore, It's still fairly good approximation to assume that you don't know the spade distribution at all. In the other thread talking about the probabilities, one post was very well said: what's the average number of your heart suit if you open 1H showing 5 or more? I guess it can't be 6 and should be in the middle of 5-6, because frequency of 5, 6, 7 hearts holdings are very different. In this case, it's the same logic. When they hold 8 card, even if you know the distribution can't be 7-1 or 8-0, you still don't know the exact distribution and 7-1 8-0 are very rare events, so you can't really take those out of your approximation easily. Therefore, you'd just stick to what you have and consider the distribution of spades is unknown. The real percentage can be calculated by computer simulations in a quite accurate way, which should be slightly different from the vacant space calculation. However, at the table, vacant space counting is still your best friend. Still, in many situations, vacant space counting actually gives the accurate result.

Thanks for the effort. I want to believe the above, but I wonder how far off both methods are:

[hv=n=skjt9hkqjxdxxcxxx&s=saxhxxxxxdakqcakq]133|200|[/hv]

♠ is trumps.

LHO leads a small ♥, RHO return a ♥ and LHO ruffs.

LHO plays a random ♦ and RHO follows that suit.

You cash ♠A and lead a second round. All follow with small ♠'s.

 

Probability that LHO has ♠Q according to vacant space theory:

13-1-3:13-6-1=9:6=0.6

 

Probability if WRONGLY taking the ♦ play into account:

13-1-3-1:13-6-1-1=8:5=0.615385

 

If Playing ♣/♦ before the 2nd trump then:

- for Vacant Spaces this will always be 0.6

- for the WRONG calculation this will be:

1 minor: 0.615385

2 minors: 0.636364

3 minors: 0.666667

4 minors: 0.714286

5 minors: 0.8

==> Is anyone able to calculate the real probabilities?

BTW: Does it make a difference for vacant space theory if opps always play their small cards from low to high?

[hanp]

Yes you can compute the exact probabilities. It is easy but takes a considerable amount of work.

[junyi_zhu]

....

You can count the vacant space only when the suit distribution is clear. ....

What is the logic about this?

(Before I read this I thought making a post to ask this & I was almost sure that you could do it always.)

Logically you would think that if LHO leads a Heart then vacant spaces are 12:13..?

I can only remember this if I understand it.

Because vacant space calculation is an approximation in some situations. For example, suppose opps have 8 spades, you draw two rounds, both follow. Therefore, spades can't be 8-0 or 7-1. However, 6-2, 5-3, 4-4 are all possible, so you just don't really know the exact distribution of that suit. In this sense, you can't include this suit into your calculation. All you know is that spades distribution is rather unknown and it's impossible to be 8-0 or 7-1, which is still a very small percentage comparing with all the possible spades distributions. Therefore, It's still fairly good approximation to assume that you don't know the spade distribution at all. In the other thread talking about the probabilities, one post was very well said: what's the average number of your heart suit if you open 1H showing 5 or more? I guess it can't be 6 and should be in the middle of 5-6, because frequency of 5, 6, 7 hearts holdings are very different. In this case, it's the same logic. When they hold 8 card, even if you know the distribution can't be 7-1 or 8-0, you still don't know the exact distribution and 7-1 8-0 are very rare events, so you can't really take those out of your approximation easily. Therefore, you'd just stick to what you have and consider the distribution of spades is unknown. The real percentage can be calculated by computer simulations in a quite accurate way, which should be slightly different from the vacant space calculation. However, at the table, vacant space counting is still your best friend. Still, in many situations, vacant space counting actually gives the accurate result.

Thanks for the effort. I want to believe the above, but I wonder how far off both methods are:

[hv=n=skjt9hkqjxdxxcxxx&s=saxhxxxxxdakqcakq]133|200|[/hv]

♠ is trumps.

LHO leads a small ♥, RHO return a ♥ and LHO ruffs.

LHO plays a random ♦ and RHO follows that suit.

You cash ♠A and lead a second round. All follow with small ♠'s.

 

Probability that LHO has ♠Q according to vacant space theory:

13-1-3:13-6-1=9:6=0.6

 

Probability if WRONGLY taking the ♦ play into account:

13-1-3-1:13-6-1-1=8:5=0.615385

 

If Playing ♣/♦ before the 2nd trump then:

- for Vacant Spaces this will always be 0.6

- for the WRONG calculation this will be:

1 minor: 0.615385

2 minors: 0.636364

3 minors: 0.666667

4 minors: 0.714286

5 minors: 0.8

==> Is anyone able to calculate the real probabilities?

BTW: Does it make a difference for vacant space theory if opps always play their small cards from low to high?

You made a mistake in your vacant space count, spade situation is still unknown.

You can apply vacant space counting only when the situation of a suit is clear, or missing one key card in your key suit. Here, you just don't know the spade distribution yet. It could be 3-4, 4-3, 5-2, 6-1.

[dkharty]

Dealer: ????? Vul: ???? Scoring: Unknown

♠ KJT9 ♥ KQJx ♦ xx ♣ xxx ♠ Ax ♥ xxxxx ♦ AKQ ♣ AKQ

♠ is trumps.

I would love to see this auction... :(

[bucky]

♠ is trumps.

LHO leads a small ♥, RHO return a ♥ and LHO ruffs.

LHO plays a random ♦ and RHO follows that suit.

You cash ♠A and lead a second round. All follow with small ♠'s.

 

Probability that LHO has ♠Q according to vacant space theory:

13-1-3:13-6-1=9:6=0.6

 

Probability if WRONGLY taking the ♦ play into account:

13-1-3-1:13-6-1-1=8:5=0.615385

How did you derive this?

 

What is "13-1-3-1:13-6-1-1"? I can see "13-1-3-1" as "13 cards minus one H minus 3 S minus 1D", but what is "13-6-1-1"? And why taking ♦ play into account is WRONG?

[ceeb]

... I wonder how far off both methods are:

Dealer: ????? Vul: ???? Scoring: Unknown

♠ KJT9 ♥ KQJx ♦ xx ♣ xxx ♠ Ax ♥ xxxxx ♦ AKQ ♣ AKQ

♠ is trumps.

LHO leads a small ♥, RHO return a ♥ and LHO ruffs.

LHO plays a random ♦ and RHO follows that suit.

You cash ♠A and lead a second round. All follow with small ♠'s.

 

Probability that LHO has ♠Q according to vacant space theory:

13-1-3:13-6-1=9:6=0.6

 

Probability if WRONGLY taking the ♦ play into account:

13-1-3-1:13-6-1-1=8:5=0.615385

 

==> Is anyone able to calculate the real probabilities?

Interesting case. Unlike Fred's original hand, we have not seen the entire spade-spot suit (the pseudo- or sub-suit consisting of small spades) so cannot accurately compute via vacant spaces. But we've seen a fair fraction of it, so how inaccurate is ignoring it altogether? Plugging into my probability calculator:

 

First calculation: Using only that hearts are 1=3, the ♠Q is 6:5 to be on the left. That I think is the vacant spaces calculation.

 

Second calculation: Using also the observation that LHO has at least 2 small spades, the ♠Q is 7:6 to be on the left.

 

Third calculation: Taking into account as well that RHO has at least 1 small spade, the ♠Q is 33:28 to be on the left

 

As decimal probabilities the above are 0.545, 0.538, 0.541.

 

In principle the ♦ plays tell us something as well -- no ♦ void about. That increases the last probability to 0.5413, but of course making a computation like that while ignoring inferences from the bidding and play is entering the twilight zone.

 

From the above computations it seems that the vacant space rule is quite accurate.

 

Note that at the opposite extreme if you misapply vacant spaces by counting every card seen, then you would always, half-way through a trick, think it is k+1:k that the player yet to play has any given card.

If Playing ♣/♦ before the 2nd trump then:

- for Vacant Spaces this will always be 0.6

- for the WRONG calculation this will be:

1 minor: 0.615385

2 minors: 0.636364

3 minors: 0.666667

4 minors: 0.714286

5 minors: 0.8

Sorry, don't understand that part.

BTW: Does it make a difference for vacant space theory if opps always play their small cards from low to high?

It can make big difference. Perhaps you could adjust the vacant space theory to account for it.

[ceeb]

... I wonder how far off both methods are:

Dealer: ????? Vul: ???? Scoring: Unknown

♠ KJT9 ♥ KQJx ♦ xx ♣ xxx ♠ Ax ♥ xxxxx ♦ AKQ ♣ AKQ

♠ is trumps.

LHO leads a small ♥, RHO return a ♥ and LHO ruffs.

LHO plays a random ♦ and RHO follows that suit.

You cash ♠A and lead a second round. All follow with small ♠'s.

 

Probability that LHO has ♠Q according to vacant space theory:

13-1-3:13-6-1=9:6=0.6

 

Probability if WRONGLY taking the ♦ play into account:

13-1-3-1:13-6-1-1=8:5=0.615385

 

==> Is anyone able to calculate the real probabilities?

Interesting case. Unlike Fred's original hand, we have not seen the entire spade-spot suit (the pseudo- or sub-suit consisting of small spades) so cannot accurately compute via vacant spaces. But we've seen a fair fraction of it, so how inaccurate is ignoring it altogether? Plugging into my probability calculator:

 

First calculation: Using only that hearts are 1=3, the ♠Q is 6:5 to be on the left. That I think is the vacant spaces calculation.

 

Second calculation: Using also the observation that LHO has at least 2 small spades, the ♠Q is 7:6 to be on the left.

 

Third calculation: Taking into account as well that RHO has at least 1 small spade, the ♠Q is 33:28 to be on the left

 

As decimal probabilities the above are 0.545, 0.538, 0.541.

 

In principle the ♦ plays tell us something as well -- no ♦ void about. That increases the last probability to 0.5413, but of course making a computation like that while ignoring inferences from the bidding and play is entering the twilight zone.

 

From the above computations it seems that the vacant space rule is quite accurate.

 

Note that at the opposite extreme if you misapply vacant spaces by counting every card seen, then you would often, half-way through a trick, think it is k+1:k that the player yet to play has any given card.

If Playing ♣/♦ before the 2nd trump then:

- for Vacant Spaces this will always be 0.6

- for the WRONG calculation this will be:

1 minor: 0.615385

2 minors: 0.636364

3 minors: 0.666667

4 minors: 0.714286

5 minors: 0.8

Sorry, don't understand that part.

BTW: Does it make a difference for vacant space theory if opps always play their small cards from low to high?

It can make big difference. Perhaps you could adjust the vacant space theory to account for it.

[bucky]

Interesting case. Unlike Fred's original hand, we have not seen the entire spade-spot suit (the pseudo- or sub-suit consisting of small spades) so cannot accurately compute via vacant spaces. But we've seen a fair fraction of it, so how inaccurate is ignoring it altogether? Plugging into my probability calculator:

 

First calculation: Using only that hearts are 1=3, the ♠Q is 6:5 to be on the left. That I think is the vacant spaces calculation.

 

Second calculation: Using also the observation that LHO has at least 2 small spades, the ♠Q is 7:6 to be on the left.

 

Third calculation: Taking into account as well that RHO has at least 1 small spade, the ♠Q is 33:28 to be on the left

 

As decimal probabilities the above are 0.545, 0.538, 0.541.

Please note that LHO has at least THREE spades: ruffed once and followed trump suit twice.

[ceeb]

Oh. Then 6/11, 43/83, 51/98, 5001/9608 or

0.545 0.518 0.5204 0.5205.

[rfp]

I agree with Fred's mathematical point. Essentially, the spade spots

become a pseudo suit since their play involved no choice or decision,

i.e., the missing queen would always be withheld. Therefore, at the

moment of decision vacant spaces are equal if RHO has exactly two

more known cards in another suit.

 

However, there is a restricted choice element (I didn't see mentioned)

relating to _how_ the ruff was obtained. If leader was simply leading his

partner's suit, there is nothing here. But if he _chose_ to lead a singleton,

a trump holding of xxx is more likely than Qxxx. That is, a singleton lead

is automatic with xxx, but hardly so with Qxxx.

 

Under these circumstances, it's certainly better to play for the drop in

the 50-50 situation. And it may be better to extend playing for the drop

even further.

 

--

Richard Pavlicek

Web site: www.rpbridge.net

[dake50]

Are you expanding on Roudi's rules?

 

How much side knowledge to sway alternate play? EG. weak2 by LHO swings a 1-4 safety play to "best"