New Top Level Area: Carding

[richardpaulhall]

Carding: 1] the 'best' way to utilize the cards in your hand in relation to a specific suit 2] Inferences that can be drawn from the card played to a trick

There are many situations that may show up. Some a perfect to show novices while others are plays only experts can profit from.

Here is an example of #2. (Notice: Inferences are suggestions, not proof. Players can play the wrong card by accident or ignorance, or play a card deceptively to fool you.)

During our card session 2013-7-11 Rak's partner took the bid. Declarer played off a few aces and exited with a side card. Rak won the trick. As a good partner is wont to do, he played a trump Ace . Lefty played small, Declarer played a King, and I, the other defender played an Ace under his Ace.

In playing a side suit my drop of an Ace would show suit control, that I had the outstanding aces. Or that I had a singleton Ace.

This was trump. The distribution would have to be grotesque for a defender to be able to show suit control. I was not showing control.

So my Ace must be singleton or stiff double Ace ..
I cannot figure out the odds of a defender holding a stiff Ace or a double stiff ace. But dropping that ace was a signal to Rak to take notice of the lay of the cards in trump.

If is a more common holding than , Rak should play a second Ace to smother my now unguarded Ace, or another trump so that Declarer can play the 3rd Ace to capture my Ace, also a smother play. And if the other defender has the third ace, he may play it, correctly or not. And this too will smother my Ace.

If is a more common holding than , Rak should play a small trump to finesse the other defender. Declarer's trump are sitting over that defender's trump. Leading though one defender, knowing the other defender has no trump has to be the right play.

If Partner plays a second Ace and Right shows out, nothing is lost. Now partner must play a thrid trump to finesse the other defender.


I was able to see this play because I was very afraid of the vulnerability of my trump holding. Whan Rak played a trump ace, eating one of mine, I could clearly see the situations that might occur where my side would not get the benefit of my second trump Ace.

---

The discussion last week about the correct order to play off your aces needs to be codified, turned into an article and cross posted in the Carding thread.

AND we have to get back to that thread and get some specifics on determining the correct order AND the inferences the Ace player's partner can make. [Playing shows a much different hand than . Partner can read the difference in length in his partner's suits.]

[rakbeater]

Good post Rick.

[richardpaulhall]

We need someone to tell us if stiff Ace is more or less common than double Ace stiff.

[FLACKprb]

(07-19-2013, 08:35 PM)richardpaulhall Wrote:  We need someone to tell us if stiff Ace is more or less common than double Ace stiff.

rph--
Can more specifics be supplied here? Like the complete initial deal showing all cards in each hand. I've not digested the entire discussion yet, but I've been in similar situations fearing loss of my bare Aces of Trumps (which is what I think you're getting at ... or any bare Aces for that matter, singletons/or doubles). I've got a Monte Carlo Simulation program that could be modified to maybe analyze this situation by detecting specific configurations and frequency counting them to get desired probabilities, but in order to do that I'd need to understand the problem to get it right. Beware, if it is a very small probability (<0.00001) the simulation will not produce meaningful results (can't get enough samples in a reasonable run time ... 10M Trials took 6 hours and 20 minutes on my iPad 4th generation, don't want to run that very often).

Not to be rude or disparaging, I have to find the problem/question interesting to me before I commit to spending time finding the/an answer.

I can tell you the following now if this helps at all:

Interesting note. It is more likely to have 5 (6.9%) or 6 (6.1%) cards of the same rank Around than it is to have just 4 (3.5%) or more than 6 (4.1%, the sum of probabilities of 7 to 16 add to the total 4.1%). This means when Aces are declared Around expect that player to most likely have have 5 or 6 Aces on average! (Actually, this goes for Arounds of any rank.) This is a manifestation of Pascal's Triangle. See any book on basic probability/combinatorics or Wikipedia for information about Pascal's Triangle.

I'll also throw out this teaser to rakbeater, I've just finished a spreadsheet summary of the paper I gave you that I'm going to let percolate (it's late and I want to make sure I didn't make any mistakes), have my brother review, then do any final revising, before I send it to you. I'm hoping it might find a home on the website.

I keep finding great nuggets of discussions peaking my curiosity that I need to follow up on ... Bare with ... Bare with ... (a lollipop to anyone who recognizes what British Sitcom this phrase comes from).

I'm slow, and apologize in advance, I'll get there eventually and methodically!

Ta!
--FLACKprb

[richardpaulhall]

For this problem the rest of the hand is immaterial. And the other three hands don't matter much either.

A good start would be the probability that a hand has a singleton and the probability that a hand has a doubleton.

Then restricted choice would let us calculate the probabilities for a stiff ace and for double stiff ace.

The numbers will help in the analysis of what to do if the first stiff ace falls second on Partners lead of a trump ace or if the stiff ace falls fourth. Especially if single stiff is more common than double stiff.

[FLACKprb]

(07-20-2013, 10:07 AM)richardpaulhall Wrote:  For this problem the rest of the hand is immaterial. And the other three hands don't matter much either.

A good start would be the probability that a hand has a singleton and the probability that a hand has a doubleton.

Then restricted choice would let us calculate the probabilities for a stiff ace and for double stiff ace.

The numbers will help in the analysis of what to do if the first stiff ace falls second on Partners lead of a trump ace or if the stiff ace falls fourth. Especially if single stiff is more common than double stiff.

Rick--

I'll reserve the right to comment on whether or not the rest of this hand and the contents of the other hands matter or not. The Engineer in me tells me every detail matters.

The following table is data gathered in a 10M trial run of a Monte Carlo Simulation program. The data is without regard to suit. (The formatting is not perfect in the preview, extra spaces get removed, but I think it's readable.)

Table 42 Probability of Number of Cards of Specific Rank
# Aces Tens Kings Queens Jacks
0 0.0055 0.0056 0.0055 0.0055 0.0056
1 0.0395 0.0395 0.0394 0.0394 0.0394
2 0.1223 0.1222 0.1223 0.1223 0.1224
3 0.2183 0.2185 0.2185 0.2186 0.2185
4 0.2516 0.2516 0.2515 0.2513 0.2513
5 0.1971 0.1971 0.1972 0.1970 0.1971
6 0.1084 0.1084 0.1083 0.1085 0.1085
7 0.0425 0.0426 0.0425 0.0425 0.0425
8 0.0120 0.0119 0.0119 0.0119 0.0119
9 0.0024 0.0024 0.0024 0.0024 0.0024
10 0.0003 0.0003 0.0003 0.0003 0.0003
11 0.0000 0.0000 0.0000 0.0000 0.0000
12 0.0000 0.0000 0.0000 0.0000 0.0000
13 0.0000 0.0000 0.0000 0.0000 0.0000
14 0.0000 0.0000 0.0000 0.0000 0.0000
15 0.0000 0.0000 0.0000 0.0000 0.0000
16 0.0000 0.0000 0.0000 0.0000 0.0000
Monte Carlo simulation run of 10M Trials from 27Jun2013 9 : 48 : 23

Can you please define/elaborate on your terminology of "stiff" and "restricted choice". I'm not familiar with these terms. I'd like to fully understand the concept(s) you're discussing without having to guess at meaning. The reference to "restricted choice" sounds like a methodology of analysis that I'd really like to understand fully. I'm always willing to learn new things.

"The numbers will help in the analysis of what to do if the first stiff ace falls second on Partners lead of a trump ace or if the stiff ace falls fourth. Especially if single stiff is more common than double stiff."

I apologize, but I'm not following this train of thought. I don't understand the "falls second" or "falls fourth". If I am correct in assuming your hand contains a doubleton bare pair of Trumps Aces, then are you saying your partner declared Trumps and also has 2 Aces in Trumps? If your partner then tries to draw down Trumps outstanding (i.e., remove Trumps from the game especially looking for the other two Aces) by playing both Aces in his hand then your two bare Aces will have to be played. What is the question? Your partner played his hand following a playing strategy specifically looking for the two Trumps Aces not in his hand and found them. His goal was accomplished. Unfortunately it is most likely a bad situation since you're now out of Trumps and opponents have all remaining outstanding Trumps against your partner. This may not matter to your partner depending on the distribution of Trumps in his starting hand. If he has a small count Trumps suit why is he playing his Trumps Aces? Leading a Queen (forces a pointer from player to left) or Jack would get to your hand (of course he doesn't know that to begin with, but will be pleasantly surprised to find you in the lead) and remove Trumps from the game. Your play will hopefully remove the garbage cards (losers) from your partners hand. If he has a large count Trumps suit with a great backup suit then all is well in Pinochle heaven! Play, watch, and enjoy!

Details matter. No disparagement intended, just trying to understand your question and would like to provide an intelligent response. Please be patient, I'll get there eventually.

[tony ennis]

I wrote a quick program.

Judging by the numbers, I am not sure it makes a lot of sense to suffer too much worrying about a singleton aces.

number of hands: 1000000
number with one void: 4738 (0.4738%)
number with more than one void: 1 (1.0E-4%)
number with at least one singleton ace: 9242 (0.9242%) <-- stiff ace
number with at least one doubleton ace: 6200 (0.62%) <-- stiff aces
number with at least once suit 1 cards long: 46256 (4.6256%) <-- common singleton
number with at least once suit 2 cards long: 198894 (19.8894%) <-- common doubleton
number with at least once suit 3 cards long: 499213 (49.9213%)
number with at least once suit 4 cards long: 819234 (81.9234%)
number with at least once suit 5 cards long: 933838 (93.3838%)
number with at least once suit 6 cards long: 761964 (76.1964%)
number with at least once suit 7 cards long: 452999 (45.2999%)
number with at least once suit 8 cards long: 198844 (19.8844%)
number with at least once suit 9 cards long: 65207 (6.5207%)
number with at least once suit 10 cards long: 15581 (1.5581%)
number with at least once suit 11 cards long: 2816 (0.2816%)
number with at least once suit 12 cards long: 374 (0.0374%)
number with at least once suit 13 cards long: 35 (0.0035%)
number with at least once suit 14 cards long: 5 (5.0E-4%)
number with at least once suit 15 cards long: 0 (0.0%)
number with at least once suit 16 cards long: 0 (0.0%)
number with at least once suit 17 cards long: 0 (0.0%)
number with at least once suit 18 cards long: 0 (0.0%)
number with at least once suit 19 cards long: 0 (0.0%)
number with at least once suit 20 cards long: 0 (0.0%)

[FLACKprb]

(07-19-2013, 11:57 AM)richardpaulhall Wrote:  Carding: 1] the 'best' way to utilize the cards in your hand in relation to a specific suit 2] Inferences that can be drawn from the card played to a trick

There are many situations that may show up. Some a perfect to show novices while others are plays only experts can profit from.

Here is an example of #2. (Notice: Inferences are suggestions, not proof. Players can play the wrong card by accident or ignorance, or play a card deceptively to fool you.)

During our card session 2013-7-11 Rak's partner took the bid. Declarer played off a few aces and exited with a side card. Rak won the trick. As a good partner is wont to do, he played a trump Ace . Lefty played small, Declarer played a King, and I, the other defender played an Ace under his Ace.

In playing a side suit my drop of an Ace would show suit control, that I had the outstanding aces. Or that I had a singleton Ace.

This was trump. The distribution would have to be grotesque for a defender to be able to show suit control. I was not showing control.

So my Ace must be singleton or stiff double Ace ..
I cannot figure out the odds of a defender holding a stiff Ace or a double stiff ace. But dropping that ace was a signal to Rak to take notice of the lay of the cards in trump.

If is a more common holding than , Rak should play a second Ace to smother my now unguarded Ace, or another trump so that Declarer can play the 3rd Ace to capture my Ace, also a smother play. And if the other defender has the third ace, he may play it, correctly or not. And this too will smother my Ace.

If is a more common holding than , Rak should play a small trump to finesse the other defender. Declarer's trump are sitting over that defender's trump. Leading though one defender, knowing the other defender has no trump has to be the right play.

If Partner plays a second Ace and Right shows out, nothing is lost. Now partner must play a thrid trump to finesse the other defender.


I was able to see this play because I was very afraid of the vulnerability of my trump holding. Whan Rak played a trump ace, eating one of mine, I could clearly see the situations that might occur where my side would not get the benefit of my second trump Ace.

---

The discussion last week about the correct order to play off your aces needs to be codified, turned into an article and cross posted in the Carding thread.

AND we have to get back to that thread and get some specifics on determining the correct order AND the inferences the Ace player's partner can make. [Playing shows a much different hand than . Partner can read the difference in length in his partner's suits.]

Rick--
I just reread this post and think I understand better now.

You had two aces in Trumps, you were sitting to left of declarer, Rak was sitting on your left.

Did declarer lay down a run meld or declare aces around? Did anyone declare aces around (you obviously didn't meld aces, hence declarer either got lucky to play a suit to get through you or took an educated risk based in his hand as to what suit might get to Rak)? If so, when Rak to your left played his Trumps ace, then you knew where all four Trumps aces were. Declarer Run or aces had one, you had two, and Rak had one. Rak could not play a second Trumps ace.

If declarer did not reveal an ace of Trumps during meld, then Rak's Ace play left one outstanding from your point of view. Presumably, declarer must have the fourth ace, else he declared Trumps without one ... What was the bidding sequence? Did it get dumped on declarer or did it appear the declarer really wanted to name Trumps? From just the limited knowledge provided (i.e., not knowing declared meld nor bidding sequence), if I was sitting in Rak's seat then because my partner did not play Trumps by leading Trumps ace(s) then I'd be leery about running Trumps out of the hand. Declarer obviously did not want to play Trumps, because he played off a side suit and got through you to Rak, his partner. If declarer had wanted Trumps run, he'd have played a queen or jack of Trumps looking for the outstanding aces of Trumps. He didn't. (A caveat here, if Rak declared aces around in meld and declarer held three aces in one suit, then declarer would have a guaranteed path to Rak and his side suit play was almost a no brainier play ... almost because distribution could cut Rak's ace in that side suit). Now once Rak is in the lead, I agree he correctly played his first Trumps Ace. If Rak did have a second Trumps ace, he should have played it regardless to show his partner where it was also. Again, based on what was melded and bid sequence, Rak would assume his partner had at least one ace of Trumps. Your second ace would get caught and assuming everyone had at least two Trumps then 8 Trumps would be gone leaving 12 outstanding with tens golden in Trumps. If I was Rak, I'd have been extremely happy to catch both of your Trumps aces ... But worried since now it is a certainty that declarer had no Trumps ace. (What was he bidding then?)

Okay, back to the play of Rak from Rak's point of view. Rak's first Trumps Ace catches your first. All smiles and pleasant feelings! Four Trumps gone, 12 outstanding, with 2 Trumps aces unaccounted for. If Rak had a second Trumps ace, no brainier play it, your partner is looking for it. If Rak doesn't have another Trumps ace (I assume he didn't), then if I was Rak, I'd take my cue from declarers side play that got to me. I'd play a queen or king in that side suit hoping to get back to my partner after I've first played all my other aces to get in first and remove garbage from declarers hand and make my aces good. My partner played that side suit for a reason and successfully got to me. Declarer either has control in that suit or is trying to short suit himself in that suit. It's his hand and I'd try to get back to him so he could play his game strategy. Once declarer is back in control, he can decide how to find the fourth ace.

I have one quibble with your suggestion of Rak playing low Trumps, his partner declared Trumps, but did not play any aces of Trumps nor lead low Trumps after declarer played his other aces, clearly, declarer is trying to husband his Trumps, Rak should not play low Trumps thereby running Trumps out of the game, remember, both declarer and Rak have played their winners, their opponents have not led yet, running Trumps makes opponents hands have a superior position lessening declarers opportunity to Trump tricks.

####################
Update: 23Jul13 @1400 Eastern
The following is for a specific suit (i.e., one suit).
See more in update below.
####################

As to the probability of a bare singleton ace:
Let C(n,r) = the number of combinations of r things chosen from n items without replacement = n! / ( r! • (n-r)! ) where "!" is factorial symbol meaning multiply all integers from 1 to n together and 0! = 1

Probability of a singleton ace is C(4,1) • C(60,19) / C(80,20) = 0.002313 = 0.231%
Probability of a doubleton ace is C(4,2) • C(60,18) / C(80,20) = 0.00156 = 0.156%

C(4,1) = 4 ways to choose 1 ace from 4 aces in the suit
C(4,2) = 6 ways to choose 2 aces from 4 aces in the suit
The cards needed to fill out the player 20 card hand are:
C(60,19) = 2,044,802,197,953,900 ways to choose 19 cards from the remaining deck of 60 cards (an entire suit removed to ensure no cards in this suit are dealt to the hand except the single and double aces)
C(60,18) = 2,044,802,197,953,900 ways to choose 18 cards from the remaining deck of 60 cards
C(80,20) = 3,535,316,142,212,170,000 total possible double deck player hands.

Thus, the singleton ace is more likely than the doubleton on average.

I'm going to search the forum for the carding thread, it sounds like an interesting discussion is on-going. I like mind exercises.

Hope this added to the discussion.

---

(07-20-2013, 01:48 PM)tony ennis Wrote:  I wrote a quick program.

Judging by the numbers, I am not sure it makes a lot of sense to suffer too much worrying about a singleton aces.

number of hands: 1000000
number with one void: 4738 (0.4738%)
number with more than one void: 1 (1.0E-4%)
number with at least one singleton ace: 9242 (0.9242%) <-- stiff ace
number with at least one doubleton ace: 6200 (0.62%) <-- stiff aces
number with at least once suit 1 cards long: 46256 (4.6256%) <-- common singleton
number with at least once suit 2 cards long: 198894 (19.8894%) <-- common doubleton
number with at least once suit 3 cards long: 499213 (49.9213%)
number with at least once suit 4 cards long: 819234 (81.9234%)
number with at least once suit 5 cards long: 933838 (93.3838%)
number with at least once suit 6 cards long: 761964 (76.1964%)
number with at least once suit 7 cards long: 452999 (45.2999%)
number with at least once suit 8 cards long: 198844 (19.8844%)
number with at least once suit 9 cards long: 65207 (6.5207%)
number with at least once suit 10 cards long: 15581 (1.5581%)
number with at least once suit 11 cards long: 2816 (0.2816%)
number with at least once suit 12 cards long: 374 (0.0374%)
number with at least once suit 13 cards long: 35 (0.0035%)
number with at least once suit 14 cards long: 5 (5.0E-4%)
number with at least once suit 15 cards long: 0 (0.0%)
number with at least once suit 16 cards long: 0 (0.0%)
number with at least once suit 17 cards long: 0 (0.0%)
number with at least once suit 18 cards long: 0 (0.0%)
number with at least once suit 19 cards long: 0 (0.0%)
number with at least once suit 20 cards long: 0 (0.0%)

Tony--

Hmmm ... You analysis doesn't match my combinatorics answer.

You've got singleton ace at 0.92% whereas I've got 0.23%
Doubleton ace - yours at 0.62% mine 0.15%

We both agree singleton more likely, but I don't like numbers not matching ... Can I get a copy of your program to look at? See my other post for my analysis. (I'd give you a link but don't know how).

Don't ya love it ... Dueling probabilities!!

If I'm wrong, I'd like to know where and understand why.

####################
Update: 23Jul13 @1400 Eastern

Tony--

Turns out we are both correct! You more than me.
I only got a quarter of the answer. My bad.

My Closed Form Combinatorics analysis was for one suit.
Your simulation must consider all four suits.

Yours = 0.0092 == Mine 0.0023 per suit * 4 suits = 0.0092
Yours = 0.0062 == Mine 0.0015 per suit * 4 suits = 0.006

Whew! I feel much better. Thanks.
####################

[richardpaulhall]

There is no carding thread yet. This discussion will form the basis for it.

Stiff: Bridge-speak for a singleton, one card in the suit
Stiff Ace: a single Ace in the suit
Stiff Double Ace: AA in the suit and nothing else

Wikipedia:
In contract bridge, the principle of restricted choice states that play of a particular card decreases the probability its player holds any equivalent card. For example, South leads a low spade, West plays a low one, North plays the queen, East wins with the king. The ace and king are equivalent cards; East's play of the king decreases the probability East holds the ace – and increases the probability West holds the ace.

Declarer was to my right. Rak to my left. I held AA of trump only.



"I apologize, but I'm not following this train of thought. I don't understand the "falls second" or "falls fourth"."
Rak leads trump and the others follow:
A-J-K-A = A falls fourth
vs
A-A-K-J = A falls second

If Rak holds a second trump Ace and MUST play it to not confuse Declarer, then it does not matter.
If the trump A fell second and is singletion: Rak should NOT play a second round of trump, he is finessing his partner.
If the trump A fell fourth and is singletion: Rak SHOULD play a second round of trump, he is finessing his left hand opponent.

(The analysis of this situation will be different if the suit is not trump.)

[rakbeater]

I'm giving everyone in this discussion a +2 for reputation points, because I am loving this and looking forward to the resolution.