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New Top Level Area: Carding
#11
(07-20-2013, 03:38 PM)FLACKprb Wrote:  Tony--

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

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

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

I banged the code out so I could have a bug, or we could have different definitions. Lemme debug it tomorrow.

I didn't use any combinatorics. Mine is a pure Monte Carlo simulation.
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#12
(07-20-2013, 06:34 PM)richardpaulhall Wrote:  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.)

Wow, my brain hurts. I spent time reading about Bridge, restricted choice, and thinking. I thank you for that because you made me recall why I like Pinochle over Bridge.

Experienced player's please bare with this discussion. I'm intentionally being verbose for those less familiar with Pinochle out of curtesy to them.

  1. In Bridge, while one has to follow suit, one is not required to beat what's been played (please correct me if I'm wrong) ... In Pinochle, one has to follow suit AND beat what is played.
  2. In Bridge, one can play any suit if they are void in the suit played (again correct me if I'm wrong) ... In Pinochle, one has to play Trumps if void.
  3. In Bridge, points are tricks (books in Bridge lingo) count ... In Pinochle, points are pointers (Aces, Tens, & Kings) and last trick.

What makes Pinochle better to me is that you can force opponents to play specific cards. That means given the right set of card distribution one can control the play. This is why I love the game. The enjoyment of watching a playing strategy fall as you planned it is very satisfying.

I think I understand now where your thinking was about your "stiff double aces" in Trumps. Your focus is specifically on that singleton ace after you lost the first ace. You are trying to puzzle out where the fourth ace is to determine Rak's best play.

May I offer the following observations.

Again without knowing the bidding sequence and melded cards, you were in danger from the get go with respect to your two bare aces when that suit got named Trumps. (Aside: Having a hand with just two cards in one suit almost guarantees it'll be named Trumps if you don't take the bid! ... years of experience.)

Everything depended on declarer's hand and the strategy of play he decided on. If he was strong and long in Trumps with say 2 aces 2 tens 4 others and decided to draw down outstanding Trumps by playing his aces of Trumps, you lose yours. If he is not that strong or decides to play a middle-to-end game strategy where tricks are richer in pointers and doesn't immediately play his Trumps aces then you have only two hopes.

First is he wants to draw out Trumps by leading into you with a Trumps Queen where you can get to play your two aces then. If he does this, you are lucky to gain control and win two tricks with those two bare aces. (Newbies take note.) You MUST play that second ace immediately.

Second hope is he doesn't get past you in a side suit that you have an ace in. If he doesn't and you get control. (Newbies take note.) Those two bare Trumps aces MUST be played immediately.

Once declarer got through you to Rak, your two bare aces were almost certainly lost. (I cannot see how to prove this mathematically, just experience talking.) You've most likely (experience again) become 4th in-line to get control of the table. Declarer controlled then Rak controlled. Once Rak had to relinquish control, he has to go through your partner and declarer before you.

I've noted this before and repeat it here for completeness. Rak took one of your aces because he had a Trumps ace and played it (Good play Rak!). If he had a second Trumps ace he should play it to show his partner where it is and remove Trumps from the game.

"If Rak holds a second trump Ace and MUST play it to not confuse Declarer, then it does not matter."

It's not a point of confusing Declarer ... Rak's role is to support Declarer. Playing his Trumps Aces 1) shows Declarer where they are (were) and 2) removes Trumps from the game strengthening Declarer's hand for later play.

"If the trump A fell second and is singletion: Rak should NOT play a second round of trump, he is finessing his partner."

Finesse means win or beat, yes? Declarer and Rak are Partners. Their points garnered in play score together. I'll argue, if and only if Rack had a second Trumps ace, he should play it. He should NOT play lesser Trumps rank. It's Declarer's strategy to decide when and if Trumps should be drawn down.

"If the trump A fell fourth and is singletion: Rak SHOULD play a second round of trump, he is finessing his left hand opponent."

Play what in Trumps? Ace yes. Lesser rank no.

Whew! I'm exhausted. Brain hurts thinking.

You've been puzzling on vulnerability of your doubleton bare Trumps aces.
I've got to know now the following (curiosity is bubbling):

  1. What the initial hands where.
  2. Who was the dealer (to attribute bid sequence).
  3. What the bidding sequence was.
  4. What was melded by each player.
  5. Finally, did you lose both of your two aces or not?

Consider the following for discussion. Pinochle is about garnering points, not necessarily tricks. One four banger (4 point trick) is equal to four single point tricks! I'd be less focused on a single trick won and more focused on strategy of controlling the play. Putting a discussion of card distribution aside for one moment, is it better to let defenders get into control and play their aces winning lesser value tricks at the beginning of the game, then take control, remove Trumps from defenders, and run the end game (with a strong Trumps suit and/or with a commanding backup suit) garnering more valuable tricks (3 and 4 pointers)? Taking last trick worth 6 points is a lot of points!

(07-20-2013, 09:06 PM)tony ennis Wrote:  
(07-20-2013, 03:38 PM)FLACKprb Wrote:  Tony--

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

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

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

I banged the code out so I could have a bug, or we could have different definitions. Lemme debug it tomorrow.

I didn't use any combinatorics. Mine is a pure Monte Carlo simulation.

Tony--

I empathize. I'm a curious person by nature and as an Engineer I always want to understand how things work.

The Monte Carlo Simulation I put together did not initially provide compatible answers with the closed form combinatorics mathematical solutions (CFCMS) that I worked out. I puzzled over this for weeks, running tests on the random number generator, double checking combinatorics, and scratching my head. It wasn't until I changed the dealing algorithm that everything fell into line. I haven't figured out why yet. I moved on once I was satisfied I was getting good numbers.

A friend of mine did a Monte Carlo Simulation (short Basic program) to calculate probabilities of Pinochles. His implementation surprised me. I'd never would have thought to do it the way he did. Two arrays: one containing numbers 1..80, second filled with random numbers on each trial. Bubble sort the arrays. JD numbers 1..4, QS numbers 5..8. Count occurrences of 1..8 in first 20 cells of first array. His results matched mine and the CFCMS.

I'd really like to see your program for my own education. How long did the 1M trials take to run?
Ta!
--FLACKprb
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#13
Flack, let me look at it one more time :-) I didn't write any test cases for the new code. It was originally designed to calculate melds using various systems, but it was easy enough to extend to count statistics.

It will count 1,000,000 hands in about 11 seconds. Fast enough.
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#14
Most information about the hand was not recorded.

5. Finally, did you lose both of your two aces or not?
Rak led a Trump Ace to grab one of my aces. He did not lead a second trump.
I do not remember who had the 4th Trump Ace. I got in several tricks later with a side-suit ace and immediately cashed my Trump Ace.

From the numbers being bandied about, the opponent who plays an Ace under the lead of a Trump Ace by Declarer or Partner initially held AGAG in less than 1% of the hands. This situation is so rare as to not figure in one's calculation of how to play the trump suit, let alone play the hand. So there is no card coup to look for.

Declarer should re-evaluate his plans once that Ace drops if he is on lead.

If Declarer's Partner is on lead, holding a second Trump Ace must play it to clarify the trump position.

All four players should be aware that the layout of the 10s might be very important. Anyone who normally does not count Trump 10s should endeavor to do so on this hand.

If the remaining Ace falls under Partner's lead of his second Trump Ace, it is almost as if trump had not been lead at all. 12 trump remain, and with the 4 Aces gone, 10s are Boss and those Kings are scheming to win trump tricks.
Should Partner now continue with a third round of Trump if he has a Trump 10 in his hand? He had to play Aces for Declarer's benefit. The Trump 10s are now winners in a 12 card suit.
Rick Hall
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#15
I ran another test, this time with 2 billion hands.

number of hands: 2000000000
number with one void: 9489285 (0.47446425%)
number with more than one void: 448 (2.24E-5%)
number with at least one singleton ace: 18504193 (0.92520965%)
number with at least one doubleton ace: 12553456 (0.6276728%)
number with at least once suit 1 cards long: 92540343 (4.62701715%)
number with at least once suit 2 cards long: 397699253 (19.88496265%)
number with at least once suit 3 cards long: 998945128 (49.9472564%)
number with at least once suit 4 cards long: 1640236225 (82.01181125%)
number with at least once suit 5 cards long: 1866213481 (93.31067405%)
number with at least once suit 6 cards long: 1521426071 (76.07130355%)
number with at least once suit 7 cards long: 906349176 (45.3174588%)
number with at least once suit 8 cards long: 398918888 (19.9459444%)
number with at least once suit 9 cards long: 130248067 (6.51240335%)
number with at least once suit 10 cards long: 31516062 (1.5758031%)
number with at least once suit 11 cards long: 5613987 (0.28069935%)
number with at least once suit 12 cards long: 730962 (0.0365481%)
number with at least once suit 13 cards long: 67575 (0.00337875%)
number with at least once suit 14 cards long: 4417 (2.2085E-4%)
number with at least once suit 15 cards long: 181 (9.05E-6%)
number with at least once suit 16 cards long: 3 (1.5E-7%)
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%)

Flack, I looked the code over, I can't find a flaw. I wrote a more stripped down version which uses none of the same code. It produces results similar to the above. If you want the code PM me your email address.

Begin Mon Jul 22 20:33:59 EDT 2013
Trials: 10000000
Stiff singleton aces: 92533 (0.92533%)
Stiff doubleton aces: 62700 (0.627%)
End Mon Jul 22 20:36:47 EDT 2013
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#16
Here's a run with 50,000,000 deals.

Begin Mon Jul 22 20:40:19 EDT 2013
Trials: 50000000
Stiff singleton aces: 461814 (0.923628%)
Stiff doubleton aces: 313203 (0.626406%)
End Mon Jul 22 20:54:18 EDT 2013
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#17
I hate to ruin this whole topic, but now that I think about it, I'm pretty sure I didn't lead an ace of trump, because I almost never do this unless I have all the aces. I think I exited with the queen of trump and either West or North took the trick with a trump ace, and RichardPaulHall had an ace of trump caught. Somehow I got the lead again before Richard did, and I led another low trump to get the final ace out and that was when RichardPaulHall told me that if I had led my ace I would have caught his and we started talking about the odds and that we should post it on the forums.

So the fundamental question is, if the first Ace of trump catches an opponent's ace of trump, how often will that opponent have another ace of trump?
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#18
(07-23-2013, 11:09 PM)rakbeater Wrote:  I hate to ruin this whole topic, but now that I think about it, I'm pretty sure I didn't lead an ace of trump, because I almost never do this unless I have all the aces. I think I exited with the queen of trump and either West or North took the trick with a trump ace, and RichardPaulHall had an ace of trump caught. Somehow I got the lead again before Richard did, and I led another low trump to get the final ace out and that was when RichardPaulHall told me that if I had led my ace I would have caught his and we started talking about the odds and that we should post it on the forums.

So the fundamental question is, if the first Ace of trump catches an opponent's ace of trump, how often will that opponent have another ace of trump?

Wow! Another Challenge/Mission!

Does the tape in the player burn up with smoke now after promising to disavow any association, Mr. Phelps?

Seriously,

tony ennis Wrote:In any suit: <-- FLACKprb clarifying addition
Stiff singleton aces: 461814 (0.923628%)
Stiff doubleton aces: 313203 (0.626406%)
FLACKprb Wrote:In one suit:
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%
Tony's simulation and my combinatorics calculation * 4 agree that bare Aces in any suit occur as stated above.

Ignore whether or not it's a Trumps Ace played and another Trumps Ace caught. An opponent played that caught Ace because he had to play it (at least any reasonable rational Player).

Given this, you can expect the following to occur on average with four Player Hands per Deal assuming games last an average 8 Deals per game (I'd like to collect the data to validate this assumption):
Code:
Condition    Probability   Hands     Deals    Games (Average 8 deals per game)
Singleton    0.00923628    108.27    27.07    3.38
Doubleton    0.00626406    159.64    39.91    4.99

Thus, you'll see one person having Singletons on average every 3-4 games and not necessarily the same lucky person doubletons on average every 5 games.

So you've got to say to yourself is this the 3rd, 4th, or 5th game I've played and ask yourself: "Do you feel lucky today Rakbeater?" And then decide if you'll pull the trigger or not!

(We now know from your own admission you're risk adverse! I'll remember that and now everyone knows!)

Yeah, I know, these are probabilities and not certainties. Everyone should remember that. I realize a lot depends on what your holdings are and playing strategy is. You may not want to go after that "Last Ace out" (from your perspective last) with your Ace for good reason.

Does this clarify it for you now?
Ta!
--FLACKprb
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#19
(07-24-2013, 04:37 AM)FLACKprb Wrote:  Given this, you can expect the following to occur on average with four Player Hands per Deal assuming games last an average 8 Deals per game (I'd like to collect the data to validate this assumption):
Code:
Condition    Probability   Hands     Deals    Games (Average 8 deals per game)
Singleton    0.00923628    108.27    27.07    3.38
Doubleton    0.00626406    159.64    39.91    4.99

Are these odds on a singleton and doubleton of any two cards of a suit or specifically just aces?
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#20
(07-24-2013, 08:41 AM)rakbeater Wrote:  
(07-24-2013, 04:37 AM)FLACKprb Wrote:  Given this, you can expect the following to occur on average with four Player Hands per Deal assuming games last an average 8 Deals per game (I'd like to collect the data to validate this assumption):
Code:
Condition    Probability   Hands     Deals    Games (Average 8 deals per game)
Singleton    0.00923628    108.27    27.07    3.38
Doubleton    0.00626406    159.64    39.91    4.99

Are these odds on a singleton and doubleton of any two cards of a suit or specifically just aces?

Great question. Made me think. Thanks!

Short Answer

Any suit any rank.

Longer Answer
  • The combinatorics math formulation identifies four identical cards without specifying rank or suit. The formulator interprets the formulation. Therefore, any rank any suit.
  • The Monte Carlo simulation probably assigned the data collection accumulator to be Aces. The result can be generalized and extended by inference to be any rank any suit. As a player just imagine playing the game with inverted card rank values where Jacks beat Queens, Queens beat Kings, etc. Then singleton/doubleton Pseudo Jacks (i.e., Aces in the inverted game) would have those probabilities.

Does that help?
Ta!
--FLACKprb
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