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Reaching partner
#1
richardpaulhall posted this hand over in the bidding discussion:

TCJCADJDASASTSTSKSQSQSQSQSJSJSJSTHKHJHJH

For this discussion, let's say we got this hand in 1st seat, bid 65, and took it right there. Let's consider this strictly as a play problem. At trick 1, AD is a no-brainer. After that, what you would desperately like to do is get partner in while any club/diamond aces might still be worth tricks. So, we'll also assume partner does NOT have the other 3 diamond aces, and therefore doesn't signal. We'll also assume neither LHO nor parnter melded aces...but we'll consider the implications if RHO melded aces.

Options:

a) JD
b) JH
c) QS
d) spades from the top

There's no reason to ever lead a club, based on your hand.

Quick comment about the defense, for the first 3 cases. From their perspective...a 65 bid, cash only 1 side ace...either a great trump suit or a big sandbag with a 2-suiter (say, 9-7 with an AGAGTGTGJGJGJG second suit or so). In either case, the defense has to worry that we have short suits. Ergo: they may have to cash in the side suits ASAP.

a) JD can work in one of two ways. First, it suggests you're looking for partner to ruff diamonds; perhaps you have a hand with

ASASASTSTSTSKSQSJSADTDTDTDQDQDJD

where you can strip trumps when you've got the secondary diamonds ready to go. This MIGHT induce LHO to duck the diamond, which greatly increases the chance you'll get partner in. Second, without any trickery, it hits partner when LHO has no AD, and partner has at least 1.

What are the odds of this? I'll use the notation X=Y=Z to denote who has the missing aces, *in clockwise order* from you. So 1=2=0 would mean LHO has one, and partner the other 2.

Each ace can go fundamentally equally to each player, so there are 3^n (n == number of aces missing) possible X=Y=Z triples, each equally likely (not strictly mathematically true, but close enough for our purposes).

So with 3 aces missing, there are 27 aces. Our 'success' cases require LHO to have no aces, and partner to have 1 or 2 aces. (If he has all 3, we'll assume he signals.) This is 2 of the 27 aces...or roughly a 7% chance. UGH.

b) JH suggests hearts is our second suit...it's actually telling a better story. The whole hand might be this:
XCXCADJDASASASTSTSKSKSQSJSAHTHTHTHQHQHJH

AD, JH is playing aggressively, but it is more logical than the story I offered for case a). My TH's are much more valuable if I retain my AH.

The chance that LHO has no AH, and partner has at least 1...there are 4 such cases, out of the (now) 81 possible. That's 1 in 20...5%. BAD. So really, you'd be hoping to mislead the defenders and induce a mistake.

c) QS. The straightforward lead trying to reach partner. There are 9 cases...you reach partner when the trump aces are 0=1=1 or 0=2=0. 2 cases, 22%. Not great, but way better than the other 2.

Now we can consider the impact if RHO melds aces. In case a) there are now 2 missing diamond aces, and partner *probably* doesn't have both (no signal). We're down to 9 cases, with 0=1=1 being the successful one. The diamond play rises to 11%. The heart play similarly rises to 3 cases out of 27, or 11%. The QS...with only 1 missing ace, that rises to 33%. Pretty nice improvement!

d) Spades from the top is a strategy to reduce your trump losers to 1. If you can, you now have 12 tricks, so your estimate is pulling 30. Give partner 1 heart ace, and you're at 31. If the trumps are 3-3-2 (the - separator means 'in any order', while the = means 'in that order)...you never lose more than 1 trump trick.

So...what's better? There are 8 cards missing, which means 3^8 or 6561 cases.

ASAS (and QS wins) loses when partner has A or Ax, and RHO the other A. Partner has singleton ace, RHO has ace...there are 6 cards distributed between the opponents. This can happen 64 ways. Partner has Ax, RHO A...5 missing cards, 32 cases. Therefore, ASAS is WRONG, 96 times.

QS loses (and ASAS wins) when *LHO* has A or Ax, and partner has at least 1 more card in the suit with the required A. If LHO has the bare A, partner must have Ax, and there are 5 cards between partner and RHO...32 cases. If LHO has Ax, partner must have Axx, so there are only 3 cards be be spread around...8 cases. Therefore QS loses only in 40 cases. Oh, and LHO can have AA, with partner having Txx or longer. That's another 8 cases. (Yes, partner might play the T on your first ace...but with the holding you have, the ONLY non-point missing, is 1 J. He can play a K to give the point.)

Still, *in the cases that matter*...QS is effective twice as often as ASAS, when the goal is to reach partner, and that really needs to be your first priority. QS at trick 2 may make you lose 2 trump tricks, when ASAS would hold the loss to 1...but getting partner in, before either defender, is worth more than 2 tricks in clubs and diamonds. Partner will have, on average, 1 diamond ace, and 1 1/3 club aces.
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#2
http://www.powerpinochle.com/forum/showt...php?tid=99

Above is the link with the stats to prove that option d) is not optimal with a 9 card suit.

I'm so tired I can't focus on the rest of this post.
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#3
I said, hold the trump losers to 1. That's a significantly different question.

There are only 3 significant cards in this question, AS, TS, TS. If you see 2 of the 3 fall, you're guaranteed to have only 1 trump loser barring someone having 7+ trump. So even if someone has ASXSXSXSXSXS, if the 5 remaining spades are 3-2, both TSs fall under your aces, and you still lose only 1 trump trick.

In this question here, or in the question in the other thread (when should I play all my trump aces from the top, trying to drop an opponent's ace)...one thing to remember is that just because someone has 5 or 6 trumps, does NOT automatically make that a failing case.
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#4
Agreed on that point. Reaching your partner (as declarer) is an overlooked strategy/skill? Would be nice to see more posts and examples, or perhaps a basics post on this.
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#5
I accept Rak's challenge.

Probabilities of Reaching your Partner

Acronyms & Terminology:
  • L = Left Hand Opponent, or Lose depending on context
  • P = Partner
  • R = Right Hand Opponent
  • Y = You (That is from your prospective seat at the Table)
  • W = Win, got successfully to Partner
  • L = Lose, didn't get to Partner
  • ?Look@Tens = You hold all 4 Aces, could interpret these configurations as Tens, but don't know why anyone would consider doing that

A priori knowledge can be garnered from:
  1. Declaration of Aces (either inference from bid, in meld, or by vernacular custom)
  2. Declaration of Run in meld
  3. Aces played in the suit, Right could play an Ace then relinquish control possibly sandbagging an Ace (i.e., holding an Ace for later).

Caveats:
  1. The card distribution (i.e., the number of cards in the suits within a hand) is not taken into account in this analysis. I don't think this will affect the result too much although another simulation will have to be conducted to determine the probabilities of suit distribution.
  2. The play of Left, Partner, and Right is assumed to be Win not "Slough off" (i.e., take the trick with an Ace or winning Ten if they have a commanding position with Aces).

Methodology:

A Monte Carlo Simulation dealt all 80 cards to a four hand table (L, P, R, Y). The Aces distribution was examined in each suit of each hand giving 4 samples (one for each suit within one deal). The 35 possible configurations were frequency counted (see Table 5). The frequency counts were divided by the number of trials (deals) times four (four samples per trial) to get the probabilities of each configuration. The following tables were constructed by summing the probabilities for winning (i.e., getting to your partner) given the number of Aces in "Y" and dividing by the sum of the probabilities of the number of total possible configurations (Win and Loss) given the number of aces in "Y". The Lose is found from 1 minus the Win in that category.

A calculation example:

Table 1, Win (You, Y, successfully getting to your Partner, P) for You having 3 Aces in 1 suit is found by dividing Table 5's configuration number 3's probability by the sum of the probabilities for configurations 2, 3, and 4.


Observations:
  1. With no a priori knowledge odds are not good in getting to your partner given any configuration.
  2. When only Partner revealed Aces odds are good you'll get to Partner.
  3. When only Right revealed Aces odds are against you getting to Partner.
  4. When Partner and Right revealed Aces only You holding 2 Aces are good odds to getting to Partner.
  5. If Left revealed Aces you can't get to Partner unless Left lets you through.

Table 1 No A Priori Knowledge
You Win Lose
3 0.3342 0.6658
2 0.3329 0.6671
1 0.2552 0.7448
0 0.1773 0.8227

Table 2 Partner Revealed Aces
You Win Lose
3 1.0000 0.0000
2 0.5952 0.4048
1 0.6963 0.3037
0 0.2181 0.7819

Table 3 Right Revealed Aces
You Win Lose
3 0.0000 1.0000
2 0.4029 0.5971
1 0.3121 0.6879
0 0.2059 0.7941

Table 4 Partner & Right Revealed Aces
You Win Lose
2 1.0000 0.0000
1 0.4870 0.5130
0 0.2634 0.7366

Table 5 Aces Configurations and Their Probabilities
(Ordered on Y column in descending order of Number of Aces held.)
(This Table has 35 items and can be scrolled to see all 35)

Code:
ConfigID Left  Partner  Right   You  Win/Lose     Probability of Occurance
1         0       0       0      4   ?Look@Tens    0.0031005
2         0       0       1      3      L          0.01442
3         0       1       0      3      W          0.0144382
4         1       0       0      3      L          0.0143495
5         0       0       2      2      L          0.0228173
6         0       2       0      2      W          0.0228943
7         2       0       0      2      L          0.022724
8         0       1       1      2      W          0.0479373
9         1       0       1      2      L          0.0482327
10        1       1       0      2      L          0.048182
11        0       0       3      1      L          0.0144845
12        0       3       0      1      W          0.0143597
13        3       0       0      1      L          0.0144058
14        0       1       2      1      W          0.0479798
15        1       0       2      1      L          0.0480382
16        1       2       0      1      L          0.0481545
17        0       2       1      1      W          0.0480545
18        2       0       1      1      L          0.047987
19        2       1       0      1      L          0.0479497
20        1       1       1      1      L          0.101158
21        0       0       4      0      L          0.003095
22        0       4       0      0      W          0.00309375
23        4       0       0      0      L          0.003078
24        0       1       3      0      W          0.0143552
25        1       0       3      0      L          0.0144248
26        0       3       1      0      W          0.0144528
27        1       3       0      0      L          0.0144852
28        3       1       0      0      L          0.014461
29        3       0       1      0      L          0.014477
30        0       2       2      0      W          0.022752
31        2       0       2      0      L          0.0227025
32        2       2       0      0      L          0.0227912
33        1       1       2      0      L          0.0480778
34        1       2       1      0      L          0.0479785
35        2       1       1      0      L          0.0481077

Please comment on anything. If I've made any mistake in methodology or math, I'll not be offended to be corrected. I just want it right and to learn from mistakes in logic.

For my fellow Engineers, this 1M trials Monte Carlo simulation ran for 1 hour 26 minutes. A Basic program on my iPad 4th generation.

Rick/ToreadorElder is this anything like Restricted Choice?
Ta!
--FLACKprb
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#6
Ewww...

Hate doing this, but...

The random number generator included with most implementations of Basic, is simply inadequate for real statistical analysis/Monte Carlo simulation. The 'classic' RNG that's included, is an almost-trivial linear congruential generator that has well-known non-random behavior, and a very short period. I don't know what RNG you're using, so this may not be a problem. I'm pointing it out because I'm seeing several fresh posts incorporating simulation results. That's great; I love to see the effort taken...but rendered incorrect because you get bitten by this. Statistically good RNGs are readily available, but almost never incorporated. Microsoft Excel's had serious issues with this for *years*...and that's a product that is used for modeling and simulation. I can't offhand see that a Basic version on an iPad, is likely to be anything more than a toy.

(What I use for these, is Java. It's free, it's got a good IDE in Eclipse which is also free, and I've been doing it professionally for 10 years now, so I'm used to it. And Java's got a heckuva lot of free stuff. I use the Apache Commons

Also, this is subject to accurate, direct calculation. Look at the 35 rows. The key is the aces distribution. When you have N aces, there are (4-N) ^ 3 possible distributions of the remaining aces. There's some error in here, based on vacant places, but with 20 cards in hand, the effect is minor.

So, if you have 2 aces, what's the probability you can reach partner? There are 8 cases: he's got both, and he's got 1 and LHO has the other...so it's a straightforward 25%, before you play any card. The odds *change* when you play your 2nd ace...does partner signal? If NOT, it doesn't mean he doesn't have both, but that will happen only with a fairly small subset of all the possible holdings he has. And some of these will actually be intentional NON-encouragements. Here's an example, with spades as trump, and you're declaring. Partner's hand:

ASKSQSJSJSADADTDKDKDKDJDJDJDXCAHAHXHXHXH

At tricks 1 and 2, you play AD,AD. Put yourself in dummy's position...do you *want* a 3rd diamond? With a 9 card suit? There's 2 significant risks: an opponent started with a doubleton and will ruff the 3rd round, and a sneaky one: partner (declarer) only has 3. More diamond leads will expose him to a forcing game in trump.

Beyond that: you have ruffing value in clubs, so you'd like to see that suit started ASAP. Your heart aces don't look *too* prone to being ruffed. And you have a trump ace...if partner plays QS at trick 3, you don't have to fear it running to the defender behind you.

It's probably more common that you just don't have a clear signaling card...AAKK, AAKQ, AAKKQ and so on. Still, all told, it's not many cases. *Generally speaking*...if I'm declaring and I've played 2 aces, and I *don't* see a signal on the 2nd ace, I'll assume my partner *doesn't* have them both.

OK, so one of the two cases gets knocked out. That means there's only 1 left: he's got 1 and LHO has the other. That's a 1 in 8 chance. When you as declarer have 1 ace, the numbers change but the approach still works: if you don't get a signal, the working cases are that partner has 2, LHO has 1, OR that partner has 1, LHO has 2. There's 27 total cases, so leading that suit will work 2 out of 27 times, or a bit over 7%.
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#7
(07-23-2013, 02:40 PM)ToreadorElder Wrote:  Ewww...

Hate doing this, but...

The random number generator included with most implementations of Basic, is simply inadequate for real statistical analysis/Monte Carlo simulation. The 'classic' RNG that's included, is an almost-trivial linear congruential generator that has well-known non-random behavior, and a very short period. I don't know what RNG you're using, so this may not be a problem. I'm pointing it out because I'm seeing several fresh posts incorporating simulation results. That's great; I love to see the effort taken...but rendered incorrect because you get bitten by this. Statistically good RNGs are readily available, but almost never incorporated. Microsoft Excel's had serious issues with this for *years*...and that's a product that is used for modeling and simulation. I can't offhand see that a Basic version on an iPad, is likely to be anything more than a toy.

(What I use for these, is Java. It's free, it's got a good IDE in Eclipse which is also free, and I've been doing it professionally for 10 years now, so I'm used to it. And Java's got a heckuva lot of free stuff. I use the Apache Commons

Also, this is subject to accurate, direct calculation. Look at the 35 rows. The key is the aces distribution. When you have N aces, there are (4-N) ^ 3 possible distributions of the remaining aces. There's some error in here, based on vacant places, but with 20 cards in hand, the effect is minor.

So, if you have 2 aces, what's the probability you can reach partner? There are 8 cases: he's got both, and he's got 1 and LHO has the other...so it's a straightforward 25%, before you play any card. The odds *change* when you play your 2nd ace...does partner signal? If NOT, it doesn't mean he doesn't have both, but that will happen only with a fairly small subset of all the possible holdings he has. And some of these will actually be intentional NON-encouragements. Here's an example, with spades as trump, and you're declaring. Partner's hand:

ASKSQSJSJSADADTDKDKDKDJDJDJDXCAHAHXHXHXH

At tricks 1 and 2, you play AD,AD. Put yourself in dummy's position...do you *want* a 3rd diamond? With a 9 card suit? There's 2 significant risks: an opponent started with a doubleton and will ruff the 3rd round, and a sneaky one: partner (declarer) only has 3. More diamond leads will expose him to a forcing game in trump.

Beyond that: you have ruffing value in clubs, so you'd like to see that suit started ASAP. Your heart aces don't look *too* prone to being ruffed. And you have a trump ace...if partner plays QS at trick 3, you don't have to fear it running to the defender behind you.

It's probably more common that you just don't have a clear signaling card...AAKK, AAKQ, AAKKQ and so on. Still, all told, it's not many cases. *Generally speaking*...if I'm declaring and I've played 2 aces, and I *don't* see a signal on the 2nd ace, I'll assume my partner *doesn't* have them both.

OK, so one of the two cases gets knocked out. That means there's only 1 left: he's got 1 and LHO has the other. That's a 1 in 8 chance. When you as declarer have 1 ace, the numbers change but the approach still works: if you don't get a signal, the working cases are that partner has 2, LHO has 1, OR that partner has 1, LHO has 2. There's 27 total cases, so leading that suit will work 2 out of 27 times, or a bit over 7%.

Thank you Sir for taking the time to comment. I appreciate it.

First, I totally agree. My iPad is a toy. A fun toy for me right now. I currently do not own a "real" computer so one makes do with what one has. I shared the same concerns regarding the pseudo random number generator contained within the only available language I could find that would host on the iPad. (I have since found a J language app, but have never used J before and have not yet spent time learning it.) At $2.99, I have a hard time complaining. I queried the developer regarding the RNG and was told it is random and doesn't repeat. Of course I didn't believe that. I did run a simple test of 100M trials looking for repetition and didn't find any. I thought about delving deeper into getting into truly significant tests to determine the uniformity and viability of this implementation because I had serious doubts myself when initial results obtained did not match the closed form combinatorics solutions for meld calculations. When I changed the dealing algorithm, the probabilities for everything fell into place to 6-7 significant digits. Based on that I'm confident in the predictions for my purposes. I'd be willing to run any test you'd provide to further verify the RNG. I'd appreciate any help that will enhance confidence in results.

I thank you for the recommendation regarding the IDE. If I ever decide to purchase a "real" computer, I'll give your recommendations serious weight.

Now to your analysis.

"When you have N aces, there are (4-N) ^ 3 possible distributions of the remaining aces."

I'm a very slow plodder, it takes me awhile to understand, but I eventually get there. Please bare with. I'm not sure where your formula comes from, my plodding says:
For "You" holding 2 Aces, then the possible arrangements for the other two is found from C(3,1) + C(3,2) = 3 + 3 = 6 possible configurations, where the first term is for one player holding two aces, 3 possibilities Left or Partner or Right (L or P or R) and the second term is for two players having 1 ace each, three possibilities (LP or PR or LR). These are enumerated in Table 5 of my post, rows 5 through 10. The Win/Loss column (W/L) of Table 5 defines the "Win" of getting to your partner successfully which is 2 (rows 6 and 8) out of the 6 possible configurations. I constructed Tables 1 through 4 to specifically elaborate the entire set of outcomes for reference. If my presentation is unclear, I'm willing to listen to how to make it better. I tried hard to make the presentation of Table 5 better but the software behind these posts removes extra spaces.

Leaving aside the correctness of the Monte Carlo Simulation results for the moment, if my methodology in constructing these tables is wrong please tell me how to fix it. I'm not proud, I'm an Engineer by education and experience. I just want to know how things work.

I wanted for completeness and for a reference for others, the entire situation of getting to your partner documented in an easy to follow form which is why I spent time putting the five tables together in my post. I have no problem, if someone with a fast computer and "proper" tools wants to perform the computations and update my post. I'd like the information to be correct/accurate. I did not take into account distribution (i.e., number of cards in each suit held in each hand) opening up the entire discussion regarding the probability of a suit getting cut. I think this is a second order factor to consider after determining how to reach your partner. I'm thinking about how to formalize the "Probability of being Cut" study and would appreciate any input into how this might be done.

No disparagement towards you intended but my mind gets confused easily reading your analysis. I wish I was quicker on the uptake. I've enjoyed reading your posts and find myself liking your play.

"There's some error in here, based on vacant places, but with 20 cards in hand, the effect is minor."

Sorry, don't follow what you mean by "vacant places". Please elaborate so that I can understand.

"*Generally speaking*...if I'm declaring and I've played 2 aces, and I *don't* see a signal on the 2nd ace, I'll assume my partner *doesn't* have them both."

I only know of two "signals" Partner can make.
  1. First, what I know as a "Direct Leadback" where one plays an Ace on his partners second ace play in that suit to signal "I've got another Ace and probably a Ten backing it". If I held AAKK or AAKQ, on my Partners play of 2 Aces, I probably would not play a Direct Leadback onto my Partner's second Ace because a) I don't have a Ten and b) I now have a "locked" suit putting aside it might be cut.
  2. Second, what I know as an "Indirect Leadback" where one plays a Jack to signal "I might have command in the suit". Might because it's not a certainty.

"OK, so one of the two cases gets knocked out. That means there's only 1 left: he's got 1 and LHO has the other. That's a 1 in 8 chance. When you as declarer have 1 ace, the numbers change but the approach still works: if you don't get a signal, the working cases are that partner has 2, LHO has 1, OR that partner has 1, LHO has 2. There's 27 total cases, so leading that suit will work 2 out of 27 times, or a bit over 7%."

I'm terrible sorry, but I'm lost. My Brain hurts.

In my analysis, I think I considered all of the situations. If I didn't, please show me what I missed. I'll add them happily. I left open the methods during play of gaining information. I pointed out where information can be garnered prior to play. This is important for anyone to pay attention to what is melded. I'll stick by the probabilities presented until provided with proof that I'm wrong. Based on prior work calculating meld probabilities and the complete agreement with closed form combinatorics solutions, I believe the Monte Carlo Simulation results are accurate (with the exception of very small probabilities where not enough representative samples can be obtained in a reasonable run time). Fortunately, the combinatorics solutions for the small probabilities are much easier to work; they verify and compliment the Monte Carlo results.

Always willing to learn new things. Thanks for commenting.
Ta!
--FLACKprb
Reply
#8
<sigh> forget the (4-N)^3. That's a brain fart.

What I was thinking...halfway...was, with 2 aces to be dealt, the first can go to any one of 3, the second can go to any one of 3...so there are 9 outcomes, 3x3. Let's just use A, B, and C. We get AA, AB, AC, BA, BB, BC, CA, CB, CC. This is the complete enumeration of all possible ways those 2 aces can be dealt to the three players. It's 9 cases because I'm including order in there...a combination calculation never does. And it matters. I'm falling all the way back to the original a priori dealing, using ONLY the fact that you have 2 aces. The cards get dealt; at some point, the first ace can go to A, B, or C. Same with the second. So there's 9 cases from this original situation. The case where A has one, and B has 1, can happen in *2* ways...A gets the first, B gets the second, or vice versa. So your combinations don't necessarily occur with equal frequency...and that is likely something I wasn't taking into account either.

Sometimes Monte Carlo sims have the great advantage that they give a crosscheck on a runaway thought process...this kind of stuff can be tricky... Smile

Gotta run here, but a quick comment on vacant places. Vacant places means the number of unknown cards in a hand. Given that we have 2 aces, what's the probability partner has both of the others? Once he gets the first ace...that's now a known card. The probability he'll get the 2nd ace, is NOT 1 in 3...it's slightly less. LHO and RHO have 20 and 20 vacant places...partner only has 19.
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#9
(07-23-2013, 07:20 PM)ToreadorElder Wrote:  Given that we have 2 aces, what's the probability partner has both of the others? Once he gets the first ace...that's now a known card. The probability he'll get the 2nd ace, is NOT 1 in 3...it's slightly less. LHO and RHO have 20 and 20 vacant places...partner only has 19.

This gives me that same feeling I experienced when I watch the movie "21" featuring Kevin Spacey. Specifically when Professor Micky Rosa (Kevin Spacey) is questioning MIT student Ben Campbell on probability.

This is my declaration that my mathematical aptitude for probability will never blossom. My brain always tells me that 1 in 3 is always 1 in 3; despite the more than sufficient support by well-studied people proving to me that I am not right. In the end, I will simply accept the facts and figures as truth, and compartmentalize the processes as "magic/voodoo" in my mind.

FYI:
As far as prettying-up posts with tables using the handicap that is mybb tags, it can be done ...kinda.
I have posted advice on using the code tags to retain spaces.

REMINDER:
If you are going to quote someone else's forum post, please wrap every quote occurrence in quote tags which include the member's name, pid, and dateline data. This helps by highlighting their text, and separates it from your own text that you have wrapped in quotes.
It's unbelievable how much you don't know about the game you've been playing all your life. -- Mickey Mantle
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#10
(07-23-2013, 07:20 PM)ToreadorElder Wrote:  <sigh> forget the (4-N)^3. That's a brain fart.

What I was thinking...halfway...was, with 2 aces to be dealt, the first can go to any one of 3, the second can go to any one of 3...so there are 9 outcomes, 3x3. Let's just use A, B, and C. We get AA, AB, AC, BA, BB, BC, CA, CB, CC. This is the complete enumeration of all possible ways those 2 aces can be dealt to the three players. It's 9 cases because I'm including order in there...a combination calculation never does. And it matters. I'm falling all the way back to the original a priori dealing, using ONLY the fact that you have 2 aces. The cards get dealt; at some point, the first ace can go to A, B, or C. Same with the second. So there's 9 cases from this original situation. The case where A has one, and B has 1, can happen in *2* ways...A gets the first, B gets the second, or vice versa. So your combinations don't necessarily occur with equal frequency...and that is likely something I wasn't taking into account either.

Sometimes Monte Carlo sims have the great advantage that they give a crosscheck on a runaway thought process...this kind of stuff can be tricky... Smile

Gotta run here, but a quick comment on vacant places. Vacant places means the number of unknown cards in a hand. Given that we have 2 aces, what's the probability partner has both of the others? Once he gets the first ace...that's now a known card. The probability he'll get the 2nd ace, is NOT 1 in 3...it's slightly less. LHO and RHO have 20 and 20 vacant places...partner only has 19.

Brain expulsions are allowed ... I have lots of senior moments myself ...

I fully understand: "Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter)."
http://betterexplained.com/articles/easy...binations/

That said, I looked at defining/enumerating all of the possible card distributions with respect to the four Aces of one suit. I initial don't care how I got to the given distribution. That is why I used combinations to enumerate the configurations. This is the 35 rows in my Table 5. Table 5 is sorted on the holding of "You" perspective in descending order from all 4 Aces to 0 Aces.

If "You" are holding 2 then Table 5 shows the possible configurations for the other 2.
These configurations are as stated before (I won't repeat here). I believe I am correct in defining the configurations. (Forget formulas, just take the 2 Aces and put them into the positions).

I used the Monte Carlo Simulation to tell me the probabilities of all of the 35 configurations. I think this handles/covers your "order" and "vacancies" issues.

<<<Someone take note!>>>
I'd appreciate someone verifying these probabilities. It'll give confidence to the answers.

We now have the conditional probability, given "You" are holding 2 Aces, what is the probability of getting to partner. I believe I am correct in taking the configurations where Y=2 and summing the "Wins" then dividing by the ("Wins" + "Lose") to get probability of getting to Your Partner.

QED
Ta!
--FLACKprb
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