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Questions for the Statisticians here:
#1
This may have been touched on in a prior thread, but I wanted to dedicate a thread to my questions:

What are the odds or percentages of declarer's partner having an ace of trump and West (declarer's left hand opponent) not having an ace of trump when declarer holds:

0 aces of trump?
1 ace of trump?
2 aces of trump?
3 aces of trump?

Thanks.
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#2
The 3 aces can be dealt in 27 different ways, each of which is roughly equally likely. Listing them, they may have been dealt out:

WWW
WWN
WWE
WNW
WNN
WNE
WEW
WEN
WEE

NWW
NWN
NWE
NNW
NNN
NNE
NEW
NEN
NEE

EWW
EWN
EWE
ENW
ENN
ENE
EEW
EEN
EEE

Count the number of Ws...that's the number of aces for West. So:

No aces: 8 out of 27
1 ace: 12 out of 27
2 aces: 6 out of 27
3 aces: 1 out of 27
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#3
(07-24-2013, 06:17 PM)ToreadorElder Wrote:  1 ace: 12 out of 27

I know enough about probability to know that chances are great that I am misunderstanding, so I will express what I am interpreting and ask questions.

Rak's question was odds of no W's in a deal-row, and at least one N in a deal row. His question was asked 4 different ways, right, if S is holding 1, 2, 3, or 4 Aces?

Why did you only choose to display the work for the scenario of when S has 1 Ace (distributing the remaining 3 Aces)?
If S=0 then you'd need to have 4 seats letters in each deal-row. e.g. "WWNE"
If S=2 then 2 letters in each deal-row. e.g. "NE"
If S=3 then only 1 letter would be displayed for each deal-row. e.g. "N"

When I counted the W's and N's in your work, I incremented each time W=0 & N>0 and came up with 7 instead of 12.
1 ace: 7 out of 27

Please explain or re-explain.
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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#4
(07-24-2013, 02:49 PM)rakbeater Wrote:  This may have been touched on in a prior thread, but I wanted to dedicate a thread to my questions:

What are the odds or percentages of declarer's partner having an ace of trump and West (declarer's left hand opponent) not having an ace of trump when declarer holds:

0 aces of trump?
1 ace of trump?
2 aces of trump?
3 aces of trump?

Thanks.

(07-22-2013, 11:42 PM)FLACKprb Wrote:  RE: Reaching partner
Probabilities of Reaching your Partner

I thought I answered this challenge ... See above post.

MickMackUSA--

Please take note of Table 5 that enumerates all possible holdings configurations.

RakBeater--

I think Tables 1 through 4 in my referenced post answer the "real" question you're asking, "What is the probability of getting your partner into the lead."

If you are narrowing your question to asking about the probabilities of your partner having an Ace and Left not having one, then I'd have to think about how to present that. I believe the data is in my post.

Please note my Caveats, specifically, I did not consider suit length distribution. I know this will change the answer. By how much? I'm not sure until the complete analysis is done.
Ta!
--FLACKprb
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#5
The reason for my specific question is because most of the time with a strong hand and trump suit, you will exit with a queen of trump to force out the trump aces, but I would like to know what the exact odds are of getting to your partner in trump based on the number of trump aces you hold. These are important number for my research.
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#6
My answer doesn't depend on declarer's suit length. It's not relevant because we're ONLY discussing the distribution of 3 specific cards in the suit, not the rest of the cards. We're also quite specifically limiting the situation to the case that S has exactly 1 ace.

rak's question also implied, to me, that he wasn't asking "1 or more" aces...he was asking "exactly 1". The question of "1 or more" is easily obtained. Either
a) add the probabilities of having 1, 2, or 3, or
b) do 1 - the probability of NOT having 1, 2, or 3...that is to say, 1 - prob(having 0)

Same thing; sometimes one approach is easier to recognize or compute. For example, if you can easily derive "probability of having 0 aces" then you don't need the rest to know "probability of 1 or more."
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#7
(07-24-2013, 08:05 PM)mickmackusa Wrote:  
(07-24-2013, 06:17 PM)ToreadorElder Wrote:  1 ace: 12 out of 27

I know enough about probability to know that chances are great that I am misunderstanding, so I will express what I am interpreting and ask questions.

Rak's question was odds of no W's in a deal-row, and at least one N in a deal row. His question was asked 4 different ways, right, if S is holding 1, 2, 3, or 4 Aces?

Why did you only choose to display the work for the scenario of when S has 1 Ace (distributing the remaining 3 Aces)?
If S=0 then you'd need to have 4 seats letters in each deal-row. e.g. "WWNE"
If S=2 then 2 letters in each deal-row. e.g. "NE"
If S=3 then only 1 letter would be displayed for each deal-row. e.g. "N"

When I counted the W's and N's in your work, I incremented each time W=0 & N>0 and came up with 7 instead of 12.
1 ace: 7 out of 27

Please explain or re-explain.

rak's question was specifically for S having *1* ace, and I was keeping the answer focused...while at the same time, showing the general method that works for S having a different number.

Ohh...I misread rak's question, my bad there. He does want the 'chance to reach partner.' As you note, it's easy to do from that list...W=0, N>=1.

Sorry for the confusion.
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#8
Somebody shout if any of this seems to be incorrect. I wrote it all out (no equations), and then picked all the combinations that satisfied rak's situations.

South holding 3 aces: 1 out of 3
...satifying combination (N)

South holding 2 aces: 3 out of 9
...satifying combinations (NN,NE,EN)

South holding 1 ace: 7 out of 27
...satifying combinations (NNN,NNE,NEN,NEE,ENN,ENE,EEN)

South holding 0 aces: 15 out of 51
...satifying combinations (NNNN,NNNE,NNEN,NNEE,NENN,NENE,NEEN,NEEE,ENNN,ENNE,ENEN,ENEE,EENN,EENE,EEEN)


NOW, with these figures, what kind of thinking can be applied to actual pinochle play? Anyone? Are there any mistakes that I made or flaws in this thinking?
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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#9
(07-27-2013, 06:04 AM)mickmackusa Wrote:  Somebody shout if any of this seems to be incorrect. I wrote it all out (no equations), and then picked all the combinations that satisfied rak's situations.
...
NOW, with these figures, what kind of thinking can be applied to actual pinochle play? Anyone? Are there any mistakes that I made or flaws in this thinking?

Oi!!
(That was a shout as you asked for ... Not intended as disrespect.)

Your elaboration is commendable if order mattered. I do not believe that the order a player receives the cards is relevant in this case. What matters is the final state. Table 5 of my presentation elaborates all of the final states possible.

Consider, if Right/East holds 4 Aces and You/South want to get to your partner, it doesn't matter the order Right/East received the cards, just the holding. You can't get to your Partner unless Partner is void in that suit and is cutting it. I grant you that is a valid situation I did not include in my analysis. I expressly caveated this in my presentation by disclosing I did not consider card distribution. I went for a quick analysis rather than a fully comprehensive one which would have to include whether or not Left, Partner, OR Right is void in that suit.

Apologies. I'll try to do better in future.

Until I'm shown the error in my logic/analysis, I still contend I've answered Rak's question (albeit sans void suit second order effect which I think is small, but the analysis should be amplified to prove it).

Tables 1-4 of my analysis summarize the probabilities of getting to your Partner where "Win" is successfully getting there for each of the Aces holding by You/South. (Rak take note, please.)

Tables 1-4 take into account whether information about the distribution is known or not known (e.g., Did someone declare Aces or lay down a run showing an Ace.).

I contend my analysis is good for any suit including Trumps. (Rak take note, please.). I did make one unstated assumption that I thought was obvious. In order to get to your Partner, one has to lead a card in the suit that is not a sure winner. So, Rak, usual practice is leading a Queen in Trumps, but any rank other than an Ace would work in the "Win" situation. (Much discussion to erupt, I'm certain!)

If my presentation is unclear, I'll gladly translate it into any form desired, if one provides me with a template. Show me how you'll better understand and I'll reformat the results to suit (pun intended!).

RTFQ/RTFA
Ta!
--FLACKprb
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#10
(07-27-2013, 10:53 AM)FLACKprb Wrote:  Table 5 of my presentation elaborates all of the final states possible.

Sorry to not reference the table data you provided <here>.

To be honest, I am not able to convert your numbers into understanding.
I also thought rak's question in this thread may have been simpler than what you were answering in the other thread.

I was able to participate in TE's discussion because his presentation was easier for me to understand and included a simple summary.

Can you textually formulate a few observations from your data? Something more tangible for new players ...and me? (please post over in the appropriate thread.)
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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