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Pinochle Percentages Trivia - Win A Free Book! - rakbeater - 08-19-2012
You look at your hand and see a 5 card suit with four aces and one ten (AAAAT). No trump has been played and you have the lead. You decide to play your four aces consecutively and follow with the ten unless one of the cards is trumped. The first two aces are winners and not trumped. What percentage of the time will: the third ace be a winner? and then the fourth ace? and then the ten? The first person to guess the three correct percentages will win a free copy of The Brick Bidding System. RE: Pinochle Percentages Trivia - Win A Free Book! - ToreadorElder - 08-19-2012
Not hard to get very close by analyzing the distributions of the outstanding cards. On the play of the 3rd ace, there are 9 cards outstanding. The 3rd ace will be ruffed happens only when the distribution is 9-0-0, 8-1-0, 7-2-0, or 6-3-0. There are 1533 cases, out of 19683 hands, so there are 18150 cases where it wins. 92.2%, approximately. NOTE: this has a slight error. The odds that someone actually has the remaining 9 or even 8 or 7 cards in the suit is overstated; it's a vacant places situation. But it doesn't throw the computation off by enough to matter. For the 4th ace to win, the distributions must be 5-2-2, 4-3-2, or 3-3-3. These are the most common cases, tho, with 11508 total. Thus, the probability the 4th ace will win, is ~58.5%. For the 5th round to win, the suit has to be 3-3-3. 1680 cases, therefore 8.5% chance. As a sidebar, why are you cashing the aces first? RE: Pinochle Percentages Trivia - Win A Free Book! - vettester90 - 08-19-2012
WHY would you play ANY "locked" aces unless you had no choice??? You play the TEN to show pure power then push your other aces then push trump... RE: Pinochle Percentages Trivia - Win A Free Book! - ToreadorElder - 08-19-2012
There are occasional times when you might want to, but this isn't intended to show best play; it's just intended to set up the context for the probability discussion. RE: Pinochle Percentages Trivia - Win A Free Book! - rakbeater - 08-20-2012
(08-19-2012, 12:02 PM)ToreadorElder Wrote: There are occasional times when you might want to, but this isn't intended to show best play; it's just intended to set up the context for the probability discussion. That is correct. Wanted to offer up another opportunity for members to win a free book. RE: Pinochle Percentages Trivia - Win A Free Book! - rakbeater - 08-22-2012
(08-19-2012, 07:25 AM)vettester90 Wrote: WHY would you play ANY "locked" aces unless you had no choice??? You play the TEN to show pure power then push your other aces then push trump... If it makes you feel better, you can play the cards in any order you want, and the questions would be what percentage of the time will the 3rd, 4th, and 5th card not be trumped? The answers stay the same regardless of the order of cards played. RE: Pinochle Percentages Trivia - Win A Free Book! - FLACKprb - 08-05-2013
Another challenge! The Specific Challenge Asks: Rakbeater Wrote:You look at your hand and see a 5 card suit with four aces and one ten (AAAAT). No trump has been played and you have the lead. You decide to play your four aces consecutively and follow with the ten unless one of the cards is trumped. The first two aces are winners and not trumped. What percentage of the time will: Definitions Acronyms & Terminology L = Left Hand Opponent P = Partner R = Right Hand Opponent U = Unknown Y = You X = Used when specific L, P, R position is not important for discussion Table Configuration. A specific card distribution in one suit of one hand dealt to all four Players of the Double Deck Pinochle 80 cards where each player receives 20 cards Y-L-P-R. When Order Matters. A specific card distribution in one suit of one hand dealt to all four Players of the Double Deck Pinochle 80 cards where each player receives 20 cards is described with the quadruple Y-L-P-R where order matters. This fully defines the specific Table Configuration for one suit. An example of Y holding 18 cards of one suit with the remaining 2 cards of the suit distributed to the other three Players, L-P-R, can be fully described by the six configurations, using the notation Y-L-P-R as follows: - 18-0-0-2
- 18-0-2-0
- 18-2-0-0
- 18-0-1-1
- 18-1-0-1
- 18-1-1-0
- Group 3 items (1, 2, and 3 from above Y-L-P-R) into 18-2-0-0, and
- Group 3 items (4, 5, and 6 from above Y-L-P-R) into 18-1-1-0.
Combinations formula. C(n,r) = n! / ( r! • (n-r)! ) where you are choosing r items from n items (or more commonly n choose r) without regard to order and without replacement. When order matters then the Permutations formula should be used. Discussion There are 136 total Table Configurations for the Y-U configuration of 5-15. These can be summarized into the 27 5-X-X-X configurations listed in Table 1. Column 1 contains an Identification Number, ID, for reference. Column 2 provides specific configurations. Column 3 contains the number of specific configurations of each Y-X-X-X Type. For example, 5-14-1-0 has 6 possible specific configurations: 5-14-1-0, 5-14-0-1, 5-1-14-0, 5-0-14-1, 5-1-0-14, and 5-0-1-14. Table 1 Summarized Table Configurations for 5-X-X-X Code: `Note this Table scrolls in the code window` For those interested, the number of L-P-R configurations given any Y value can be found using C( (n + k - 1), k ) where n = 3 Players (L-P-R) and k = number of cards to distribute among the 3 Players. The probabilities for each specific configuration are computed using the following methodology. The Conditional Probability of a specific L-P-R configuration given a specific Y configuration is found by multiplying the Probability of L times Probability of P times Probability of R together. Pr( L-P-R | Y ) = Pr( L ) • Pr( P ) • Pr( R )
The summation of all individual probabilities for each L-P-R configuration given Y must add to 1 to be a probability distribution. Σ Pr( L-P-R | Y ) = 1 for all L-P-R configurations given Y
The Probabilities Pr( L ), Pr( P ), and Pr( R ) are found from the following formulas. - Pr( L ) = ( C( (20-Y), L) • C( (60-(20-Y)), (20-L) ) ) / C(80,20)
- Pr( P ) = ( C( (20-(Y+L) ), P) • C( (40-(20-(Y+L) ) ), (20-P) ) ) / C(40,20)
- Pr( R ) = ( C( (20-(Y+L+P) ), R) • C( (20-(20-(Y+L+P) ) ), (20-R) ) ) / C(20,20)
Table 2 Summarized 5-X-X-X Configurations Probabilities Code: `Note this Table scrolls in the code window` Now we can begin to answer Rakbeater's challenge. To be complete all Plays are shown. First Card Probability of the 1st card of AAAAT being played and not being Trumped is found from adding the Probabilities of Table 2 that do not contain a 0 in any X position of 5-X-X-X. These are configurations: 4, 6, 8, 9, 11, 12, 14, 15, 16, and 18 through 27. Conversely, one can add all those with a 0 in the configuration and then subtract this from 1. Probability of 1st card NOT Trumped = 0.9977 = 99.77% Second Card Probability of the 2nd card of AAAAT being played and not being Trumped is found from reducing the 27 member set configurations of non-zero 5-X-X-X members by 1 card each. This gives a reduced set of 19 members as shown in Table 3. The ID from Table 1 is maintained for reference. Table 3 Summarized 4-X-X-X Configurations Probabilities After 1st Card Played Code: `Note this Table scrolls in the code window` Probability of the 2nd card of AAAAT being played and not being Trumped is found from adding the Probabilities of Table 3 that do not contain a 0 in any position of 4-X-X-X and dividing by the sum of all of the probabilities in Table 3. The division by the sum of all the Probabilities in Table 3 must be done to normalize the distribution and make it a proper Probability Distribution that adds to 1. These are configurations: 9, 12, 14, 15, 16, 19, 20, and 22 through 27. Conversely, one can add all those with a 0 in the configuration and then subtract this from 1. Probability of 2nd card NOT Trumped after 1st card played = 0.9738 = 97.38% Third Card Probability of the 3rd card of AAAAT being played and not being Trumped is found from reducing Table 3's 19 member set configurations of non-zero members by 1 card each. This gives a reduced set of 12 members as shown in Table 4. Table 4 Summarized 3-X-X-X Configurations Probabilities After 2nd Card Played Code: `Note this Table scrolls in the code window` Probability of the 3rd card of AAAAT being played and not being Trumped is found from adding the Probabilities of Table 4 that do not contain a 0 in any position of 3-X-X-X and dividing by the sum of all of the probabilities in Table 4. The division by the sum of all the Probabilities in Table 4 must be done to normalize the distribution and make it a proper Probability Distribution that adds to 1. These are configurations: 16, 20, and 23 through 27. Conversely, one can add all those with a 0 in the configuration and then subtract this from 1. Probability of 3rd card NOT Trumped after 2nd card played = 0.8677 = 86.77% Fourth Card Probability of the 4th card of AAAAT being played and not being Trumped is found from reducing Table 4's 12 member set configurations of non-zero members by 1 card each. This gives a reduced set of 7 members as shown in Table 5. Table 5 Summarized 2-X-X-X Configurations Probabilities After 3rd Card Played Code: `ID 2-X-X-X Probability (Not Normalized)` Probability of the 4th card of AAAAT being played and not being Trumped is found from adding the Probabilities of Table 5 that do not contain a 0 in any position of 2-X-X-X and dividing by the sum of all of the probabilities in Table 5. The division by the sum of all the Probabilities in Table 5 must be done to normalize the distribution and make it a proper Probability Distribution that adds to 1. These are configurations: 24, 26, and 27. Conversely, one can add all those with a 0 in the configuration and then subtract this from 1. Probability of 4th card NOT Trumped after 3rd card played = 0.5944 = 59.44% Fifth Card Probability of the 5th card of AAAAT being played and not being Trumped is found from reducing Table 5's 7 member set configurations of non-zero members by 1 card each. This gives a reduced set of 3 members as shown in Table 6. Table 6 Summarized 1-X-X-X Configurations Probabilities After 4th Card Played Code: `ID 1-X-X-X Probability (Not Normalized)` Probability of the 5th card of AAAAT being played and not being Trumped is found from adding the Probabilities of Table 6 that do not contain a 0 in any position of 1-X-X-X and dividing by the sum of all of the probabilities in Table 6. The division by the sum of all the Probabilities in Table 6 must be done to normalize the distribution and make it a proper Probability Distribution that adds to 1. There is only one configuration: 27. Conversely, one can add all those with a 0 in the configuration and then subtract this from 1. Probability of 5th card NOT Trumped after 4th card played = 0.1398 = 13.98% Summary The following probabilities are conditional probabilities given Y holding the AAAAT 5 cards of one suit after the previous cards are played and not Trumped.
Before 1st card is played Conditional Probability of 1st card NOT Trumped = 0.9977 = 99.77% After 1st card is played and NOT Trumped Conditional Probability of 2nd card NOT Trumped = 0.9738 = 97.38% After 2nd card is played and NOT Trumped Conditional Probability of 3rd card NOT Trumped = 0.8677 = 86.77% After 3rd card is played and NOT Trumped Conditional Probability of 4th card NOT Trumped = 0.5944 = 59.44% After 4th card is played and NOT Trumped Conditional Probability of 5th card NOT Trumped = 0.1398 = 13.98% Interestingly, if the question is asked as to what the probabilities are before any cards are played, the conditional probabilities of the 5 cards played and not being Trumped given Y holding AAAAT are then: Conditional Probability of 1st card NOT Trumped = 0.998 = 99.8% Conditional Probability of 2nd card NOT Trumped = 0.972 = 97.2% Conditional Probability of 3rd card NOT Trumped = 0.843 = 84.3% Conditional Probability of 4th card NOT Trumped = 0.501 = 50.1% Conditional Probability of 5th card NOT Trumped = 0.07 = 7.0% The above numbers are all calculated from Table 2 with the methodology shown above WITHOUT normalizing the distributions at each step. For those interested the probability of AAAAT occurring is: Pr(AAAAT) = 0.00024 = 0.024% More understandable (1 / Pr(AAAAT) ) on average 1 in 4,154 hands. One player at the Table (4 hands dealt per deal) on average will see this 1 in 1,039 deals. If one assumes an average 8 deals per game then one player at the Table on average will see this 1 in 130 games. Rakbeater Wrote:The first person to guess the three correct percentages will win a free copy of The Brick Bidding System. Where's my free book? I still get it even though I didn't "guess", Yes? RTFQ/RTFA RE: Pinochle Percentages Trivia - Win A Free Book! - rakbeater - 08-06-2013
Yes, you will get it. |