Multiple outranking pairs
November 10th, 2017 at 8:51:54 AM
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Is this the proper place to ask a probability question? It isn't obvious where on the site this type of question should be asked.
Specifically, I am interested in the mathematics involved in the following questions.
If 2 cards are dealt to each of 10 players from one deck, what is the probability that two players will be dealt pairs of different ranks?
If one of 9 players is dealt a pair of queens what is the probability that another players has been dealt a pair of aces OR king?
The Wizard already answered the question as to the probability of two players being dealt pairs of equal rank e.g. two players both being dealt pairs of ACES; however, I cannot find any discussion of these questions.
I am more interested in the logic of the mathematical calculation rather than just the probability.
Thank you
John
Specifically, I am interested in the mathematics involved in the following questions.
If 2 cards are dealt to each of 10 players from one deck, what is the probability that two players will be dealt pairs of different ranks?
If one of 9 players is dealt a pair of queens what is the probability that another players has been dealt a pair of aces OR king?
The Wizard already answered the question as to the probability of two players being dealt pairs of equal rank e.g. two players both being dealt pairs of ACES; however, I cannot find any discussion of these questions.
I am more interested in the logic of the mathematical calculation rather than just the probability.
Thank you
John
November 10th, 2017 at 1:40:35 PM
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Quote: JohnMinAZ
If 2 cards are dealt to each of 10 players from one deck, what is the probability that two players will be dealt pairs of different ranks?
If one of 9 players is dealt a pair of queens what is the probability that another players has been dealt a pair of aces OR king?
I am more interested in the logic of the mathematical calculation rather than just the probability.
Welcome to the forum.
I think these are hard probabilities to calculate, which may be why you have not yet had any respondents.
Your 2nd question was:
If one of 9 players is dealt a pair of queens what is the probability that another players has been dealt a pair of aces OR king?
Given the player's hand is QQ, I calculate that the probability of there being one or more pairs of (AA) or (KK) in the other 8 hands is 0.07690838.
I calculate that the probability of there being one and only one AA/KK pair amongst the other 8 hands is 0.07521471.
I can explain the logic and show you the calculation but it will be a significant amount of work. Let me ask questions: do you still want to know the logic and methodology? Do you understand combination math?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
November 10th, 2017 at 4:04:38 PM
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Yes. I am actually more interested in the math and logic than the actual probabilities. I do understand combinatorial math; however, applying it to multiple hands that are also OR situations is an uncertain territory for me. I enjoy the mathematics and am looking for guidance in the logic in these kinds of situations.
I appreciate your interest and help.
I am 82 and just enjoy extending my knowledge. I am working my way through a C# course at the moment. I play poker on line and that prompts me to think about things like these questions.
I appreciate your interest and help.
I am 82 and just enjoy extending my knowledge. I am working my way through a C# course at the moment. I play poker on line and that prompts me to think about things like these questions.
November 10th, 2017 at 6:16:29 PM
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Well then, JohnMinAZ, welcome to the forum! I think you've posted it in the most appropriate place, and it's certainly within the scope of the board. I'm going to delete your blog post for redundancy, and I hope you generate a good discussion.
If the House lost every hand, they wouldn't deal the game.
November 10th, 2017 at 6:28:42 PM
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Thank you
November 11th, 2017 at 9:21:27 AM
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Thanks for introducing yourself, John. It will be my pleasure to try to help you.
How frequently will “n” opponents have AA or KK pair, when you hold QQ pair? To show you the logic, I will explain how to calculate this for n=1 (1 opponent), n=2 and then n=8. Please have patience and bear with me.
One Opponent
Notation: I will use c(x,y) to mean combin(x,y).
We have 2 queens, so there are 50 cards remaining. We have one opponent with a 2 card hand.
The total number of two-card combinations that the opponent can have is c(50,2) = 1225.
AA Pair. The number of AA pair combinations that the opponent can have is c(4,2)=6. There are 4 aces, and you are choosing 2 of 4. The 6 combinations look like this A♠A♥, A♠A♦, A♠A♣, A♥A♦, A♥A♣, A♦A♣,
Obviously, the probability of the opponent having an AA pair is c(4,2)/c(50,2)= 6/1225.
KK Pair. The situation with KK pair is identical to AA pair, that is, there are 4 cards within the rank from which to make a pair. So KK pair also has 6 combinations and a frequency of occurrence = 6/1225.
AA Pair OR KK Pair Nothing could be easier. The total number of combinations in your opponent's hand that are either AA or KK are 12, i.e., 6+6.
Probability of either AA or KK pair (1 opponent): 12/1225 = 0.009796. Slightly less than one percent.
Given 8 opposing hands, a rough estimate of the probability of AA or KK pair appearing would be 8 * 0.9796% = 7.8367%, or about 8%. Honestly, that’s how I would estimate it when I’m playing poker. But that is not rigorously correct and a bit too high, because, as we shall see, it double counts (or triple counts or worse) the occasions when there are multiple AA/KK pairs in your opponents' 8 hands.
Two Opponents
With 2 opposing hands comprised of 4 cards the first concept to wrestle with is the number of possible combinations. A rookie mistake is to calculate this as c(50,4)=230300, because there are 4 cards chosen from 50. But that is not correct. To illustrate consider these four cards: A♠A♥ Q♦4♣. This 4-card combination will be one of the 230300 combinations that we calculated as c(50,4). But it matters how those 4 cards are arranged. Note the different implications in these arrangements of those 4 cards into two 2-card hands: A♠A♥ + Q♦4♣ versus A♠ 4♣ + A♥ Q♦.
The best way to define the combinations is c(50,2)*c(48,2) which is equal to 1,225*1,128 = 1,381,800. The first hand is any two cards from 50, and the 2nd hand is any two cards from the remaining 48. As a way to check our logic, we note that 1,381,800 is 6 *c(50,4) or 6*230,300. That makes sense because given four random cards, the number of ways to arrange them into two 2-card hands is c(4,2)*c(2,2) =6.
In order to properly analyze the two opposing hands, I explicitly calculate the number of combinations for each of these cases. (Notation: I use XX to mean any two card hand that is NOT AA or KK pair.)
Case#1: Opponents have AA+AA
Case #2: Opponents have KK+KK
Case #3: Opponents have AA+KK
Case #4 Opponents have AA+XX
Case #5 Opponents have KK+XX
Case#1: Opponents have AA+AA
We saw before that the number of combinations for the 1st AA pair was c(4,2)=6. Given that the first pair has already been dealt, the number of combinations for the 2nd AA pair is 1, or c(2,2). So, I will define C1 as the total number of combinations for AA+AA and C1 = c(4,2)*(c2,2)=6. The probability of this case, P1, =6/1381800.
Case#2: Opponents have KK+KK This is identical to the AA+AA case. C2= 6 combinations, and the probability P2 = 6/1381800.
Case#3: Opponents have AA+KK The number of combinations for the AA pair will be 6, and The number of combinations for the KK pair also will be 6. The total combinations for AA+KK, defined as C3, is 2*6*6. The factor of 2 arises from the fact that the hands can be dealt as first AA then KK or in the reverse order. I will formalize that a bit more when we get to 8 hands. So C3= 72 and P3= 72/1381800.
Case#4: Opponents have AA+XX We start out by noting that the number of combinations for the AA+XX case, C4, will be C4 =2*c(4,2)*c(48,2) where 2 accounts for the combinations of which player has which hand , c(4,2) for the AA pair and c(48,2) is for the XX hand. But c(48,2) is not strictly correct, because allowing the XX hand to be any 2 cards of the remaining 48 means that we are allowing it to be the remaining AA pair or a KK pair and we have already defined AA+AA to be case#1 and AA+KK to be case #3. There are several ways to correct for this, but the way I prefer to write the algorithm for C4 is is:
This equation takes advantage of the work we have already done in defining the previous cases. Plugging the numerical values into the above equation, we get C4 = 2*6 *1128 – 6 – 72 or C4 =13458. Thus the probability of AA + XX, (P4) = 13458/1,381,800 = 0.9739%.
Case#5: Opponents have KK+XX This case is analogous to the previous case. The equation for calculating the number of combinations for this case is:
Note that here we subtract C2 rather than subtracting C1 as we did in the prior case. C5=13458 and P5 = 0.9739%.
Adding together all the combinations for all five cases, we get 6+6+72+13458+13458 = 27000. Thus the probability of at least one of your 2 opponents having either an AA or KK pair is 27000/1381800 or 1.9544%. Note that simply doubling the probability of one opposing hand having AA or KK would yield a probability of 2*0.9796% = 1.9592%. The actual value of 1.9544% is slightly lower because we have avoiding double counting the cases where both opponents have either AA or KK pair.
I’ll write up the case of 8 opponents in a separate post. It extends the methodology introduced in the 2 opponent case, but I need to be more formal about a few factors that I glossed over in these simpler cases.
How frequently will “n” opponents have AA or KK pair, when you hold QQ pair? To show you the logic, I will explain how to calculate this for n=1 (1 opponent), n=2 and then n=8. Please have patience and bear with me.
One Opponent
Notation: I will use c(x,y) to mean combin(x,y).
We have 2 queens, so there are 50 cards remaining. We have one opponent with a 2 card hand.
The total number of two-card combinations that the opponent can have is c(50,2) = 1225.
AA Pair. The number of AA pair combinations that the opponent can have is c(4,2)=6. There are 4 aces, and you are choosing 2 of 4. The 6 combinations look like this A♠A♥, A♠A♦, A♠A♣, A♥A♦, A♥A♣, A♦A♣,
Obviously, the probability of the opponent having an AA pair is c(4,2)/c(50,2)= 6/1225.
KK Pair. The situation with KK pair is identical to AA pair, that is, there are 4 cards within the rank from which to make a pair. So KK pair also has 6 combinations and a frequency of occurrence = 6/1225.
AA Pair OR KK Pair Nothing could be easier. The total number of combinations in your opponent's hand that are either AA or KK are 12, i.e., 6+6.
Probability of either AA or KK pair (1 opponent): 12/1225 = 0.009796. Slightly less than one percent.
Given 8 opposing hands, a rough estimate of the probability of AA or KK pair appearing would be 8 * 0.9796% = 7.8367%, or about 8%. Honestly, that’s how I would estimate it when I’m playing poker. But that is not rigorously correct and a bit too high, because, as we shall see, it double counts (or triple counts or worse) the occasions when there are multiple AA/KK pairs in your opponents' 8 hands.
Two Opponents
With 2 opposing hands comprised of 4 cards the first concept to wrestle with is the number of possible combinations. A rookie mistake is to calculate this as c(50,4)=230300, because there are 4 cards chosen from 50. But that is not correct. To illustrate consider these four cards: A♠A♥ Q♦4♣. This 4-card combination will be one of the 230300 combinations that we calculated as c(50,4). But it matters how those 4 cards are arranged. Note the different implications in these arrangements of those 4 cards into two 2-card hands: A♠A♥ + Q♦4♣ versus A♠ 4♣ + A♥ Q♦.
The best way to define the combinations is c(50,2)*c(48,2) which is equal to 1,225*1,128 = 1,381,800. The first hand is any two cards from 50, and the 2nd hand is any two cards from the remaining 48. As a way to check our logic, we note that 1,381,800 is 6 *c(50,4) or 6*230,300. That makes sense because given four random cards, the number of ways to arrange them into two 2-card hands is c(4,2)*c(2,2) =6.
In order to properly analyze the two opposing hands, I explicitly calculate the number of combinations for each of these cases. (Notation: I use XX to mean any two card hand that is NOT AA or KK pair.)
Case#1: Opponents have AA+AA
Case #2: Opponents have KK+KK
Case #3: Opponents have AA+KK
Case #4 Opponents have AA+XX
Case #5 Opponents have KK+XX
Case#1: Opponents have AA+AA
We saw before that the number of combinations for the 1st AA pair was c(4,2)=6. Given that the first pair has already been dealt, the number of combinations for the 2nd AA pair is 1, or c(2,2). So, I will define C1 as the total number of combinations for AA+AA and C1 = c(4,2)*(c2,2)=6. The probability of this case, P1, =6/1381800.
Case#2: Opponents have KK+KK This is identical to the AA+AA case. C2= 6 combinations, and the probability P2 = 6/1381800.
Case#3: Opponents have AA+KK The number of combinations for the AA pair will be 6, and The number of combinations for the KK pair also will be 6. The total combinations for AA+KK, defined as C3, is 2*6*6. The factor of 2 arises from the fact that the hands can be dealt as first AA then KK or in the reverse order. I will formalize that a bit more when we get to 8 hands. So C3= 72 and P3= 72/1381800.
Case#4: Opponents have AA+XX We start out by noting that the number of combinations for the AA+XX case, C4, will be C4 =2*c(4,2)*c(48,2) where 2 accounts for the combinations of which player has which hand , c(4,2) for the AA pair and c(48,2) is for the XX hand. But c(48,2) is not strictly correct, because allowing the XX hand to be any 2 cards of the remaining 48 means that we are allowing it to be the remaining AA pair or a KK pair and we have already defined AA+AA to be case#1 and AA+KK to be case #3. There are several ways to correct for this, but the way I prefer to write the algorithm for C4 is is:
C4=2*c(4,2)*c(48,2) - C1 - C3
This equation takes advantage of the work we have already done in defining the previous cases. Plugging the numerical values into the above equation, we get C4 = 2*6 *1128 – 6 – 72 or C4 =13458. Thus the probability of AA + XX, (P4) = 13458/1,381,800 = 0.9739%.
Case#5: Opponents have KK+XX This case is analogous to the previous case. The equation for calculating the number of combinations for this case is:
C5=2*c(4,2)*c(48,2) – C2 - C3
Note that here we subtract C2 rather than subtracting C1 as we did in the prior case. C5=13458 and P5 = 0.9739%.
Adding together all the combinations for all five cases, we get 6+6+72+13458+13458 = 27000. Thus the probability of at least one of your 2 opponents having either an AA or KK pair is 27000/1381800 or 1.9544%. Note that simply doubling the probability of one opposing hand having AA or KK would yield a probability of 2*0.9796% = 1.9592%. The actual value of 1.9544% is slightly lower because we have avoiding double counting the cases where both opponents have either AA or KK pair.
I’ll write up the case of 8 opponents in a separate post. It extends the methodology introduced in the 2 opponent case, but I need to be more formal about a few factors that I glossed over in these simpler cases.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
November 11th, 2017 at 1:52:28 PM
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Thank you very much. I had worked some of those partially but was getting lost somewhere. I knew the answer I was getting couldn't be correct. you make it the thought process very clear. It is a pleasure to see how clearly you write. It is easy to find people who know a subject and it is easy to find people who can write well but persons who know the answers and can write well are much rarer.
I will work through your examples and see if I can extend the logic before I read your next post arrives
I will work through your examples and see if I can extend the logic before I read your next post arrives
November 12th, 2017 at 10:41:41 AM
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Thank you for your kind words.
Eight Opponents
How frequently will 8 opponents have AA or KK pair, when you hold QQ pair?
First, me introduce the term “High Pair” to indicate AA or KK.
Now, lets calculate the total number of combinations with 8 opponents, each having two cards:
Total = c(50,2) * c(48,2) * c(46,2) * c(44,2) * c(42,2) * c(40,2) * c(38,2) * c(36,2)
That will be a very large number and Excel only calculates large numbers to 15 digits. What I do is define:
B = c(42,2)*c(40,2) * c(38,2)*c(36,2),
thereby allowing me to define:
Total = c(50,2)*c(48,2) * c(46,2) * c(44,2) * B
Total = 1,352,934,198,000 * B.
This works well in this problem because when we calculate probabilities, every numerator and denominator will have B as a multiplicative factor, so B will cancel out and we won’t need to actually multiply anything by it.
Cases to Analyze
It is a funny aspect of combination math that we cannot directly calculate the probability that “nothing bad happens.” We can only calculate the probability of every bad scenario we can think of and then subtract from one.
Case#1: AA+AA+KK+KK+4*XX
Case#2: AA+AA+KK+5*XX
Case#3: AA+KK+KK+5*XX
Case#4: AA+AA+6*XX
Case#5: KK+KK+6*XX
Case#6: AA+KK+6*XX
Case#7: AA+7*XX
Case#8: KK+7*XX
As before, we need to evaluate all these cases so we avoid “double-counting” the cases with multiple high pairs.
There is a multiplicative factor in the combinations formula that we will need to use to account for the number of arrangements of the high pairs and "non-high pair hands" amongst the 8 opponents. in each case. This factor is:
c(k,m)*c(m,n)
k = total number of opponents' hands
m= number of high pair hands
n = number of AA pairs
Note that this algebraic expression has the same value if we instead define n as n = number of KK pairs. Now let’s write the equations for the number of combinations for each 8-hand case.
Case#1: AA+AA+KK+KK+4*XX
C1 = c(8,4)*c(4,2)*(6*1+6*1)*B
C1 = 5040*B P1 = 3.72324 E-09
Case#2: AA+AA+KK+5*XX
C2 = c(8,3)*c(3,2)*(6*1+6)*c(44,2)*B - C1
C2 = 1,902,096*B P2 = 1.4059 E-06
Case#3: AA+KK+KK+5*XX
C3 = c(8,3)*c(3,1)*(6+6*1)*c(44,2)*B - C1
C3 = 1,902,096*B P3 = 1.4059 E-06
Case#4: AA+AA+6*XX
C4 = c(8,2)*c(2,2)*(6*1+0)*c(46,2)*c(44,2)*B - C1 - C2
C4 = 162,583,344*B P4 = 0.000120171
Case#5: KK+KK+6*XX
C5 = c(8,2)*c(2,0)*(0+6*1)*c(46,2)*c(44,2)*B - C1 - C3
C5 = 164,490,480*B P5 = 0.000120171
Case#6: AA+KK+6*XX
C6 = c(8,2)*c(2,1)*(6+6)*c(46,2)*c(44,2)*B - C1 - C2 - C3
C6 = 654,152,688*B P6 = 0.000483507
Case#7: AA+7*XX
C7 = c(8,1)*c(1,1)*6*c(48,2)*c(46,2)*c(44,2)*B - C1 - C2 - C3 - C4 - C6
C7 = 52,192,386,576*B P7 = 0.03857718
Case#8: KK+7*XX
C8 = c(8,1)*c(1,0)*6*c(48,2)*c(46,2)*c(44,2)*B - C1 - C2 - C3 - C5 - C6
C8 = 52,192,386,576*B P8 = 0.03857718
So, if you add up the combinations for all the cases and divide by the total combinations, you should get the probability value for "one or more high pairs (AA or KK) in 8 opponent hands (given QQ)." Right now I calculate the value 0.077881025 -which is different than the value I quoted in my 1st post in this thread. Its easy to make errors in these kind of calculations, and it seems I have made an error somewhere. Perhaps someone can check me.
Eight Opponents
How frequently will 8 opponents have AA or KK pair, when you hold QQ pair?
First, me introduce the term “High Pair” to indicate AA or KK.
Now, lets calculate the total number of combinations with 8 opponents, each having two cards:
Total = c(50,2) * c(48,2) * c(46,2) * c(44,2) * c(42,2) * c(40,2) * c(38,2) * c(36,2)
That will be a very large number and Excel only calculates large numbers to 15 digits. What I do is define:
B = c(42,2)*c(40,2) * c(38,2)*c(36,2),
thereby allowing me to define:
Total = c(50,2)*c(48,2) * c(46,2) * c(44,2) * B
Total = 1,352,934,198,000 * B.
This works well in this problem because when we calculate probabilities, every numerator and denominator will have B as a multiplicative factor, so B will cancel out and we won’t need to actually multiply anything by it.
Cases to Analyze
It is a funny aspect of combination math that we cannot directly calculate the probability that “nothing bad happens.” We can only calculate the probability of every bad scenario we can think of and then subtract from one.
Case#1: AA+AA+KK+KK+4*XX
Case#2: AA+AA+KK+5*XX
Case#3: AA+KK+KK+5*XX
Case#4: AA+AA+6*XX
Case#5: KK+KK+6*XX
Case#6: AA+KK+6*XX
Case#7: AA+7*XX
Case#8: KK+7*XX
As before, we need to evaluate all these cases so we avoid “double-counting” the cases with multiple high pairs.
There is a multiplicative factor in the combinations formula that we will need to use to account for the number of arrangements of the high pairs and "non-high pair hands" amongst the 8 opponents. in each case. This factor is:
c(k,m)*c(m,n)
k = total number of opponents' hands
m= number of high pair hands
n = number of AA pairs
Note that this algebraic expression has the same value if we instead define n as n = number of KK pairs. Now let’s write the equations for the number of combinations for each 8-hand case.
Case#1: AA+AA+KK+KK+4*XX
C1 = c(8,4)*c(4,2)*(6*1+6*1)*B
C1 = 5040*B P1 = 3.72324 E-09
Case#2: AA+AA+KK+5*XX
C2 = c(8,3)*c(3,2)*(6*1+6)*c(44,2)*B - C1
C2 = 1,902,096*B P2 = 1.4059 E-06
Case#3: AA+KK+KK+5*XX
C3 = c(8,3)*c(3,1)*(6+6*1)*c(44,2)*B - C1
C3 = 1,902,096*B P3 = 1.4059 E-06
Case#4: AA+AA+6*XX
C4 = c(8,2)*c(2,2)*(6*1+0)*c(46,2)*c(44,2)*B - C1 - C2
C4 = 162,583,344*B P4 = 0.000120171
Case#5: KK+KK+6*XX
C5 = c(8,2)*c(2,0)*(0+6*1)*c(46,2)*c(44,2)*B - C1 - C3
C5 = 164,490,480*B P5 = 0.000120171
Case#6: AA+KK+6*XX
C6 = c(8,2)*c(2,1)*(6+6)*c(46,2)*c(44,2)*B - C1 - C2 - C3
C6 = 654,152,688*B P6 = 0.000483507
Case#7: AA+7*XX
C7 = c(8,1)*c(1,1)*6*c(48,2)*c(46,2)*c(44,2)*B - C1 - C2 - C3 - C4 - C6
C7 = 52,192,386,576*B P7 = 0.03857718
Case#8: KK+7*XX
C8 = c(8,1)*c(1,0)*6*c(48,2)*c(46,2)*c(44,2)*B - C1 - C2 - C3 - C5 - C6
C8 = 52,192,386,576*B P8 = 0.03857718
So, if you add up the combinations for all the cases and divide by the total combinations, you should get the probability value for "one or more high pairs (AA or KK) in 8 opponent hands (given QQ)." Right now I calculate the value 0.077881025 -which is different than the value I quoted in my 1st post in this thread. Its easy to make errors in these kind of calculations, and it seems I have made an error somewhere. Perhaps someone can check me.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
November 12th, 2017 at 7:05:26 PM
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gordon,
Outstanding set of posts. Thanks for taking the time.
Outstanding set of posts. Thanks for taking the time.
If the House lost every hand, they wouldn't deal the game.
November 18th, 2017 at 12:50:28 PM
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You are an amazing resource.
I was slow to respond because I was in the hospital (not serious but required Inpatient s.tatus. I will work carefully through al of this and expect to have a much deeper understanding of combinatorial probabilities when I am through. (Unless I bog down = but you have provided an exceptional guide.
Many thanks. I'll let you know hoe I make out.
I was slow to respond because I was in the hospital (not serious but required Inpatient s.tatus. I will work carefully through al of this and expect to have a much deeper understanding of combinatorial probabilities when I am through. (Unless I bog down = but you have provided an exceptional guide.
Many thanks. I'll let you know hoe I make out.
