- #1
Democritus Blantayre
- 3
- 0
So here's the basic situation:
1. I have combined two Poker decks, with a total of 104 cards
2. There's no limit to the size of the hand (using Poker-style hands). Thus for example one could have a Straight that is of 6,7, or more cards in a sequence.
3. There's "wraparound", as in: a Straight could run from Q-K-A-2-3.
Now, my method of combinatorical calculation extrapolated from existing and proven formulas of calculating probability from Tom Ramsey of the University of Hawaii, available here. His formula for a Straight is:
With wraparound, 10 in the formula becomes 13. In my situation there are two sets of 13 cards (2,3,4...K,A) in each suit, instead of one in conventional Poker, so I accommodate my formula by simply changing the 4C1 (4 choose 1) to 8C1. So I suppose my immediate question would be if this method is legit or not - it seems so to me (just treating the Spades for instance as Spades type 1 or Spades type 2). My formula is thus: 13 * (8C1)^5, for a Straight of size 5 (a normal Poker Straight). I calculate this number as the total # of Straights, but then comes the part where I need to subtract the # of Straight Flushes and Royal Flushes...
It's easy to count the # of Royal Flushes in conventional Poker: there's only 4 possibilities. But when you combine 2 Poker decks, the number of these possibilities increases enormously. This because for every card in the sequence of 5, you would need to choose between say one 10 of Spades, or the other 10 of Spades. The simple combinatorics formula is (2C1). Since you need to do this with each of the 5 cards in the Royal Flush, thus the formula is (2C1)^5. This is appended to the conventional Royal Flush Formula of (4C1)[choosing which of the four suits will be used] * (2C1)^5 = 128 Royal Flush possibilities. The Straight Flush formula (which includes Royal Flushes) is easy: just append (13C1) to account for all the different positions where the Straight begins. Thus the formula is: (4C1) * (13C1) * (2C1)^5 = 1,664, which is the number of Straight & Royal Flushes that I subtract from the Straight Formula. The denominator is simple: 104C5 (91,962,520). The total # of Straights is 13 * (8C1)^5 - ((4C1) * (13C1) * (2C1)^5) = 424,320. The probability of drawing it if you draw 5 cards at random from the well shuffled double-deck is thus: 424,320/91,962,520 = .004614053638.
That's all fine and kosher, so following this method, I then calculated the probabilities of getting Straights of size 6,7,8,9, and 10, by simply adjusting the exponents in the formula to have numbers corresponding to the draw size (ie: it would for instance be (8C1)^6 instead of (8C1)^5, and likewise for the (2C1)^5 element of the subtracting Straight & Royal Flushes element of the formula). Of course, I would adjust the denominator to be 104Cn for whatever size was the size of the Straight hand. As I was calculating the probabilities, I was however surprised to discover that the difference in probabilities between Straights 10 and 11 dropped to almost nothing, that the probabilities didn't diminish by a continuous multiplier, but steadily became less and less with an increase in the draw size, and (horror): that the probabilities of getting a Straight 12 on the first draw actually INCREASED! This latter event is highly anomalous, which leads me to suspect that there's something wrong with my formula, my combinatorical reasoning, or both. So what's going on here, to account for this strangeness?
FYI: I calculated the following probabilities for Straights size 5-12 using the above method:
1. I have combined two Poker decks, with a total of 104 cards
2. There's no limit to the size of the hand (using Poker-style hands). Thus for example one could have a Straight that is of 6,7, or more cards in a sequence.
3. There's "wraparound", as in: a Straight could run from Q-K-A-2-3.
Now, my method of combinatorical calculation extrapolated from existing and proven formulas of calculating probability from Tom Ramsey of the University of Hawaii, available here. His formula for a Straight is:
"This is five cards in a sequence (e.g., 4,5,6,7,8), with aces allowed to be either 1 or 13 (low or high) and with the cards allowed to be of the same suit (e.g., all hearts) or from some different suits. The number of such hands is 10*[4-choose-1]^5. The probability is 0.003940. IF YOU MEAN TO EXCLUDE STRAIGHT FLUSHES AND ROYAL FLUSHES (SEE BELOW), the number of such hands is 10*[4-choose-1]^5 - 36 - 4 = 10200, with probability 0.00392465"
With wraparound, 10 in the formula becomes 13. In my situation there are two sets of 13 cards (2,3,4...K,A) in each suit, instead of one in conventional Poker, so I accommodate my formula by simply changing the 4C1 (4 choose 1) to 8C1. So I suppose my immediate question would be if this method is legit or not - it seems so to me (just treating the Spades for instance as Spades type 1 or Spades type 2). My formula is thus: 13 * (8C1)^5, for a Straight of size 5 (a normal Poker Straight). I calculate this number as the total # of Straights, but then comes the part where I need to subtract the # of Straight Flushes and Royal Flushes...
It's easy to count the # of Royal Flushes in conventional Poker: there's only 4 possibilities. But when you combine 2 Poker decks, the number of these possibilities increases enormously. This because for every card in the sequence of 5, you would need to choose between say one 10 of Spades, or the other 10 of Spades. The simple combinatorics formula is (2C1). Since you need to do this with each of the 5 cards in the Royal Flush, thus the formula is (2C1)^5. This is appended to the conventional Royal Flush Formula of (4C1)[choosing which of the four suits will be used] * (2C1)^5 = 128 Royal Flush possibilities. The Straight Flush formula (which includes Royal Flushes) is easy: just append (13C1) to account for all the different positions where the Straight begins. Thus the formula is: (4C1) * (13C1) * (2C1)^5 = 1,664, which is the number of Straight & Royal Flushes that I subtract from the Straight Formula. The denominator is simple: 104C5 (91,962,520). The total # of Straights is 13 * (8C1)^5 - ((4C1) * (13C1) * (2C1)^5) = 424,320. The probability of drawing it if you draw 5 cards at random from the well shuffled double-deck is thus: 424,320/91,962,520 = .004614053638.
That's all fine and kosher, so following this method, I then calculated the probabilities of getting Straights of size 6,7,8,9, and 10, by simply adjusting the exponents in the formula to have numbers corresponding to the draw size (ie: it would for instance be (8C1)^6 instead of (8C1)^5, and likewise for the (2C1)^5 element of the subtracting Straight & Royal Flushes element of the formula). Of course, I would adjust the denominator to be 104Cn for whatever size was the size of the Straight hand. As I was calculating the probabilities, I was however surprised to discover that the difference in probabilities between Straights 10 and 11 dropped to almost nothing, that the probabilities didn't diminish by a continuous multiplier, but steadily became less and less with an increase in the draw size, and (horror): that the probabilities of getting a Straight 12 on the first draw actually INCREASED! This latter event is highly anomalous, which leads me to suspect that there's something wrong with my formula, my combinatorical reasoning, or both. So what's going on here, to account for this strangeness?
FYI: I calculated the following probabilities for Straights size 5-12 using the above method:
- Straight 5: 0.00461405363
- Straight 6: 0.00224369667
- Straight 7: 0.00128305234
- Straight 8: 0.00084670504
- Straight 9: 0.00070504325
- Straight 10: 0.00053479168
- Straight 11: 0.00050065747
- Straight 12: 0.00051680808