In dire Straights: Straights of 11+ w/ 2 Poker decks

  • #1
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:

"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
 

Answers and Replies

  • #2
pbuk
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A few observations:

Welcome to PhysicsForums!

What is the source of this problem - if it is a study assignment there is a different sub-forum for that?

It would be easier to read your post if you used LaTeX e.g. ## 13 \times {8 \choose 1} ^ 5 - {4 \choose 1} \times {13 \choose 1} \times {2 \choose 1} ^ 5 = 424,320 ##

As there is a relatively small number of possible hands you could check the results computationally.

I think it is easier to read results as the number of possible hands rather than a decimal approximation.

I think you should look again at the number of Straight Flushes - I don't think you are properly taking account of the fact that there are 2 of each card. Perhaps calculate the number of these separately.
 
  • #3
Firstly: this is not a homework assignment.

I don't see a problem with my method of calculating the # of Straight Flushes. Though I humbly admit that I may not be seeing a problem that's actually there.

Calculating the # of Straight Flushes in a standard (1) Poker deck is easy enough: ## \binom 4 1## to choose the suit, X ##\binom {13} {1}## to choose the starting position of any of the 13 cards in a suit with wraparound, which gives 52 (in normal Poker without wraparound, the number is 40). This is inclusive of Royal Flushes.

Since however there are 2 of each card, because there are 2 decks, it is possible to get a Royal Flush (one particular type of Straight Flush, which I'll use as an example) in many more ways than if there was just one deck (if there was just one deck, there would be only one way for each suit to get a Royal Flush). Thus one could get a Royal Flush in Spades if you had them all be of Spades from one deck (10S##_1##, JackS##_1##, etc), from the other deck's Spades (10S##_2##, JackS##_2##, etc), or any combination thereof (eg: 10S##_1##, JackS##_2##, etc). So the rule of using ## {\binom 2 1}^5## to determine which of the two cards will be used for each element in the Straight combination appears sound. If I am wrong about this, please overtly come out and tell me that I am, and how so, please.

Using my method, I calculate the number of Straight Flushes (including Royal Flushes) in my double-deck to be: ## {\binom 4 1}*{\binom {13} {1}} *{\binom 2 1}^5 = 1,664##. Am I wrong or right in this calculation?

Am I wrong in any wise regarding my combinatorial methodology altogether? I suspect I am, but don't know positively that I am.

Is it also possible that I'm not at all wrong in my methodology, but there is rather some mysterious Wonderland-esque peculiarity whereby it really is more probable that you'll get a Straight of 12 cards from a random draw of 12 cards, than a Straight of 11 from a random draw of 11? If so, could someone explain why such is the case, if such is the case?
 
  • #4
pbuk
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Firstly: this is not a homework assignment.
Is it a puzzle or something? As you are asking for others to contribute their time it is reasonable to explain why.

So the rule of using ## {\binom 2 1}^5## to determine which of the two cards will be used for each element in the Straight combination appears sound. If I am wrong about this, please overtly come out and tell me that I am, and how so, please.
I don't know that you are wrong, it just looked like the most likely source of error and it was YOU that suspected (and still suspect) an error! Looking at it again (the clearer explanation and formatting helps), I agree it is sound.

Is it also possible that I'm not at all wrong in my methodology, but there is rather some mysterious Wonderland-esque peculiarity whereby it really is more probable that you'll get a Straight of 12 cards from a random draw of 12 cards, than a Straight of 11 from a random draw of 11? If so, could someone explain why such is the case, if such is the case?
But you are not calculating the probability that you are getting a straight, you are calculating the probability that you are getting a straight that is not a flush, that is why I suggested you take that complication out of the equation. I have done this, and I realise that the difference is immaterial as straight flushes are so rare. I also realise that whilst a computational check is feasible for the two-pack 5 card draw, for the 12 card draw there are more than 1.7 x 1015 possibilities so that is not going to help.

This result does seem counter-intuitive, but I can't find anything wrong with your calculations. It occurred to me that if you already have an 11 Straight (e.g. A, 2 .. J) then drawing as your 12th card any rank card that you have not already drawn (Q or K) completes a 12 Straight, whereas the same does not apply going from a 10-Straight to an 11-Straight, but that doesn't seem to help explain anything.
 
  • #5
As you wish to know more specifically what this is for Mr. pbuk (and others may also wish to know), it is for a game I'm designing that combines Chess, Poker, and Go. I call it Chepogo. But in order for this game to work, I have to come up with a ranked list of what one might call "hyper-Poker" hands, which depends upon their probability of being drawn. If you or anyone wishes to know more, I can share any information I have, or perhaps the full ranked list of all the hyper-Poker hands, when I do get this completed. I was only not discussing such things because I thought there was no interest, and was only trying to reassure people that I'm not some student trying to get others to do his homework for him.


This result does seem counter-intuitive, but I can't find anything wrong with your calculations. It occurred to me that if you already have an 11 Straight (e.g. A, 2 .. J) then drawing as your 12th card any rank card that you have not already drawn (Q or K) completes a 12 Straight, whereas the same does not apply going from a 10-Straight to an 11-Straight, but that doesn't seem to help explain anything.
The same thought had occurred to me. Perhaps there's some vague outline of an explanation that I can barely discern viz: the probability of drawing from the pool of cards not already within the Straight that builds upon the Straight goes from being a possibility in a Straight 10 to 11 to an inevitability in Straight 11 to 12, which may be enough to tip the overall decimal probability to become counterintuitively Wonderland-esque. Since my statement is I think difficult to understand (forgiveness, please), let me explicate in this fashion: with a Straight 10 and another card drawn, there is of course the possibility (indeed the likelihood) that the new card will not be any of the 2-A series of 13 cards which are yet undrawn in the already built Straight. Let's say we already have a Straight 10 that runs from 2 to the J. Any new card drawn will most likely be in that range, but there's those 3 other sorts of cards: the Q, K, and A. If we keep drawing from the deck, inevitably one of those three cards will be drawn. The Q or the A will add to the Straight and make it become a Straight 11, but a drawn K will not. So there's a 2/3 chance that from that Q-K-A pool of cards not in the Straight, that a Straight 11 will be built up. Interestingly, as we proceed to have Straights that are lower and lower, the likelihood that the pool of cards not within the Straight that will add to the Straight goes lower and lower. If we for example have a normal 5-card Straight of say 4-5-6-7-8, of all the 8 cards not within the Straight, only 2 of them will add to the Straight (a 3 or a 9), thus there's a 1/4 chance. And as the likelihood of building up the Straight goes lower and lower with diminished hand size, the rarity of a Straight increases, which is to say decreases its probability. This tendency operates contrary to the general tendency of a Straight or any hand to become more rare as its size increases, and is moreover obscured & overruled methinks by that more powerful general tendency. However, with a Straight 11, any of the two remaining cards of the undrawn pool, as you observed, must add to the Straight, which diminishes its rarity...perhaps enough to behold that previously obscured reverse of the general tendency.

I'm not sure about how to design some combinatorical test of this hypothesis - it appears to involve some more advanced brainiac combinatorics that are at present out of my league. If anyone has ideas, I'd welcome them.

Another thought occurs to me. I don't know how to graph such things, but I've noticed when plugging in a bunch of numbers from ## \binom {104} {2} ## thru ## \binom {104} {20} ##, that there isn't a consistent change of committee size from one k value to the next. Further recalling how there's symmetry in k and n-k values, I'm coming under the suspicion that a graphing of a committee size across the range of k values will be a normal (bell-shaped) curve. This is just a "MWAG" (Mathematics Wild-Ass Guess) at present though, and somewhat tangential to the issue at hand, save in this wise (wish I knew how to graph stuff to show you): the Wonderland-esque phenomenon could be accounted for by how the (punnish) straight-line growth of Straights (in the numerator of probabilistic calculation) is outpacing temporarily the growth rate, if not the total numerical size, of all possible outcomes (in the numerator). That growth rate, if along a normal curve's sigmoidal tail-end, would be temporarily outpaced by the growth-rate of Straights, perhaps right where there's those "dire Straights" nos 11 & 12.
 
  • #6
pbuk
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Another thought occurs to me. I don't know how to graph such things, but I've noticed when plugging in a bunch of numbers from ## \binom {104} {2} ## thru ## \binom {104} {20} ##, that there isn't a consistent change of committee size from one k value to the next. Further recalling how there's symmetry in k and n-k values, I'm coming under the suspicion that a graphing of a committee size across the range of k values will be a normal (bell-shaped) curve.
I am not sure what you mean by "committee" here, but yes the number of combinations is given by the binomial coefficients, which in the limit approach the normal distribution.

I wouldn't worry about it too much, the maths seems right and I have verified the numbers computationally for "wrapping" straights in a double pack with hands of 5 and 6 cards.
 
  • #7
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If you've drawn 11 cards, and there were no duplicate ranks, there's a one in 6 probability that you have a straight. (there are 12 possibilities for the second rank you are missing, only 2 of them give a straight.)
There are 93 possible cards left in the deck, 16 of them will give a straight if you draw them as a twelfth card The probability that you have a straight actually increases if you draw a twelfth card. 16 / 93 > 1 / 6
 
  • #8
pbuk
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If you've drawn 11 cards, and there were no duplicate ranks, there's a one in 6 probability that you have a straight. (there are 12 possibilities for the second rank you are missing, only 2 of them give a straight.)
There are 93 possible cards left in the deck, 16 of them will give a straight if you draw them as a twelfth card The probability that you have a straight actually increases if you draw a twelfth card. 16 / 93 > 1 / 6
How is that different from the position with 10 and 11 cards? 16 / 94 is also greater than 1 / 6.
 

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