Aeroelasticity: Difference between revisions

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Aeroelasticity: Difference between revisions

Revision as of 10:12, 27 February 2014

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C2twmM Brazil 6s csts6XQV

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@@ Line 1: / Line 1: @@
-{{dablink|See also [[Poker probability (Texas hold 'em)]] and [[Poker probability (Omaha)]] for probabilities specific to those games.}}
+== dHsXVV Jordan 6 Brazil HYQhzmNU ==
-{{Unreferenced|date=April 2008}}
-In [[poker]], the [[probability]] of each type of 5-card [[poker hand|hand]] can be computed by calculating the proportion of hands of that type among all possible hands.
+Top compound AD02 performed very well in stabilization of cognitive functions in a dose dependent manner. 24 31 % of treated patients showed stabilization of clinical progression. And in the group of kids with the most intellectual impairments   IQs less than 34   there were just as many girls as [http://www.srmcolleges.in/indexss.asp??p=15658-buy-cheap-High-OG-Family-Forever-1s-running-online-2014.html buy cheap High OG Family Forever 1s running online 2014] boys. Found the same [http://tsfdirect.com/contactss.php??p=3277-jordan-18-OG-white-varsity-red-sneaker-store-2014.html jordan 18 OG white varsity red sneaker store 2014] trend but very different numbers. <br><br>There also was a spike in shooting deaths. Fifty nine federal, state and local officers were killed by gunfire in 2010, a 20 percent jump from last year figures, when 49 were killed. "There is lots of debris and fast water   there's no visibility," said Doug McLaughlin, a volunteer firefighter working as a diver. The cars and trucks were submerged in 25 feet of water, he said. <br><br>The 66. Indeed, recent studies indicate that, in addition to or instead of the 26S proteasome, the 20S proteasome could be responsible for the degradation of oxidatively damaged proteins, which occurs in an ATP  and ubiquitin independent manner28. Another issue is the fact that the company's OvaDx and  Disease still require approval for from the FDA. Any regulatory delays could have a significant impact on the company's top line growth. <br><br>Ever since doctors performed the first stem cell transplant in the late 1950s, the so called [http://www.srmcolleges.in/indexss.asp??p=27355-When-Will-gs-ntrl-gry-7-Jordan-Come-Out.html When Will gs ntrl gry 7 Jordan Come Out] blank slate cells have held a promise that seems almost too good to be true: human body cells that can heal and perhaps cure what ails us. In the intervening decades, stem cell transplants have been used on thousands of cancer patients and are being hailed as a potential cure for Parkinson's disease, paralysis, and, in a new twist, aging skin.
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+ </ul>
-==Frequency of 5-card poker hands==
+== h14sgT Jordan 6 World Cup Brazil fsUvm40N ==
-The following chart [[enumeration|enumerates]] the (absolute) [[frequency]] of each hand, given all [[combination]]s of 5 cards [[random]]ly drawn from a full deck of 52 without replacement. [[Wild Card (card games)|Wild cards]] are not considered. In this chart:
-*"Distinct Hands" is the number of different ways to draw the hand, not counting different suits.
+Shauna Jacobson and her husband, retired Detective Malcolm "Kurt" Jacobson. According to the report, prepared by Highway Patrol Trooper Michael Campos, Kurt Jacobson headed out to Lockhart to join a benefit poker run on his bike shortly before noon and made two stops before picking up Shauna at their Dripping Springs home shortly before 3pm. <br><br>Got here six weeks ago, just found out about this today. Returning April 10, so really don have time to order by mail. Things there have changed, not always for the better. In many ways it not as German as it once was, it less distinct. "I know you think things are bad, but they can get better," Briggs said. About 1,500 people have plunged from the bridge, making it one of the world s favorite suicide spots. <br><br>Interactions are particularly difficult with other children because they are less supportive. It is immensely important to provide children with SLI with the additional supports they need.. It's by design. If you encourage it, you end up with a population that is weak and afraid and even if 70% want something very badly, you can just say to them, "No. <br><br>The first thing I did when the cop wrote the ticket was demand [http://www.srmcolleges.in/indexss.asp??p=4471-outlet-jordan-7-gs-ntrl-gry-2014.html outlet jordan 7 gs ntrl gry 2014] a 'change of venue', which generally makes them drive out of their way. The officer in my case actually didn't know what the county seat was, and was not familiar with 40502(b), so I told him to call his watch commander, who [http://www.srmcolleges.in/indexss.asp??p=30403-When-Do-Champagne-6-Release.html When Do Champagne 6 Release] would explain it to him. <br><br>35% negative urine samples during weeks 1 24 for Probuphine and Suboxone, respectively). This data has been both published and presented for peer review. Technically, anything photographed by that camera [http://www.srmcolleges.in/indexss.asp??p=9613-cheap-bordeaux-7s-men.html cheap bordeaux 7s men] cannot be used as evidence in a trial. So, it a $20,000 paper weight.
-*"Frequency" is the number of ways to draw the hand, ''including'' the same card values in different suits
+相关的主题文章:
-*The "[[probability]]" of drawing a given hand is calculated by dividing the number of ways of drawing the hand ("Frequency") by the total number of 5-card hands (the [[sample space]]; <math>\, \begin{matrix} {52 \choose 5} = 2,598,960 \end{matrix}</math>). For example, there are 4 different ways to draw a Royal flush (one for each suit), so the probability is {{sfrac|4|2,598,960}}, or about 0.000154%.
+ <ul>
-*The "Cumulative probability" refers to the probability of drawing a hand as good as ''or better than'' the specified one. For example, the odds of drawing ''three of a kind'' are approximately 2.11%, while the odds of drawing a hand ''at least'' as good as three of a kind are about 2.87%. The cumulative probability is determined by adding one hand's probability with the probabilities of all hands above it.
-*The [[odds]] are defined as the ratio of the number of ways ''not'' to draw the hand, to the number of ways to draw it. For instance, with a Royal flush, there are 4 ways to draw one, and 2,598,956 ways to draw something else (2,598,960 - 4), so the odds against drawing a Royal flush are 2,598,956 : 4, or 649,739 : 1. The formula for establishing the odds can also be stated as ''(1/p) - 1 : 1'', where ''p'' is the aforementioned probability.
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-*The values given for "probability", "Cumulative probability", and "odds" are rounded off for simplicity; the "Distinct hands" and "Frequency" values are exact.
+   <li>[http://forums.gardenweb.com/forums/cacti/nph-ind.cgi http://forums.gardenweb.com/forums/cacti/nph-ind.cgi]</li>
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+ </ul>
-The ''[[Binomial coefficient|nCr]]'' function on most scientific calculators can be used to [[#anyfive|calculate hand frequencies]]; entering {{code|nCr}} with {{code|52}} and {{code|5}}, for example, yields <math>\begin{matrix} {52 \choose 5} = 2,598,960 \end{matrix}</math> as above.
+== C2twmM Brazil 6s csts6XQV ==
-{| class="wikitable" style="text-align:center;"
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-|-
+相关的主题文章:
-! Hand !! Distinct Hands !! Frequency !! Probability !! Cumulative probability !! Odds !! Mathematical expression of absolute frequency
+ <ul>
-|-
-| [[Hand rankings#Straight flush|Royal flush]]<br/>
+   <li>[http://swiftfan.com/forum.php?mod=viewthread&tid=57757 http://swiftfan.com/forum.php?mod=viewthread&tid=57757]</li>
-{{card|spade|10|40px}} {{card|spade|J|40px}} {{card|spade|Q|40px}} {{card|spade|K|40px}} {{card|spade|A|40px}}
-| 1
+   <li>[http://tbitsglobal.com/wiki/index.php?title=User:Bazfpnwxnh#qnLsYZ_Brazil_6s_xL7QZjoK http://tbitsglobal.com/wiki/index.php?title=User:Bazfpnwxnh#qnLsYZ_Brazil_6s_xL7QZjoK]</li>
-| 4
-| 0.000154%
+   <li>[http://www.ur-ysera.de/forum/viewtopic.php?f=2&t=84765 http://www.ur-ysera.de/forum/viewtopic.php?f=2&t=84765]</li>
-| 0.000154%
-| 649,739 : 1
+ </ul>
-| <math>{4 \choose 1}</math>
-|-
-| [[Hand rankings#Straight flush|Straight flush]] (not including royal flush)<br/>
-{{card|heart|4|40px}} {{card|heart|5|40px}} {{card|heart|6|40px}} {{card|heart|7|40px}} {{card|heart|8|40px}}
-| 9
-| 36
-| 0.00139%
-| 0.00154%
-| 72,192 : 1
-| <math>{10 \choose 1}{4 \choose 1} - {4\choose 1}</math>
-|-
-| [[Hand rankings#Four of a kind|Four of a kind]]<br/>
-{{card|heart|A|40px}} {{card|diamond|A|40px}} {{card|club|A|40px}} {{card|spade|A|40px}} {{card|diamond|4|40px}}
-| 156
-| 624
-| 0.0240%
-| 0.0256%
-| 4,164 : 1
-| <math>{13 \choose 1}{12 \choose 1}{4 \choose 1}</math>
-|-
-| [[Hand rankings#Full house|Full house]]<br/>
-{{card|heart|8|40px}} {{card|diamond|8|40px}} {{card|club|8|40px}} {{card|heart|K|40px}} {{card|spade|K|40px}}
-| 156
-| 3,744
-| 0.144%
-| 0.17%
-| 693 : 1
-| <math>{13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}</math>
-|-
-| [[Hand rankings#Flush|Flush]] (excluding royal flush and straight flush)<br/>
-{{card|club|10|40px}} {{card|club|4|40px}} {{card|club|Q|40px}} {{card|club|7|40px}} {{card|club|2|40px}}
-| 1,277
-| 5,108
-| 0.197%
-| 0.367%
-| 508 : 1
-| <math>{13 \choose 5}{4 \choose 1} - {10 \choose 1}{4 \choose 1}</math>
-|-
-| [[Hand rankings#Straight|Straight]] (excluding royal flush and straight flush)<br/>
-{{card|club|7|40px}} {{card|heart|8|40px}} {{card|diamond|9|40px}} {{card|heart|10|40px}} {{card|spade|J|40px}}
-| 10
-| 10,200
-| 0.392%
-| 0.76%
-| 254 : 1
-| <math>{10 \choose 1}{4 \choose 1}^5 - {10 \choose 1}{4 \choose 1}</math>
-|-
-| [[Hand rankings#Three of a kind|Three of a kind]]<br/>
-{{card|heart|Q|40px}} {{card|club|Q|40px}} {{card|diamond|Q|40px}} {{card|spade|5|40px}} {{card|diamond|A|40px}}
-| 858
-| 54,912
-| 2.11%
-| 2.87%
-| 46.3 : 1
-| <math>{13 \choose 1}{4 \choose 3}{12 \choose 2}{4 \choose 1}^2</math>
-|-
-| [[Hand rankings#Two pair|Two pair]]<br/>
-{{card|heart|3|40px}} {{card|diamond|3|40px}} {{card|club|6|40px}} {{card|heart|6|40px}} {{card|spade|K|40px}}
-| 858
-| 123,552
-| 4.75%
-| 7.62%
-| 20.0 : 1
-| <math>{13 \choose 2}{4 \choose 2}^2{11 \choose 1}{4 \choose 1}</math>
-|-
-| [[Hand rankings#One pair|One pair]]<br/>
-{{card|heart|5|40px}} {{card|spade|5|40px}} {{card|club|2|40px}} {{card|club|J|40px}} {{card|diamond|A|40px}}
-| 2,860
-| 1,098,240
-| 42.3%
-| 49.9%
-| 1.36 : 1
-| <math>{13 \choose 1}{4 \choose 2}{12 \choose 3}{4 \choose 1}^3</math>
-|-
-| [[Hand rankings#High card|No pair]] / High card<br/>
-{{card|diamond|2|40px}} {{card|spade|5|40px}} {{card|spade|6|40px}} {{card|heart|J|40px}} {{card|club|A|40px}}
-| 1,277
-| 1,302,540
-| 50.1%
-| 100%
-| 0.995 : 1
-| <math>\left[{13 \choose 5} - 10\right]\left[{4 \choose 1}^5 - 4\right]</math>
-|-
-! Total
-! 7,462
-! 2,598,960
-! 100%
-! ---
-! 0 : 1
-! <math>{52 \choose 5}</math>
-|}
-The royal flush is a case of the straight flush. It can be formed 4 ways (one for each suit), giving it a probability of 0.000154% and odds of 649,739 : 1.
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-When ace-low straights and ace-low straight flushes are not counted, the probabilities of each are reduced: straights and straight flushes each become 9/10 as common as they otherwise would be.  The 4 missed straight flushes become flushes and the 1,020 missed straights become no pair.
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-Note that since suits have no relative value in poker, two hands can be considered identical if one hand can be transformed into the other by swapping suits.  For example, the hand '''3♣ 7♣ 8♣ Q♠ A♠''' is identical to '''<span style="color:red;">3♦ 7♦ 8♦ Q♥ A♥</span>''' because replacing all of the clubs in the first hand with diamonds and all of the spades with hearts produces the second hand.  So eliminating identical hands that ignore relative suit values, there are only 134,459 distinct hands.
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-The number of distinct poker hands is even smaller.  For example, '''3♣ 7♣ 8♣ Q♠ A♠''' and '''<span style="color:red;">3♦</span> 7♣ <span style="color:red;">8♦ Q♥ A♥</span>''' are not identical hands when just ignoring suit assignments because one hand has three suits, while the other hand has only two—that difference could affect the relative value of each hand when there are more cards to come.  However, even though the hands are not identical from that perspective, they still form equivalent poker hands because each hand is an '''A-Q-8-7-3''' high card hand.  There are 7,462 distinct poker hands.
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-===Derivation of frequencies of 5-card poker hands===
+== RZL75Q cheap jordans 2014 IcWvi8j4 ==
-The following computations show how the above frequencies for 5-card poker hands were determined. To understand these derivations, the reader should be familiar with the basic properties of the [[binomial coefficient]]s and their interpretation as the number of ways of choosing elements from a given set.  See also: [[sample space]] and [[event (probability theory)]].
-*''Straight flush'' &mdash; Each straight flush is uniquely determined by its highest-ranking card.  These ranks go from '''5''' ('''A-2-3-4-5''') up to '''A''' ('''10-J-Q-K-A''') in each of the 4 suits. Thus, the total number of straight flushes is:
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-*:<math>{10 \choose 1}{4 \choose 1} = 40</math>
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-**''Royal straight flush'' &mdash; A royal straight flush is a subset of all straight flushes in which the ace is the highest card (i.e. '''10-J-Q-K-A''' in any of the four suits). Thus, the total number of royal straight flushes is
+ <ul>
-**:<math>{5 \choose 5}{8 \choose 0}{4 \choose 1} = 4</math>
-**:or simply <math>{4 \choose 1} = 4</math>. ''Note'': this means that the total number of non-Royal straight flushes is 36.
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-*''Four of a kind'' &mdash; Any one of the thirteen ranks can form the four of a kind by selecting all four of the suits in that rank. The final card can have any one of the twelve remaining ranks, and any suit. Thus, the total number of four-of-a-kinds is:
+== O5dIeS cheap jordans MaMwZO7I ==
-*:<math>{13 \choose 1}{4 \choose 4}{12 \choose 1}{4 \choose 1} = 624</math>
-*''Full house'' &mdash; The full house comprises a triple (three of a kind) and a pair.  The triple can be any one of the thirteen ranks, and consists of three of the four suits.  The pair can be any one of the remaining twelve ranks, and consists of two of the four suits.  Thus, the total number of full houses is:
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-*:<math>{13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2} = 3,744</math>
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-*''Flush'' &mdash; The flush contains any five of the thirteen ranks, all of which belong to one of the four suits, minus the 40 straight flushes.  Thus, the total number of flushes is:
-*:<math>{13 \choose 5}{4 \choose 1} - 40 = 5,108</math>
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-*''Straight'' &mdash; The straight consists of any one of the ten possible sequences of five consecutive cards, from '''5-4-3-2-A''' to '''A-K-Q-J-10'''.  Each of these five cards can have any one of the four suits.  Finally, as with the flush, the 40 straight flushes must be excluded, giving:
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-*:<math>{10 \choose 1}{4 \choose 1}^5 - 40 = 10,200</math>
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-*''Three of a kind'' &mdash; Any of the thirteen ranks can form the three of a kind, which can contain any three of the four suits.  The remaining two cards can have any two of the remaining twelve ranks, and each can have any of the four suits.  Thus, the total number of three-of-a-kinds is:
-*:<math>{13 \choose 1}{4 \choose 3}{12 \choose 2}{4 \choose 1}^2 = 54,912</math>
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-*''Two pair'' &mdash; The pairs can have any two of the thirteen ranks, and each pair can have two of the four suits.  The final card can have any one of the eleven remaining ranks, and any suit.  Thus, the total number of two-pairs is:
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-*:<math>{13 \choose 2}{4 \choose 2}^2{11 \choose 1}{4 \choose 1} = 123,552</math>
+  </ul>
-*''Pair'' &mdash; The pair can have any one of the thirteen ranks, and any two of the four suits.  The remaining three cards can have any three of the remaining twelve ranks, and each can have any of the four suits.  Thus, the total number of pair hands is:
-*:<math>{13 \choose 1}{4 \choose 2}{12 \choose 3}{4 \choose 1}^3 = 1,098,240</math>
-*''No pair'' &mdash; A no-pair hand contains five of the thirteen ranks, discounting the ten possible straights, and each card can have any of the four suits, discounting the four possible flushes. Alternatively, a no-pair hand is any hand that does not fall into one of the above categories; that is, the [[complement (set theory)|complement]] of the [[union (set theory)|union]] of all the above hands, where the [[universe (mathematics)|universe]] is any way to choose five out of 52 cards. Thus, the total number of no-pair hands is:
-*:<math>\left[{13 \choose 5} - 10\right]\left[{4 \choose 1}^5 - 4\right] = {52 \choose 5} - 1,296,420 = 1,302,540</math>
-*{{anchor|anyfive}}''Any five card poker hand'' &mdash; The total number of five card hands that can be drawn from a deck of cards is found using a [[Combinatorics|combination]] selecting five cards, in any order where n refers to the number of items that can be selected and r to the sample size; the "!" is the [[factorial]] operator:
-*:<math>{n\choose r} = {{n!} \over {r!(n - r)!}} = {52 \choose 5} = {{52!} \over {5!(52 - 5)!}} = 2,598,960</math>
-==Frequency of 7-card poker hands==
-In some popular variations of poker, a player uses the best five-card poker hand out of seven cards.  The frequencies are calculated in a manner similar to that shown for 5-card hands, except additional complications arise due to the extra two cards in the 7-card poker hand.  The total number of distinct 7-card hands is <math>\begin{matrix} {52 \choose 7} = 133,784,560 \end{matrix}</math>.  It is notable that the probability of a no-pair hand is ''less'' than the probability of a one-pair or two-pair hand.
-The Ace-high straight flush or royal flush is slightly more frequent (4324) than the lower straight flushes (4140 each) because the remaining two cards can have any value; a King-high straight flush, for example, cannot have the Ace of its suit in the hand (as that would make it ace-high instead).
-:{| class="wikitable" style="font-size: 95%;"
-|-
-!Hand         !! Frequency !! Probability !! Cumulative !! Odds
-|-
-|[[Rank of hands (poker)#Royal flush|Royal flush]]
-|align=right|        4,324 ||   0.0032%   ||   0.0032%  ||align=right|  30,939 : 1
-|-
-|[[Rank of hands (poker)#Straight flush|Straight flush]] (excl. royal flush)
-|align=right|       37,260 ||   0.0279%   ||   0.0311%  ||align=right|   3,589.6 : 1
-|-
-|[[Rank of hands (poker)#Four of a kind|Four of a kind]]
-|align=right|      224,848 ||   0.168%    ||   0.199%   ||align=right|     594 : 1
-|-
-|[[Rank of hands (poker)#Full house|Full house]]
-|align=right|    3,473,184 ||   2.60%     ||   2.80%    ||align=right|    37.5 : 1
-|-
-|[[Rank of hands (poker)#Flush|Flush]]
-|align=right|    4,047,644 ||   3.03%     ||   5.82%    ||align=right|    32.1 : 1
-|-
-|[[Rank of hands (poker)#Straight|Straight]]
-|align=right|    6,180,020 ||   4.62%     ||  10.4%     ||align=right|    20.6 : 1
-|-
-|[[Rank of hands (poker)#Three of a kind|Three of a kind]]
-|align=right|    6,461,620 ||   4.83%     ||  15.3%     ||align=right|    19.7 : 1
-|-
-|[[Rank of hands (poker)#Two pair|Two pair]]
-|align=right|   31,433,400 ||  23.5%      ||  38.8%     ||align=right|    3.26 : 1
-|-
-|[[Rank of hands (poker)#One pair|One pair]]
-|align=right|   58,627,800 ||  43.8%      ||  82.6%     ||align=right|    1.28 : 1
-|-
-|[[Rank of hands (poker)#High card|No pair]]
-|align=right|   23,294,460 ||  17.4%      || 100%       ||align=right|    4.74 : 1
-|-
-| '''Total'''
-|align=right|  133,784,560 || 100%        || 100%       ||align=right|       0 : 1
-|}
-(The frequencies given are exact; the probabilities and odds are approximate.)
-Since suits have no relative value in poker, two hands can be considered identical if one hand can be transformed into the other by swapping suits.  Eliminating identical hands that ignore relative suit values leaves 6,009,159 distinct 7-card hands.
-The number of distinct 5-card poker hands that are possible from 7 cards is 4,824.  Perhaps surprisingly, this is fewer than the number of 5-card poker hands from 5 cards because some 5-card hands are impossible with 7 cards (e.g. 7-high).
-===Derivation of frequencies of 7-card poker hands===
-''See ''"[http://www.math.sfu.ca/~alspach/comp20 7-Card Poker Hands]"'' by Brian Alspach for the article on which this explanation is based.''
-The following computations show how the above frequencies for 7-card poker hands were determined.  To understand these derivations, the reader should be familiar with the basic properties of the [[binomial coefficient]]s and their interpretation as the number of ways of choosing elements from a given [[Set (mathematics)|set]].  See also: [[sample space]] and [[event (probability theory)]].
-*''Straight flush'' &mdash; Each straight flush is uniquely determined by its highest-ranking card; these ranks go from '''5''' ('''A-2-3-4-5''') up to '''A''' ('''10-J-Q-K-A''') in each of the 4 suits. For any particular suit where the straight flush is ace-high, the extra 2 cards may be chosen from the remaining 47 cards. In the 9 remaining cases when the straight flush is not ace-high, the extra 2 cards may be chosen from the remaining 47 cards, minus the card in that suit directly above the high-card (which would change the rank of the hand). Thus, the total number of straight flushes is:
-::<math>{4 \choose 1}\left[{1 \choose 1}{47 \choose 2} + {9 \choose 1}{46 \choose 2}\right] = 41,584</math>
-*''Four of a kind'' &mdash; Any 1 of the 13 ranks can form the four of a kind, with the 3 extra cards being chosen from the remaining 48 cards. Thus, the total number of four of a kinds is:
-::<math>{13 \choose 1}{48 \choose 3} = 224,848</math>
-*''Full house'' &mdash; With 7 cards, a full house may be constructed in 1 of 3 ways:
-:'''1 triple, 1 pair and 2 kickers'''
-:The triple may be 1 of 13 ranks, and by definition 3 of the 4 of that rank are chosen. The pair may be 1 of the remaining 12 ranks, and (again, by definition) 2 of the 4 of that rank are chosen.  The ranks of the 2 kickers are chosen from the remaining 11 ranks, and 1 of the 4 of each rank are chosen.  Thus, the total number of full houses in this form is:
-::<math>{13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}{11 \choose 2}{4 \choose 1}^2 = 3,294,720</math>
-:'''1 triple and 2 pairs'''
-:The triple is chosen the same way as before, the ranks of the two pairs are chosen from the remaining 12 ranks, and the 2 of the 4 of each rank are chosen as usual.  Thus, the total number of full houses in this form is:
-::<math>{13 \choose 1}{4 \choose 3}{12 \choose 2}{4 \choose 2}^2 = 123,552</math>
-:'''2 triples and 1 kicker'''
-:The ranks of both triples are chosen from the 13, then the rank of the kicker is chosen from the remaining 11 ranks.  Thus, the total number of full houses in this form is:
-::<math>{13 \choose 2}{4 \choose 3}^2{11 \choose 1}{4 \choose 1} = 54,912</math>
-:Thus, the total number of full houses is:
-::<math>3,294,720 + 123,552 + 54,912 = 3,473,184\,</math>
-*''Flush'' &mdash; A flush may be formed with either 5, 6 or 7 cards in any of the 4 suits.  The flush cards are chosen from the 13 in that suit, and the extra cards (if any) are chosen from the other 3 suits.  The number of straight flushes must then be subtracted from the total.  Thus, the total number of flushes is:
-::<math>{4 \choose 1}\left[{13 \choose 5}{39 \choose 2} + {13 \choose 6}{39 \choose 1} + {13 \choose 7}\right] - 41,584 = 4,047,644</math>
-*''Straight'' &mdash; Significantly more complications arise when working out the frequencies for a 7-card straight due to the possibility of a straight and a flush (though not necessarily a straight flush) being formed simultaneously, and the fact that pairs and triples of cards can appear.  Therefore, the calculations must be broken down into several separate sections:
-:'''7 distinct ranks'''
-:In this type of straight, all 7 cards are of unique ranks (i.e., no pairs occur).  First, ignoring suits, the total number of possible sets (combinations) that form a straight with 7 distinct ranks is found.  As with straight flushes, a straight is defined by its high card.  With an ace-high straight, the ranks of the 2 extra cards may be chosen from any of the remaining 8 ranks, while with the 9 other possible straights, any of the ranks but the rank directly above the high card may be chosen.  Thus, the total number of sets of ranks is:
-::<math>{8 \choose 2} + {9 \choose 1}{7 \choose 2} = 217</math>
-:Next, the total number of possible sets of suits, for any of the sets of ranks, is found.  Given that each card is of a distinct rank, the total number of sets of suits is:
-::<math>{4 \choose 1}^7 = 16,384\,</math>
-:However, the instances where a flush is formed must be subtracted from the total; there are 3 ways of achieving this: There is 1 case per suit where all 7 are of the same suit.  If 6 of the 7 are in the same suit, then the remaining card is chosen from the remaining 3 suits.  If 5 of the 7 in the same suit, then 2 independent choices are made for each of the extra cards.  Thus, the total number of cases where a flush is formed with 7 distinct ranks is:
-::<math>{4 \choose 1}\left[1 + {7 \choose 6}{3 \choose 1} + {7 \choose 5}{3 \choose 1}^2\right] = 844</math>
-:Thus, the total number of sets of suits which produce a straight, but not a flush is:
-::<math>16,384 - 844 = 15,540\,</math>
-:And as each set of suits occurs for each set of ranks, the total number of straights with 7 distinct ranks is:
-::<math>217 \cdot 15,540 = 3,372,180\,</math>
-:'''6 distinct ranks'''
-:A straight can also be formed with only 6 distinct ranks (i.e., the hand contains 1 pair).  In this case, one of the extra cards will have the same rank as one of the cards forming the straight, therefore only one extra rank need be chosen.  Thus, the total number of sets of ranks is:
-::<math>{8 \choose 1} + {9 \choose 1}{7 \choose 1} = 71</math>
-:The way to proceed now is to calculate the total number of ways to form a pair, and then calculate the total number of ways to form a straight, but not a flush (given that the pair has already been chosen).  The pair can be 1 of the 6 previously chosen ranks, and 2 of the 4 of that rank form the pair.  Thus, the total number of ways for form a pair is:
-::<math>{6 \choose 1}{4 \choose 2} = 36</math>
-:The total number of sets of suits for the remaining 5 cards can be calculated in the same way as for 7 cards:
-::<math>{4 \choose 1}^5 = 1,024\,</math>
-:As with 7 distinct ranks, the instances where a flush is formed must be subtracted from the total.  The remaining 5 cards can be chosen in two different manners in order to form a flush: either they are all of the same suit, or 4 of them are in the same suit as either of the two paired cards.  If all 5 are in the same suit, 1 of the 4 suits is chosen.  If 4 of the 5 are in the same suit, 1 of the 2 suits forming the pair is chosen, and the suit of the extra card is chosen from the remaining 3 suits.  Thus the total number of ways to form a flush is:
-::<math>{4 \choose 1} + {5 \choose 4}{2 \choose 1}{3 \choose 1} = 34</math>
-:Thus, the total number of sets of suits which produce a straight, but not a flush is:
-::<math>1,024 - 34 = 990\,</math>
-:Thus the total number of straights with 6 distinct ranks equals the total number of sets of ranks, multiplied by the total number of ways to form the pair, multiplied by the total number of ways to form a straight:
-::<math>71 \cdot 36 \cdot 990 = 2,530,440\,</math>
-:'''5 distinct ranks with a triple'''
-:There are two ways to form a straight with 5 distinct ranks.  The first is using 3 cards of the same rank, and 4 of separate ranks.  There are only 10 sets of ranks in this case, as there are no extra ranks to be chosen.  The triple can be 1 of the 5 ranks, and 3 of the 4 of that rank make up the triple.  Thus, the number of ways to choose the triple is:
-::<math>{5 \choose 1}{4 \choose 3} = 20</math>
-:The total number of sets of suits for the remaining 4 cards is <math>4^4</math> and the only ways to form a flush are if all 4 cards are of the same suit as 1 of the 3 suits forming the triple.  Thus, the total number of straights form a straight, but not a flush is:
-::<math>{4 \choose 1}^4 - {3 \choose 1} = 253\,</math>
-:Thus the total number of straights with 5 distinct ranks and a triple is:
-::<math>10 \cdot 20 \cdot 253 = 50,600\,</math>
-:'''5 distinct ranks with 2 pairs'''
-:The second way to form a straight with 5 distinct ranks is to have 2 pairs and 3 other cards of separate ranks.  As before, there are 10 different sets of ranks, however, calculating the number of ways that a flush is formed is complicated, due to the fact that the two pairs can consist of either 2, 3 or 4 suits.  Firstly, the ranks for the two pairs are chosen from the 5 available.  Thus, the number of ways to choose the ranks for the two pairs is:
-::<math>{5 \choose 2} = 10</math>
-:Then the cards are chosen for each of the pairs.  Thus, the number of ways to choose the suits for the pairs is:
-::<math>{4 \choose 2}^2 = 36</math>
-:6 of these ways, the pairs consist of 2 suits, 24 of these ways the pairs consist of 3 suits, and the remaining 6 of these ways they consist of 4 suits.  Note that the total number of sets of suits for the remaining 3 cards is <math>4^3</math>.  When the pairs consist of 2 suits, a flush will be formed when the remaining 3 cards are all in either of those two suits.  There are 2 ways of this happening which must be subtracted from the total.  When there are 3 suits, a flush will be formed when the remaining 3 cards are all in the suit of the 2 cards of matching suit in the pairs.  There is 1 way of this happening.  When there are 4 suits there are no ways of making a flush.  Thus, the total number of sets of suits that do not form a flush is:
-::<math>6 \cdot \left[64 - {2 \choose 1}\right] + 24 \cdot (64 - 1) + 6 \cdot 64 = 2,268\,</math>
-:Thus, the total number of straights with 5 distinct ranks and 2 pairs is:
-::<math>10 \cdot 10 \cdot 2,268 = 226,800\,</math>
-:Thus, the total number of straights is:
-::<math>3,372,180 + 2,530,440 + 50,600 + 226,800 = 6,180,020\,</math>
-*''Three of a kind'' &mdash; A three of a kind must consist of 5 of the 13 ranks, but the 10 combinations that form straights must be subtracted, giving the total number of sets of ranks as:
-::<math>{13 \choose 5} - 10 = 1,277</math>
-:The rank of the triple is chosen from the 5 available, and 3 of the 4 of that rank are chosen.  Thus, the total number of ways of choosing the triple is:
-::<math>{5 \choose 1}{4 \choose 3} = 20</math>
-:There are <math>4^4</math> ways to choose the suits of the remaining 4 cards, minus the ways in which all 4 match one of the 3 suits in the triple (making a flush):
-::<math>{4 \choose 1}^4 - {3 \choose 1} = 253</math>
-:Thus, the total number of three of a kinds is:
-::<math>1,277 \cdot 20 \cdot 253 = 6,461,620\,</math>
-*''Two pair'' &mdash; A two pair can be formed in 2 ways:
-:'''3 pairs with 1 kicker'''
-:The 4 ranks are chosen, then 3 of the 4 are chosen for the 3 pairs, and 2 of the 4 of each rank are chosen for each pair.  The kicker is then chosen from the 4 cards in the remaining rank.  Thus, the total number of 3 pairs with 1 kicker is:
-::<math>{13 \choose 4}{4 \choose 3}{4 \choose 2}^3{4 \choose 1} = 2,471,040</math>
-:'''2 pairs with 3 kickers'''
-:A two pair hand must consist of 5 of the 13 ranks, but the 10 combinations that form straights must be subtracted.  2 of the ranks are chosen for the pairs and as with the calculations for straights with 5 ranks and two pairs, there are 2,268 sets of suits that do not form flushes.  Thus, the total number of 2 pairs with 3 kickers is:
-::<math>\left[{13 \choose 5} - 10\right]{5 \choose 2} \cdot 2,268 = 28,962,360</math>
-:Thus, the total number of two pairs is:
-::<math>2,471,040 + 28,962,360 = 31,433,400\,</math>
-*''Pair'' &mdash; A pair hand must consist of 6 of the 13 ranks, but the combinations that form straights must be subtracted.  There are 9 ways to form a 6-card straight (6- to ace-high).  With 5-card straights, when the straight is either 5- or ace-high, the remaining card may be selected from any of the 8 other ranks, minus the rank at the open end of the straight (6 and 9 respectively).  In any of the other 8 situations, the remaining card may be selected from any of the other 8 ranks, minus the two ranks at either end of the straight.  Thus, the total number of sets of ranks that do not form straights is:
-::<math>{13 \choose 6} - 9 - \left[2 \cdot {7 \choose 1} + 8 \cdot {6 \choose 1}\right] = 1,645</math>
-:There are <math>4^5</math> ways of choosing the ranks of the kickers, and as with the calculations for straights with 6 distinct suits, there are 34 sets of suits that form flushes, therefore the total number of sets of suits that do not form flushes is:
-::<math>{4 \choose 1}^5 - 34 = 990</math>
-:There are 6 different ranks to choose for the pair and the pair can be formed from 2 of the 4 cards in that rank, therefore the number of ways to choose the pair is:
-::<math>{6 \choose 1}{4 \choose 2} = 36</math>
-:Thus, the total number of pair hands is:
-::<math>1645 \cdot 990 \cdot 36 = 58,627,800</math>
-*''No pair'' &mdash; The 7 ranks are chosen, but the combinations that form straights must be subtracted.  There are 8 ways to form a 7-card straight (7- to ace-high).  With 6-card straights, as with 5-card straights in the pair hand calculations, any of the remaining ranks minus 1 may be chosen for the highest and lowest straight (6 ranks), while in the other cases, any remaining rank minus 2 may be chosen (5 ranks).  With 5-card straights, the calculations are the same as with pairs, but 2 cards must be chosen rather than 1.  Thus, the total number of sets of ranks that do not form straights is:
-::<math>{13 \choose 7} - 8 - \left[2 \cdot {6 \choose 1} + 7 \cdot {5 \choose 1}\right] - \left[2 \cdot {7 \choose 2} + 8 \cdot {6 \choose 2}\right] = 1,499</math>
-:There are <math>4^7</math> ways of choosing the suits of the cards, and as with the calculations for straights with 7 distinct suits, there are 844 sets of suits that form flushes, therefore the total number of sets of suits that do not form flushes is:
-::<math>{4 \choose 1}^7 - 844 = 15,540</math>
-:Thus, the total number of no pair hands is:
-::<math>1,499 \cdot 15,540 = 23,294,460\,</math>
-==Frequency of 5-card lowball poker hands==
-{{See also|Lowball (poker)}}
-Some variants of poker, called [[Lowball (poker)|lowball]], use a [[low hand]] to determine the winning hand.  In most variants of lowball, the ace is counted as the lowest card and straights and flushes don't count against a low hand, so the lowest hand is the five-high hand '''A-2-3-4-5''', also called a ''wheel''.  The probability is calculated based on <math>\begin{matrix} {52 \choose 5} = 2,598,960 \end{matrix}</math>, the total number of 5-card combinations.  (The frequencies given are exact; the probabilities and odds are approximate.)
-:{| class="wikitable" style="font-size: 95%;"
-|-
-!Hand        !!   Distinct hands !!            Frequency !! Probability !! Cumulative !! Odds
-|-
-|     5-high ||align=right|    1 ||align=right|    1,024 ||     0.0394% ||    0.0394% ||align=right| 2,537.05 : 1
-|-
-|     6-high ||align=right|    5 ||align=right|    5,120 ||     0.197%  ||    0.236%  ||align=right|   506.61 : 1
-|-
-|     7-high ||align=right|   15 ||align=right|   15,360 ||     0.591%  ||    0.827%  ||align=right|   168.20 : 1
-|-
-|     8-high ||align=right|   35 ||align=right|   35,840 ||     1.38%   ||    2.21%   ||align=right|    71.52 : 1
-|-
-|     9-high ||align=right|   70 ||align=right|   71,680 ||     2.76%   ||    4.96%   ||align=right|    35.26 : 1
-|-
-|    10-high ||align=right|  126 ||align=right|  129,024 ||     4.96%   ||    9.93%   ||align=right|    19.14 : 1
-|-
-|  Jack-high ||align=right|  210 ||align=right|  215,040 ||     8.27%   ||   18.2%    ||align=right|    11.09 : 1
-|-
-| Queen-high ||align=right|  330 ||align=right|  337,920 ||    13.0%    ||   31.2%    ||align=right|     6.69 : 1
-|-
-|  King-high ||align=right|  495 ||align=right|  506,880 ||    19.5%    ||   50.7%    ||align=right|     4.13 : 1
-|-
-|'''Total''' ||align=right|1,287 ||align=right|1,317,888 ||    50.7%    ||   50.7%    ||align=right|     0.97 : 1
-|}
-As can be seen from the table, just over half the time a player gets a hand that has no pairs, three- or four-of-a-kinds. (50.7%)
-If aces are not low, simply rotate the hand descriptions so that 6-high replaces 5-high for the best hand and ace-high replaces king-high as the worst hand.
-===Derivation of frequencies for 5-card lowball hands===
-The following computations show how the above frequencies for 5-card lowball poker hands were determined.  To understand these derivations, the reader should be familiar with the basic properties of the [[binomial coefficient]]s and their interpretation as the number of ways of choosing elements from a given set.  See also: [[sample space]] and [[event (probability theory)]].
-The probability for any specific low hand with 5 distinct ranks (i.e. no paired cards) is the same.  The frequency of a 5-high hand or any a specific low hand is calculated by making 5 independent choices for the suit for each rank, which is:
-:<math>{4 \choose 1}^5 = 1,024</math>
-There is one way to choose the ranks for a five-high hand:
-:<math>{5 \choose 5} = 1</math>
-To determine the number of distinct six-high hands, once the six is chosen, the other 4 ranks are chosen from the 5 ranks '''A''' to '''5''', which is:
-:<math>{5 \choose 4} = 5</math>
-This can be generalized for any non-paired low hand.  Where <math>r</math> is the highest rank in the hand (numbering jack–king as 11–13), the number of distinct low hands is:
-:<math>D = {r-1 \choose 4}</math>
-and the frequency of low hands that are <math>r</math>-high is <math>1,024 \cdot D</math>.
-'''Derivation for lowball hands without straights and flushes:'''
-In the case where straights and flushes count against a low hand, the frequency of a specific hand must subtract the 4 combinations of suits that yield a flush, and the calculation for the number of distinct hands must subtract the <math>r - 4</math> combinations of ranks that yield a straight.  This gives the following frequency for low hands of rank <math>r</math> that do not include a straight or a flush:
-:<math>\left[ {4 \choose 1}^5 - 4 \right] \cdot \left[ {r-1 \choose 4} - (r - 4) \right]</math>
-==Frequency of 7-card lowball poker hands==
-{{See also|Lowball (poker)}}
-In some variants of poker a player uses the best five-card [[low hand]] selected from seven cards.  In most variants of lowball, the ace is counted as the lowest card and straights and flushes don't count against a low hand, so the lowest hand is the five-high hand '''A-2-3-4-5''', also called a ''wheel''.  The probability is calculated based on <math>\begin{matrix} {52 \choose 7} = 133,784,560 \end{matrix}</math>, the total number of 7-card combinations.
-The table does not extend to include five-card hands with at least one pair. Its "Total" represents 95.4% of the time that a player can select a 5-card low hand without any pair.<!-- I suggest another line for "pair or better" so that the table total is 100% -->
-:{| class="wikitable" style="font-size: 95%;"
-|-
-!Hand        !!               Frequency !! Probability !! Cumulative !! Odds
-|-
-|     5-high ||align=right|     781,824 ||     0.584%  ||    0.584%  ||align=right| 170.12 : 1
-|-
-|     6-high ||align=right|   3,151,360 ||     2.36%   ||    2.94%   ||align=right|  41.45 : 1
-|-
-|     7-high ||align=right|   7,426,560 ||     5.55%   ||    8.49%   ||align=right|  17.01 : 1
-|-
-|     8-high ||align=right|  13,171,200 ||     9.85%   ||   18.3%    ||align=right|   9.16 : 1
-|-
-|     9-high ||align=right|  19,174,400 ||    14.3%    ||   32.7%    ||align=right|   5.98 : 1
-|-
-|    10-high ||align=right|  23,675,904 ||    17.7%    ||   50.4%    ||align=right|   4.65 : 1
-|-
-|  Jack-high ||align=right|  24,837,120 ||    18.6%    ||   68.9%    ||align=right|   4.39 : 1
-|-
-| Queen-high ||align=right|  21,457,920 ||    16.0%    ||   85.0%    ||align=right|   5.23 : 1
-|-
-|  King-high ||align=right|  13,939,200 ||    10.4%    ||   95.4%    ||align=right|   8.60 : 1
-|-
-|'''Total''' ||align=right| 127,615,488 ||    95.4%    ||   95.4%    ||align=right|   0.05 : 1
-|}
-(The frequencies given are exact; the probabilities and odds are approximate.)
-If aces are not low, simply rotate the hand descriptions so that 6-high replaces 5-high for the best hand and ace-high replaces king-high as the worst hand.
-===Derivation of frequencies for 7-card lowball hands===
-The following computations show how the above frequencies for 7-card lowball poker hands were determined.  To understand these derivations, the reader should be familiar with the basic properties of the [[binomial coefficient]]s and their interpretation as the number of ways of choosing elements from a given set.  See also: [[sample space]] and [[event (probability theory)]].
-To make a low hand of a specific rank four ranks are chosen that are lower than the high rank.  Where <math>r</math> is the highest rank in the hand (numbering jack–king as 11–13), the number of sets of 5 ranks that can make a low hand is:
-:<math>{r-1 \choose 4}</math>
-There are then three different ways to choose the remaining two cards that are not used in the low hand.  Each of these cases must be considered separately:
-'''7 distinct ranks'''
-In this type of hand the two additional ranks are chosen from the ranks higher than <math>r</math>, so this type of hand can only occur when there are at least two ranks greater than <math>r</math>—that is, jack-high or better hands.  The suits can be assigned by making 7 independent choices for the suit for each rank, so the number of ways to make a low hand with two distinct higher ranks is:
-:<math>{13-r \choose 2}{4 \choose 1}^7 = {13-r \choose 2} \cdot 16,384</math>
-'''6 distinct ranks'''
-In this type of hand there are 6 distinct ranks and one pair.  The additional rank is chosen from the ranks higher than <math>r</math>, so this type of hand can only occur when there is at least one rank greater than <math>r</math>—that is, queen-high or better hands.  One of the 6 ranks is chosen for the pair and two of the four cards in that rank are chosen.  The suits for the remaining 5 ranks are assigned by making 5 independent choices for each rank, so the number of ways to make a low hand with one higher ranks and a pair is:
-:<math>{13-r \choose 1}{6 \choose 1}{4 \choose 2}{4 \choose 1}^5 = {13-r \choose 1} \cdot 36,864</math>
-'''5 distinct ranks'''
-There are two ways to choose 5 distinct ranks for seven cards.  Either two pair and three unpaired ranks or three of a kind and four unpaired ranks.
-:'''Two pair'''
-:In this type of hand there are 5 distinct ranks and two pair.  Two of the 5 ranks are chosen for the pairs and two of the four cards in each rank are chosen.  The suits for the remaining 3 ranks are assigned by making 3 independent choices for each rank, so the number of ways to make a low hand with two pair is:
-::<math>{5 \choose 2}{4 \choose 2}^2{4 \choose 1}^3 = 23,040</math>
-:'''Three of a kind'''
-:In this type of hand there are 5 distinct ranks and three of a kind.  One of the 5 ranks is chosen for the three of a kind and three of the four cards in the rank are chosen.  The suits for the remaining 4 ranks are assigned by making 4 independent choices for each rank, so the number of ways to make a low hand with three of a kind is:
-::<math>{5 \choose 1}{4 \choose 3}{4 \choose 1}^4 = 5,120</math>
-Thus there are <math>23,040 + 5,120 = 28,160</math> ways to make a low hand with five distinct ranks.
-'''Derivation'''
-Thus where <math>r</math> is a rank from 5 to jack (11), the total number of <math>r</math>-high low hands is:
-:<math>{r-1 \choose 4} \cdot \left[ {13-r \choose 2} \cdot 16,384 + {13-r \choose 1} \cdot 36,864 + 28,160 \right]</math>
-The total number of queen-high low hands is:
-:<math>{11 \choose 4} \cdot \left[ {1 \choose 1} \cdot 36,864 + 28,160 \right] = {11 \choose 4} \cdot 65,024 = 21,457,920</math>
-The total number of king-high low hands is:
-:<math>{12 \choose 4} \cdot 28,160 = 13,939,200</math>
-==See also==
-{{Portal|Sports and games}}
-'''Poker topics:'''
-* [[Poker probability (Texas hold 'em)]]
-* [[Poker probability (Omaha)]]
-'''Math and probability topics:'''
-* [[Probability]]
-* [[Odds]]
-* [[Sample space]]
-* [[Event (probability theory)]]
-* [[Binomial coefficient]]
-* [[Combination]]
-* [[Permutation]]
-* [[Combinatorial game theory]]
-* [[Game complexity]]
-* [[Set theory]]
-* [[Gaming mathematics]]
-== Notes ==
-{{reflist}}
-==External links==
-* [http://www.math.sfu.ca/~alspach/ Brian Alspach's mathematics and poker page]
-* [http://mathworld.wolfram.com/Poker.html MathWorld: Poker]
-* [http://www.probabilityof.com/poker.shtml Poker probabilities including conditional calculations]
-* [http://www.suffecool.net/poker/table1.html Numerous poker probability tables]
-* [http://www.durangobill.com/Poker.html 5, 6, and 7 card poker probabilities]
-* [http://www.cardschat.com/odds-for-dummies.php Poker Odds for Dummies]
-* [http://www.suffecool.net/poker/7462.html The 7,462 and 4,824 equivalence classes]
-* [http://www.compatiblepoker.com/Poker+Odds+Table.cms.htm Preflop, After Flop and Chance of Making Hand Odds]
-* [http://www.world-of-poker.org/pokertip/holdem-odds-outs.htm Odds and Outs probability table]
-* [http://www.pokerprobability.net Poker probability calculator] 5, 6 and 7 cards
-* [http://www.cardschat.com/poker-odds-calculator.php Visual odds calculator]
-{{Poker}}
-{{Use dmy dates|date=September 2010}}
-{{DEFAULTSORT:Poker Probability}}
-[[Category:Poker probability| ]]

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