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Wieviele verschiedenen 5 Karten-Schürhakenhände würden nur Karten einer einzelnen Klage enthalten?

icon1 Posted by pocketaces in Mathematics on 01 30th, 2009 | 4 responses
poker
Wonderwoman asked:


Ich weiß, dass es 2.598.960 verschiedene 5 Karten-Schürhakenhände aber I don' gibt; t wissen, wieviele nur eine einzelne Klage enthalten würden.

Solve Your Credit

4 Comments »

  1. avatar a²+b²=c² Says:
    February 1st, 2009 at 9:44 pm

    Es gibt 4 Klagen und für jede Klage, 13 Karten. You' wählende Re5 von ihnen, so 4 (13C5) = 5148 Hände.

    Sie können dieses auch lösen, indem Sie oben die Frequenz der Arten der Hände im Schürhaken schauen. Was Sie wünschen, ist die Zahl des Errötens plus gerades Erröten plus königliches Erröten.

  2. avatar kro Says:
    February 4th, 2009 at 2:17 am

    5108 to tzw flush. 36 pokera 4 Royal flush

  3. avatar Bob K Says:
    February 5th, 2009 at 6:36 am

    Eerst, kiest u welk kostuum u 5 kaarten van wilt hebben, zodat 4C1 zou zijn, daarna, zijn er 13 kaarten in een kostuum en u wilt 5 van die, of 13C5 kiezen. En vermenigvuldig dan de twee keuzen aangezien zij onafhankelijk zijn
    4C1*13C5
    4! /3! 1! * 13! /5! 8!
    4 * 1287
    5148

  4. avatar icemetalpunk Says:
    February 5th, 2009 at 3:50 pm

    Denk aan het op deze manier: Er zijn 13 kaarten van een pak. Dus vind het aantal combinaties van 13 kaarten, geen herhaling. Dan besef dat er een 4 pakken, dus dat getal vermenigvuldigen met 4. De formule voor combinaties is n! / (R! (Nr)!) Waarbij n = aantal mogelijke kaarten (13) en r = aantal gekozen kaarten (5). Vervanging geeft ons: 13! / (5! (13-5)!) = 13! / (120 * 8!) Als we dit op te lossen, krijgen we 1287. Nu zijn we vergeten er 4 pakken, en vermenigvuldig dit met 4 te krijgen: 5148 combinaties. -IMP;):)

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