November 17th, 2012 at 3:28:23 AM
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How can I calculate the total number of combinations for the Oasis Poker? Why are they different from the total number of combinations in the Carribean Stud poker (combin(52,5)*47*combin(46,4) = 19933230517200)?
November 17th, 2012 at 5:08:32 AM
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In my opinion the total combinations should be:
(combin(52,5)+combin(52,4)*42+combin(52,3)*combin(42,2)+combin(52,2)*combin(42,3)+52*combin(42,4)+combin(42,5))
*47*combin(46,4)=420997366299510
But this is not right
(combin(52,5)+combin(52,4)*42+combin(52,3)*combin(42,2)+combin(52,2)*combin(42,3)+52*combin(42,4)+combin(42,5))
*47*combin(46,4)=420997366299510
But this is not right
November 20th, 2012 at 6:45:40 AM
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If you switch a card, the card you drop is put away by the dealer. It cannot come back in his hand.
Hence, there are less than 46 cards left for him to complete his hand.
The 46 in the last combin(46,4) in your formula should be adapted separately to each number of switches, inside the parentheses.
Actually, at first sight, the parenthesis is false. Why would you add the terms? I guess I would place C(52,5) at the beginning, and then only subdivide for switches.
C(52,5)*47 *( C(46,4) +C(5,1)*46*C(45,4)+C(5,2)*C(46,2)*C(44,4)+C(5,3)*C(46,3)*C(43,4)+C(5,4)*C(46,4)*C(42,4)+C(46,5)*C(41,4) )
(Not sure.)
Hence, there are less than 46 cards left for him to complete his hand.
The 46 in the last combin(46,4) in your formula should be adapted separately to each number of switches, inside the parentheses.
Actually, at first sight, the parenthesis is false. Why would you add the terms? I guess I would place C(52,5) at the beginning, and then only subdivide for switches.
C(52,5)*47 *( C(46,4) +C(5,1)*46*C(45,4)+C(5,2)*C(46,2)*C(44,4)+C(5,3)*C(46,3)*C(43,4)+C(5,4)*C(46,4)*C(42,4)+C(46,5)*C(41,4) )
(Not sure.)
Reperiet qui quaesiverit
November 20th, 2012 at 9:04:40 AM
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A = combin(52, 5) = number of cards in player hand
B = combin(47, 5) = number of cards in dealer hand
C = combin(05, 1) = number of cards the player could discard
D = combin(42, 1) = number of cards the discard can be replaced with
A * B * C * D = combin(52, 5) * combin(47, 5) * combin(5, 1) * combin(42, 1) = 837,195,681,722,400
B = combin(47, 5) = number of cards in dealer hand
C = combin(05, 1) = number of cards the player could discard
D = combin(42, 1) = number of cards the discard can be replaced with
A * B * C * D = combin(52, 5) * combin(47, 5) * combin(5, 1) * combin(42, 1) = 837,195,681,722,400
November 20th, 2012 at 10:45:22 AM
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Quote: JBA = combin(52, 5) = number of cards in player hand
B = combin(47, 5) = number of cards in dealer hand
C = combin(05, 1) = number of cards the player could discard
D = combin(46, 1) = number of cards the discard can be replaced with
A * B * C * D = combin(52, 5) * combin(47, 5) * combin(5, 1) * combin(46, 1) = 837,195,681,722,400
A*B*C*D = 916,928,603,791,200
D should be combin(42,1) since the replacement card cannot be an original card in either hand.
This gives A * B * C * D = combin(52, 5) * combin(47, 5) * combin(5, 1) * combin(42, 1) = 837,195,681,722,400