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#1
22-02-2006, 6:47 AM
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Another puzzle
What is the possible number of whole card combinations (where K♦6♣does not equal K♥6♦)? I believe if my math is correct it should be ((52*52) + 52)/2 - 52 = 1326. IE: 51+50....+1, where each number is one less than the number before it because for example if I've already used K♦ as my first card and run through all the possible second cards, K♦ can not be counted again in any possible combination. Can anyone confirm/deny this?
Last edited by shwingzilla : 22-02-2006 at 6:56 AM. |
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#2
22-02-2006, 3:57 PM
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You forget to carry the one, take the absolute value of the suqare root of pi, then divide it by the radius and circumference. And, if it's negative you have to mulitply by three and divide by X-2.  |
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#3
22-02-2006, 4:25 PM
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Yep, 1326 is right, and the odds of getting a particular pair (AA for example) is 220/1, because there are 6 ways of getting AA, and 1326/6 = 221
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