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This is a stupid question to ask but I need help with likeliness or probability. I dont really know what it is called in English. Basicly the likliness to hit a #6 on a dice is 1/6.

So lets say I have 4 Aces(cards) of each color on the table. What is the probability that:

Card #1 = Heart
Card #2 = Diamond
Card #3 = Club
Card #4 = Spade

Card #1 must be 1/4, #2 1/3, #4 1/2. And then what? Im stuck :(

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1 Answer 1

up vote 3 down vote accepted

You are on the right track: choice #3 has probability 1/2 to be the right one and once the three first choices are correct, the fourth one is automaically correct. So the probability you are after is 1/4 times 1/3 times 1/2, which is 1/24.

You can also consider that the whole order of the four cards is drawn at random amongst the 4! possible ones. By symmetry every possible order has the same probability hence the probability to get the one you want is 1/4!=1/24.

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Ok thank you! But I can not really understand why you take 1/4 times 1/3 and so on. How comes that? –  TomBGO May 3 '11 at 17:36
    
@user10426: The chance that two events happen is the product of the probability that the first happens, times the chance that the second happens assuming the first already did. For this, you might just try it. 1/4 of the time the first card is a heart. Assuming that it is, there are now three to choose from, so the chance that the second is a diamond is 1/3 and the chance the first two work is 1/12. –  Ross Millikan May 3 '11 at 17:58
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