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- Thread starter Cylab
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HallsofIvy

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Note that for each x the second probability differs and for each y the third does. Take the sums of the probabilities over all possible values of x and y.

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wondering if there is any alternative?

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Ray Vickson

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That is wrong. The chances associated with the second number depend on what was the first number. For example, if the first number is 3 there is just one allowed choice of second number, but it the first number is 9 there are 7 choices for the second number (8,7,6,5,4,3, or 2), etc.

RGV

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wondering if there is any alternative?

Your problem is a bit unclear. Is the procedure as follows?

You choose 3 numbers a,b,c from the set {1,2,3,4,5,6,7,8,9} such that a>b>c.

someone else puts 9 balls numbered {1,2,3,4,5,6,7,8,9} in a bag, and draws out 3 at random without replacement. You're being told that is just so happened that those 3 balls were in the same order as your pick.

(or the drawing was repeated with refilling the bag after each set of 3 draws until the balls were drawn out in descending order)

If that is the case, then the probability is just 1 divided by the number of triples (a,b,c) with a>b>c. (should be easy)

The reply of HallsifIvy is then wrong, since the probability of getting a 3 or a 9 on the first draw isn't the same. There's only one combination that starts with a 3, and many that start with 9, so with the given information, 9 is far more likely than 3.

I believe it would be correct if all balls that are bigger than a ball that's already drawn, or that couldn't produce a valid combination (1,2 on the first draw and 1 on the second)

would be thrown back in the bag.

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