What is the probability of winning with three descending numbers?

[Cylab]

you pick three numbers between 0 and 9 (no multiplication of number). You win if you pick the three numbers in the exact order that they are drawn. Say you pick 3-2-1. What is the probability of winning, given the winning numbers are three in descending order (1st>2nd>3rd).

 

[HallsofIvy]

Since the numbers do not repeat and are decreasing, the first number must be 3 or more. There are 7 such numbers, 3, 4, 5, 6, 7, 8 , 9, so the probability of picking the first number, x, correctly is 1/7. The probability of picking the second number, y, correctly is 1/(x- 1) and the probability of picking the third number correctly is 1/(y- 1).
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.

 

[Cylab]

Thanks a lot,, Yes, it makes sense, but it is going to be big big big calculation...
wondering if there is any alternative?

 

[2milehi]

Something doesn't seem right. The first number needs to be from the range of 2 through 9 and that is an 8 in 10 chance. Then there is a 1 in 9 chance for the second number and a 1 in 8 chance in the third number.

 

[Ray Vickson]

Something doesn't seem right. The first number needs to be from the range of 2 through 9 and that is an 8 in 10 chance. Then there is a 1 in 9 chance for the second number and a 1 in 8 chance in the third number.

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

 

[willem2]

Thanks a lot,, Yes, it makes sense, but it is going to be big big big calculation...
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.

 

[2milehi]

I might be misunderstanding the original question. Do the numbers have to descend by one like in your example, or just have be lower than the previously chosen number?