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Pick number between 0 and 9 
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#1
Nov211, 04:45 AM

P: 54

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 321. What is the probability of winning, given the winning numbers are three in descending order (1st>2nd>3rd).



#2
Nov211, 07:47 AM

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P: 39,345

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. 


#3
Nov1511, 03:16 AM

P: 54

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


#4
Nov1511, 11:20 PM

P: 109

Pick number between 0 and 9
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.



#5
Nov1611, 01:09 AM

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RGV 


#6
Nov1611, 01:25 PM

P: 1,395

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. 


#7
Nov1611, 08:55 PM

P: 109

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?



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