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How would I get the odds of picking 5 numbers in the right order.?

Posted on October 16, 2008

For this game, I need to pick 5 numbers, and depending on whether the numbers are matched, and in order, I win a prize.

What I'd like to know is what are the chances of picking 5 numbers in the same order as a personal.

The numbers I pick are between 1-9.

I would pick something like 25682 for example.

The personal would pick 23948 and I would have the two matched, but not the rest.

Thanks.


Answer:
Why is the number 0 not valid - is the computer prevented from selecting it? If not, the odds are 1 in 10 for each number, the chances for an all correct solution with one five figure number chosen being 1 in 100000. Why do some contributors assume “without replacement”? If a 2 is chosen (by the computer) for the first number, “without replacement” means it cannot be chosen as a later number . This is highly unlikely and seems a silly assumption. In fact this assumption is refuted by the questioner's own chosen number of 25682, which would be void on that assumption. If the personal is restricted to selecting 9 numbers only, the first answer is correct, 1 in 59049.

Answer:
If I understand you correctly, you want permutation (where order matters) without replacement. Therefore 9*8*7*6*5= 15,120.

On Excel, the formula is =PERMUT(9,5), confirming the above.


Answer:
The odds of you picking the exact same number is (1/9)^5 = 1/59049, or about 0.0017%

Answer:
I comprehend you correctly, you want permutation (where order matters) without replacement.

Answer:
my answer is

9×9x9×9x9 to 1

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