I would be grateful if anyone can help me with calculating the odds of successfully choosing 6 numbers from 10 (1 - 10) to match a predetermined result in the following three scenarios.

(1) The six numbers chosen have to match the predetermined numbers and can be chosen in any order. Each number can only be used once.

(2) The six numbers chosen have to match the predetermined numbers (any random order) and also be chosen in the same order. Each number can only be used once.

(3) The six numbers chosen have to match six predetermined numbers and also be chosen in the same order. Any number can be used more than once.

22 views and no response. Perhaps my questions are too silly, poorly stated or difficult.

I can work out the odds in (1) and (2) for each individual, successive choice, but don’t know how to convert these to overall odds.

(1) = First choice = 6 - 10, second = 5 - 9, third = 4 - 8, forth = 3 - 7, fifth = 2 - 6 and sixth = 1 - 5.

(2) = First choice = 1 - 10, second = 1 - 9, third = 1 - 8, forth = 1 - 7, fifth = 1 - 6 and sixth = 1 - 5.

I have worked out the overall odds of (3) to be 1 - 1,000,000,000. This is 1 - 10 multiplied by ten six times. Is this correct?

Last edited:
In the first case, this is just a matter of combinations. We are to chose 6 things out of 10 without replacement.

Take the simpler case of choosing 2 from 10. The first one can be chosen in 10 ways, the second in 9. Thus for these two cases there are 90 ways it can be done. But half of those ways are the same, (since ab and ba are the same) since the choosing order is not important. So the answer is 10x9/2 = 45, which can be written as the combination: $$\frac{10!}{2!8!}$$ So that should help you get started.

The second case, as I read it, involves permutations.

The third case has a real simple way of looking at it. Just suppose the ten things we are talking about are the digits 0,1.....9. That hint should get you started.

Last edited:
In the first case, this is just a matter of combinations. We are to chose 6 things out of 10 without replacement.

Take the simpler case of choosing 2 from 10. The first one can be chosen in 10 ways, the second in 9. Thus for these two cases there are 90 ways it can be done. But half of those ways are the same, (since ab and ba are the same) since the choosing order is not important. So the answer is 10x9/2 = 45, which can be written as the combination: $$\frac{10!}{2!8!}$$ So that should help you get started.

The second case, as I read it, involves permutations.

The third case has a real simple way of looking at it. Just suppose the ten things we are talking about are the digits 0,1.....9. That hint should get you started.
Thanks - So with (1) I multiply 10 by 9, then 8, then 7, then 6, then 5, to get 151,200. I then dived this by 6 to get 25,200. Does this mean the odds of (1) are 1 - 25,200?

With (2) I multiply 10 by 9, then 8, then 7, then 6, then 5, to get 151,200. As the numbers must be chosen in the correct order, are the odds 1 - 151,200?

Is my previous answer for (3) correct?

No more hints on homework kind of problems. You should find out about combinations, permutations and powers of ten.

No more hints on homework kind of problems. You should find out about combinations, permutations and powers of ten.

Thanks again. I think I have it now.

(1) Combinations . . . (10!)/[ (10 - 6)! x (6)! ] = 210

(2) Permutations . . . (10!)/(10-6)! = 151,200

(3) Powers of 10 . . . 106 = 1,000,000 or 10 x 10 x 10 x 10 x 10 x 10 = 1,000,000

Now you are cooking with gas!

Keep in mind the insight that what happened to the 6! in the second case is that that is number of ways we can arrange 6 different things, and so distinguishes permutations from combinations.

In the third case, keep in mind that we actually include the zero going from 0 to 999,999, which is 10^6 different numbers.

Last edited:
Now you are cooking with gas!

Keep in mind the insight that what happened to the 6! in the second case is that that is number of ways we can arrange 6 different things, and so distinguishes permutations from combinations.

In the third case, keep in mind that we actually include the zero going from 0 to 999,999, which is 10^6 different numbers.
So the actual odds would be . . .

(1) 209 to 1
(2) 151,199 to 1
(3) 999,999 to 1

Really appreciate your help - thanks.

That is correct.

The way my math dictionary defines odds is: Given a probability of p, the odds are p/(1-p) for example, = $$\frac{1/10^6}{999999/10^6} =\frac{1}{999,999}$$ And the dictionary goes on add this figure is "sometimes inverted."

So, that from a simple gambling standpoint, if p=1/3, the odds = 1/2, (often spoken of as 2-1). This means that if you put up $1 and win, you get two additional dollars meaning the total payoff is$3.00.

The difference between the odds and payout is often expressed, for example, as, odds are 2 TO 1 and the payout is 3 FOR 1. (Odds are frequently used in gambling and much less in statistics, which prefers probabilities.)

So what you are saying above seems completely correct.

Last edited: