If a person selects four randon numbers between one and ten, but not including one or ten, and multiplied them together one at a time, what are the odds that the answer will be a "Didital Nine" number? What percentage?
First example 2x3x4x7=168 Not a digital nine.
Second example 2x3x4x6=144 Which IS a digital nine.
I hope that I'm using the term Digital Nine correctly.
Thanks
Dave

I've never heard of the term "digital nine" before, but it looks like you mean the digits add up to 9. But that just means the result is a multiple of 9. I'll assume that's what you mean. I'll also assume you mean that the same number can't be selected twice. (2x3x4x4 would be out for instance.) If these assumptions are wrong, let me know.

So we have eight numbers (from 2 through 9), and we're selecting four at random. Because we're multiplying them, the order in which they're picked doesn't matter. (2x3x4x5 = 5x4x3x2, etc.) Math geeks would write this as 8C4 which equals 8!/4!4! which equals 70. So out of these 70 combinations, how many are multiples of 9?

Well the first (and most obvious) case is when 9 is one of the four numbers chosen. If you got a number by multiplying something by 9, you can divide that resulting number by 9. So how many combinations have a 9 in them? Since order doesn't matter, this is like always picking 9 first, and then three random numbers from the remaining seven (2 through 8). Yup, this is 7C3 = 7!/4!3! = 35. So 35 of the possible combinations have a 9 in them, and therefore result in a multiple of 9.

But having a 9 isn't the only way. If a 3 AND a 6 are both picked, you'll end up with a multiple of 9. (3x6 = 18, which is itself a multiple of 9.) So if we pick 3 and 6 and then two others ramdomly from the remaining six, that's 6C2 = 6!/4!2! = 15.

But not so fast! What about combinations that have 3 and 6 AND 9? We counted these in the number of combinations that have 9, AND we counted these in the number of combinations that have 3 and 6. That is, we counted them twice. How many did we count twice? Well, pick 3,6, and 9 and then one randomly from the remaining five numbers, which is 5C1 = 5!/4!1! = 5.

So the number of combinations that multiply together to be a multiple of 9 is 7C3 + 6C2 - 5C1, which is 35 + 15 - 5, which is 45. So the probability that four randomly chosen numbers from the set of 2 through 9 is 45 / 70 = .6429 or 64.29% The odds are almost 2 to 1 (actually 9 to 5) that you WILL end up with a multiple of 9.

:banana:

If I understand you correctly, you are asking what the odds are of ending up with a number with a digital root of 9?

First, I'll assume we're talking about 4 different numbers, between one and ten, as you mention.

Let's start with how many possibilities there are in total: 1,680 (8 * 7 * 6 * 5)

As you point out, not all of these combinations, when multiplied are divisible by 9.

If 9 is one of the numbers chosen, then obviously the digital root of the answer will be 9. There are 210 combinations (1 * 7 * 6 * 5) that include a 9 and will create the result you want.

Any of the combinations including a 3 and a 6 will also work. There are 30 of these (1 * 1 * 6 * 5), but some of these also include a 9, as well, and therefore cross partially with the above 210. 20 (1 * 1 * 5 * 4) of the 30 combinations don't include a 9, but do include a 3 and a 6 together.

20+210=230

230/1680=a roughly 13.7% chance of the digital root of the answer being 9, when 4 different numbers are chosen from 2 through 9 and multiplied together.

This assumes that the numbers are chosen truly randomly. I have no way to account for psychology in these equations, of course.

Can anyone verify that I did the math correctly?

I think that it's more like 64 Percent. Chances are almost one in three that a 3, 6, or 9 is chosen per selection. Right? With four selections chances are that a nine or a combination of a three or a six will be chosen....??? Just my regular brain thinking here.

If two people choose four numbers (Independently) and multiply them together, what will the odds be them.???? Over 90%?

By the Way.You guys ROCK...!!!!!!!
Thanks
Dave

Scott, that's how many permutations there are of four elements from an eight element set. Permutations are where order matters. The possible combinations on a lock would be an example of permutations.

But we're multiplying the selections here. Since multiplication is commutative, the order of the selection doesn't matter. So we want the number of (mathematical) combinations there are of four elements from an eight element set.

The general equation for the permutations of n elements from a set of k unique elements is kPn = k!/(k - n)!, which is what you used. The general equation for the combinations of n elements from a set of k unique elements is kCn = k!/[n!(k - n)!], which is what I used.

Dave asks:


Just to make things clear, participant A chooses four numbers from between 2 and 9. Then participant B chooses 4 numbers from between 2 and 9 (and some (or even all) of B's choices may be the same as A's). What are the odds that the eight numbers multiplied together will be a multiple of 9?

If either A or B (or both) ends up with a multiple of 9, and we multiply their numbers together, we'll have a multiple of 9. So what is the probability that neither A nor B will have a multiple of 9? Well, we know for each individually, it's 25/70 (about 35.71%). Since these are independent events, the probability that neither ends up with a multiple of 9 is 25/70 x 25/70 = 625/4900, about 12.76%

But we're not done. If A chose 3 or 6 as one of his numbers, but not the other, and not 9, then when you multiply his four numbers together, you get a multiple of 3, but not a multiple of 9. The same is true for B. Now if both A and B chose numbers so that the results are multiples of 3, and we multiply those numbers together, we have a multiple of 9. We need to take that into account.

So how many ways can A choose 3 without choosing 6 or 9? Since order doesn't matter, let's have him pick 3 first. That leaves him three more choices from five numbers (since we've eliminated the 3 he picked, and the 6 and 9 he's not allowed to pick). That's 5C3 = 5!/3!2! = 10. Similarly, there are 10 ways he can pick a 6 without a 3 or a 9. So there are 20 ways he can pick a multiple of 3 that is not a multiple of 9. The same for B. We already know there are 70 different combinations, so the probability that A will pick four numbers that result in a multiple of 3 but not 9, and that B will also pick four numbers that result in a multiple of 3 but not 9 is 20/70 x 20/70 = 400/4900. Now we subtract that from 625/4900, and we get 225/4900, approximately 4.59%

That's the probability that we won't end up with a multiple of 9. The probability that we will end up with a multiple of 9 is 1 - 225/4900 = 4675/4900, about 95.41%

:pepper:

Thanks so much!!!!!!!!
This has a great effect on how some of my Remote Radio Magic Effects will be presented. I can afford to be wrong One out of Twenty times. No problem.
This way I just play the odds and all of the selections are totally free....within my guidelines.
Thanks Again!
Dave

Scott,

You can use permutations (rather than combinations), but you must be
careful.

Yes, there are 1680 length-four permutations of the eight digits (2-9)
that don't contain any repeated digits. (8P4 = 8*7*6*5)

How many of these permutations contain a 9? There are 7*6*5 distinct-digit
length-three permutations of the digits 2-8. Given any one of these
210 permutations, the 9 can be inserted in one of 4 positions. (To the left
of all three, between the first and second, between the second and third, or
to the right of all three.) Thus there are 4*210=840 length-four
distinct-digit permutations of 2-9 that contain a 9.

Now we need to count the distinct-digit length-four permutations that
DON'T contain a 9 but DO contain both a 3 and a 6. There are 5*4=20
length-two distinct-digit permutations of the digits 2,4,5,7,8.
The 3 can be inserted in any of 3 positions. (To the left of both digits,
in between both digits, or to the right of both digits.) The 6 can now
be inserted in any of 4 positions. Thus there are 20*3*4=240 length-four
distinct-digit permutations that contain both a 3 and a 6 but no 9.

So the probability of getting a distinct-digit length-four permutation
whose digit product is divisible by nine is (840 + 240)/1680. This
fraction reduces to rgranville's 45/70 which in turn reduces to
9/14.

Are you saying 9/14 chance for one person choosing four digits or for two people choosing four digits each?
Thanks
Dave

ONE person. The probability for two people choosing four digits each and multiplying all eight digits together and getting a multiple of 9 is 95.41%

:carrot:

I guess my redundant post caused confusion. I agree with
rgranville's 100% correct analysis of both situations.

I only wanted to show that you can use combinations OR permutations
to arrive at the answers.

I am a small boat floating on an Ocean of Mathematical Principles.....My only question...Will I float?
And I will 95% of the time. Now to figure an "Out" for the extra 5% and I'm there!
Thanks a Lot
Dave