# Puzzle: Random Blocks In the first section of George Lenchner’s Creative Problem Solving in School Mathematics, right after his obligatory obeisance to George Polya (see the third quote here), Lechner poses this problem. If you have seen it before, be patient — his point was much more than simply counting blocks.

A wooden cube that measures 3 cm along each edge is painted red. The painted cube is then cut into 1-cm cubes as shown above. How many of the 1-cm cubes do not have red paint on any face?

And then he challenges us as teachers:

Do you have any ideas for extending the problem?
If so, then jot them down.

This is strategically placed at the end of a right-hand page, and I was able to resist turning to read on. I came up with a list of 15 other questions that could have been asked — some of which will be used in future Alexandria Jones stories. Lechner wrote only seven elementary-level problems, and yet his list had at least two questions that I had not considered. How many can you come up with?

For hints on creating your own list, read How to modify problems — an example. (Yes, I have linked to this before. It is a good article!)

## My favorite puzzles

Several of the puzzles I created grew from my experience with MathCounts last school year — many, many problems about combinatorics and probability. The amazing thing to me was that I have learned enough over the last several months to actually solve the problems I made up. Well, at least, I was able to get answers that seemed reasonable to me. I would love to hear what you come up with, as a “reality check” on my own calculations.

• If I choose one of the small cubes at random and toss it in the air, what is the probability that it will land red-painted side up?
• If I tossed all the small cubes in the air, so that they landed randomly on the table, how many cubes should I expect to land with a painted face up?
• If I put all the small cubes in a bag and randomly draw out 3, what is the probability that at least 3 faces on the cubes I choose are painted red?
• If I put the small cubes in a bag and randomly draw out 3, what is the probability that exactly 3 of the faces are painted red? Want to help your kids learn math? Claim your free 24-page problem-solving booklet, and sign up to hear about new books, revisions, and sales or other promotions.

## 15 thoughts on “Puzzle: Random Blocks”

1. Nick says:

I think I’ve got these figured out, but I’m not positive on the last one.

*SPOILER?*

1. 1/3
2. 9
3. 578/585
4. 192/2925

2. Luke says:

Hmm. I agree with Nick on the first two and the last, but I got 584/585 for 3 (the only case that doesn’t have at least 3 is if I draw 1, 1, 0 which has a 30/17550=1/585 chance of happening).

3. Denise says:

Thanks, Nick and Luke! I got the same answers for the first two problems, also. The first problem was just a way of randomly choosing one face out of the 27×6=162 possible faces—a relatively easy probability problem. The second problem makes a good introduction to expected value, which I want to teach my junior high students this year.

On the third problem, I agree that the easiest method is to calculate the probability of NOT getting 3 painted faces and then subtract that number from one. And I came up with the same answer as Luke on the chance of drawing the 1,1,0 combination. But couldn’t it also be 1,0,1 or 0,1,1? That would give us a probability of 3×1/585=1/195 to NOT get 3 painted faces, so my answer for the question came out 194/195. Have I missed something here?

I was the least confident in my answer to that last question. I came up with 436/1075, so I will have to go back and reconsider my calculations. Anyone else want to take a stab at it?

4. Luke says:

You are, of course, correct; I neglected to account for ordering. That changes my answer on 3 to 1-90/17550 = 194/195, and my answer on 4 to 552/17550 = 92/2925 (not 192/2925).

5. Denise says:

OK, I have 92/2925 for number 4, also. The first time around, I made an overcounting error AND a mental arithmetic error (being too lazy to walk across the room for my calculator). Thank you guys so much for helping me think this through!

6. jd2718 says:

#3 is still not right. Let me try a few ways.

The probability of at least 3 faces being painted = 1 – P(exactly 2 are painted). Note that it is not possible for none or only one to be painted.

A. So we have 1 – 3*(1*3*2)/(27*26*25) = 1 – 1/975 = 974/975. Let’s identify the role of each number. The first 3 is for the 3 orderings. The 1/27 is for grabbing the middle first. The 3/26 is for grabbing a face second. The 2/25 is for grabbing another face third. Looks good to me. Let’s try another way:

B. 1 – (All combos that give sum=2)/(all possible combos)
= 1 – 1*C(3,2)/C(27,3)
= 1 – 3/2925. Which is 974/975.

C. let’s try one more approach. There’s only four cubes we are allowed to draw from. So 23/27 we fail on the first draw, 4/27 we succeed times 23/26 we fail on the second, 4/27*3/26 we succeed times 23/25 we fail on the third. 23/27 + (4*23)/(27*26) + (4*3*23)/(27*26*25) = (23*26*25 + 4*23*25 + 4*3*23)/(27*26*25) = (14950 + 2300 + 276)/17550 = 17526/17550. Of course, we also fail if we draw a face, a face, and a face or 3/27*2/26*1/25 = 6/17550. So total P(failure) = (17526+6)/17550, or, once again, 974/975.

This stuff is tricky.

7. jd2718 says:

Yegads, I was counting 3 faces instead of 6. No wonder I got the same wrong answer three times!

8. jd2718 says:

For #1, there are two approaches. The more often used takes the probability of a corner and multiplies it by 3 out of 6, of an edge and multiplies it by 2 out of 6, of a face and multiplies by 1 out of 6, and of a center and multiplies by 0 out of 6:
8/27 * 3/6 + 12/27 * 2/6 + 6/27 * 1/6 + 1/27 * 0/6 = 54/162 = 1/3

But we can also divide the number of painted sides by the total number of sides:
27 cubes * 6 sides each = 162. 9 painted sides on a face * 6 faces = 54. 54/162

It’s a little atonement for the lousy post, 2 up.

9. Denise says:

I did #1 the hard way first, adding up all those probabilities, before I realized that all I needed was to find the ratio of painted sides to total sides.

As for #2, well, maybe you read it in the spirit of the old joke, “All we know is that there is at least one sheep in Scotland, and that at least one side of that one sheep is black.” After all, the picture only shows 3 sides being painted, and the story didn’t specify otherwise.

10. Eric says:

I know this is a fairly old post, but I only just discovered this blog, and I’m hoping it’s still okay to respond.

I’m getting different results from all of you for #3 and #4. I’ll try to explain my reasoning, and maybe you can tell me if I’ve gone off track.

My basic approach to discrete probability is to consider the ratio of how many outcomes satisfy what we’re looking for to how many total outcomes there are. In both #3 and #4, there are 27 cubes, and we’re drawing 3, so the total number of outcomes is C(27,3)=2925, which will serve as the denominator once we discover the total number of successful (or, in the case of #3, unsuccessful) outcomes. So…

#3
The only way to get fewer than 3 red faces requires us to draw two cubes with one red face each, plus the blank cube. There are six cubes with one red face each, and we have to draw two of them, so that’s C(6,2)=15 outcomes that satisfy that portion, multiplied by the three positions into which we can also draw the blank cube, which yields 3*15/2925=1/65.
Thus, the probability of getting at least three red faces is 1-1/65=64/65.

#4
The only outcomes that give exactly three red faces are permutations of 0,1,2 or 1,1,1 (where the numbers represent how many red faces each cube has). The number of possible ways to achieve 0,1,2 is 3!*1*6*12=432, and the number of ways to achieve 1,1,1 is C(6,3)=20. Since these outcomes are disjoint, we can add them together, and so the final probability should be (432+20)/2925=452/2925.

-Eric

11. Denise says:

Hi, Eric!
Most of the old posts are still open for comments, except for a few that for some reason got heavily targeted by spammers.

In your solutions, you are mixing combinations and permutations. One of the things I find most confusing about probability is trying to figure out when C(x,y) works and when it doesn’t, but I know that drawing things out of a bag is the type of puzzle in which it doesn’t. You have to look at drawing each cube as an individual event, so order matters, which means you have P(27,3)=17,550 for your denominator.

#3 — Each pair of red-faced cubes can be drawn in 2 orders.

#4 — You need permutations for the (1,1,1) set.

12. Eric says:

I used combinations because, as far as I could tell, order didn’t matter here. That’s not to say it’s right for certain, just why I did what I did.

After some research, I think these problems represent multivariate hypergeometric distributions.

In a hypergeometric distribution, you have two types of objects in a set, and you want to find the probability of selecting a certain number of each. If A and B represent the total numbers of each type, and if a and b represent the selected numbers of each respective kind, the probability of that selection is C(A,a)*C(B,b)/C(A+B,a+b).

This can be extended to any number of items as the multivariate hypergeometric distribution.

We have the following cases:

#3
27 cubes, of which 1 has no red faces, 6 have one red face, and 20 have more than one red face. Using (the complement of) the multivariate hypergeometric distribution, we get that the probability of choosing 1 no-red, 2 one-red, and 0 other is
1-C(1,1)*C(6,2)*C(20,0)/C(27,3)=1-1*15*1/2795=194/195
And it was a whole lot easier to compute than what I was trying to do before.

#4
We have 1 no-red, 6 one-red, 12 two-red, and 8 three-red. If we make a list, in order, of the numbers of each that we’d like to get, the useful lists are 1,1,1,0 and 0,3,0,0. The probability of the first is C(1,1)*C(6,1)*C(12,1)*C(8,0)/C(27,3)=6*12/2925=72/2925
The probability of the second is C(1,0)*C(6,3)*C(12,0)*C(8,0)/C(27,3)=20/2925
So together, we get 92/2925.

I think that should do it.

13. Eric says:

@#\$%! My English is messed up in what I said about #3. The complement clause should apply to the probability portion.
The final result, however, is fine.

14. Denise says:

I am forced once again to plead ignorance. I don’t know what multivariate hypergeometric distributions are.

But it did seem to me that the probability for drawing a given set must be the same whether the set is drawn one block at a time or as a whole handful. In other words, order really should not matter, but I could never get the answers to come out right except when I took it into account. I didn’t know the combinations shortcut you gave here. I’m glad to learn it!

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