# Dice problem

Discussion in 'Math' started by Mark44, Apr 8, 2008.

1. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
Roll a pair of ordinary dice.
1. Multiply the values of the two top faces.
2. Multiply the values of the two bottom faces.
3. Multiply the value on the top face of the first die by the value on the bottom face of the second die.
4. Multiply the value on the top face of the second die by the value on the bottom face of the first die.
5. Add the four numbers from the previous four steps.
If you multiplied and added correctly you will always get the same number.

Some questions:
1. What's the number?
2. Why do you always get that number? (Again, assuming that your arithmetic is correct.)
Mark

2. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
Hint: The spots (pips?) on a die follow certain rules. If I can see three faces of a die, I can tell you how many spots are on the remaining faces.

3. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
I haven't thrashed out a set of answers, but the first thing that springs to mind is the fact that opposite faces of a die always sum to 7. Is this a factor?

Dave

4. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
It most certainly is.

5. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Greetings Mark44,

Before I blurt out my answer, I should ask if this is a homework assignment. I don't want to deprive you of the pleasure of deriving the answer yourself.

In either case, what do you believe the answers to be.

hgmjr

PS. Dave, I think you are on to the key.

6. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
Nope, hgmjr, it's not a homework assignment, but I did give out a lot of homework assignments in the 21 years I was a teacher. It's actually a fairly simple problem once you understand how the spots are arranged on a die.

7. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
It's half-past midnight here so I won't crack open my "Games-set". Pen and paper tomorrow lunchtime I think!

I've come across many-a-maths problem with dice, and the root is always the fact that the opposite sides sum to 7 - something I discovered by accident when a curious youngster!

Dave

8. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Since that is the case then I attach my answer below.

hgmjr

File size:
19.4 KB
Views:
30
9. ### MusicTech Active Member

Apr 4, 2008
144
0
Yep, a simple case in point example of factor by grouping...

Good example for an algebra 2 class.

Mark22, do you teach Algebra 2 or Pre-Calc, by any chance?

10. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
I taught all levels of high-school math for two years, and I taught all levels that we offered in the community college, near Seattle, where I worked for 18 years. There was also some student teaching before I got my job in the high school, plus a couple of quarters as a teaching assistant (I actually taught the class) in the university I went to.

While I was working at the community college I taught everything from just plain arithmetic to algebra and precalculus with trig, engineering calculus, linear algebra, and differential equations. I also taught classes in BASIC, Pascal, one class in Modula-2, many, many C classes, a couple of classes in C++, and several classes in FORTRAN. During that time I taught myself Java and x86 assembler.
Mark

11. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
Yep, it's 49. You can determine this with only two variables, though.

Let x = no. of spots on the top face of die 1
Let y = no. of spots on the top face of die 2

Carrying out the multiplications described in the problem gives:
xy + x (7 - y) + y (7 -x) + (7 - x)(7 - y)
= xy + 7x - xy + 7y -xy + 49 - 7x - 7y + xy
= 49

Mark

12. ### MusicTech Active Member

Apr 4, 2008
144
0
Wow, very impressive, that's all there is to say (to the reply two posts back, of course)

13. ### uPC New Member

Apr 8, 2008
1
0
Variable designation is incorrect in solution paper...although the logic is clear.

Better to correct it....

14. ### drewlas New Member

May 12, 2008
7
0
Well,
The numbers on the first die are x and (7-x) and,
The numbers on the second die are y and (7-y).
Just go from there.....

drewlas

15. ### recca02 Senior Member

Apr 2, 2007
1,211
0
any idea why that is the case?
Of course the dice won't become biased if we change it a little.

Last edited: May 14, 2008
16. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
It is either a) convention (i.e. the way it is always done with a reason lost in annals of time), or b) there is some statistical reason which ensures that by having opposite faces summing to 7 statistically we have a "fair dice".

I honestly can't say which of the above it is, but a quick Google search doesn't shed any light on it.

Dave