Spinners
How do scores on dice and factors of polynomials relate to each
other?
Age
16 to 18
Challenge level
(a) Evaluate $(x + x^2 + x^3 + x^4)^2$.
(b) Imagine you have two spinners labelled $1, 2, 3$ and $4$ and spin them together. The score is the sum of the results from the two spinners. Find the theoretical frequency distribution of the scores.
(c) What do you notice about this frequency distribution and the coefficients in the polynomial expansion from (a)?
(d) Notice that the powers in $(x^1 + x^2 + x^3 + x^4)$ correspond to the labels on the spinners. Can you factorize the expression $(x + x^2 + x^3 + x^4)^2$ into two different polynomials which correspond to a re-labelling of the spinners, so each has four non-negative integer labels, giving new pairs of spinners with the same frequency distribution of scores? This re-labelling can be done
in more than one way.
(e) Find other pairs of spinners which can be re-labelled in more than one way to give the same frequency distribution of scores.
What about a $2$-spinner and a $3$-spinner? What about two ordinary dice?
For this solution we have to thank Andrei from Tudor Vianu
National College, Bucharest, Romania.
(a) Evaluating $(x + x^2 + x^3 + x^4)^2$, I obtained $$x^2
+2x^3+3x^4+4x^5+3x^6+2x^7+x^8$$
(b) The sum of two spinners labelled 1, 2, 3 and 4 varies from
2 to 8. There are 16 possibilities in total, giving the following
probabilities: \begin{eqnarray} &\text{Scores} &2 &3
&4 &5 &6 &7 &8 \\ &\text{Frequencies}
&1 &2 &3 &4 &3 &2 &1 \\
&\text{Probabilities} &0.0625 &0.125 &0.1875
&0.25 &0.1875 &0.125 &0.0625 \end{eqnarray}
(c) I observe that the frequency distribution is the same as
the coefficients from the expansion of the polynomial from
(a).
(d) Using the computer simulation, I took two spinners
labelled from 1 to 4, obtaining the following table of
frequencies:
| Exp.$\quad$ | Relative frequency$\quad$ |
| 0 | 0 |
| 1 | 0 |
| 2 | 0.0631 |
| 3 | 0.1253 |
| 4 | 0.1872 |
| 5 | 0.2493 |
| 6 | 0.188 |
| 7 | 0.1248 |
| 8 |
0.062
|
The results obtained are very close to the theoretical
frequency distribution of the scores.
(e) $(x + x^2 + x^3 + x^4)^2$ could be factorised as follows:
$x^2 (1+x)^2 (1+x^2)^2$ and these factors can be re-written
as:
1) $(1+x)(x+x^2) (1+x^2)(x+x^3)$
2) $(x+x^2)^2 (1+x^2)^2$
3) $(1+x)^2(x+x^3)^2$
The first case corresponds to 4 spinners (0,1), (1,2),
(0,2),(1,3), the second to 4 spinners: (1,2), (1,2), (0,2), (0,2)
and the third to another 4 spinners (0,1), (0,1), (1,3), (1,3).\par
(f) Using the simulation, I obtained similar frequency
distributions as with two 1,2,3,4 spinners.
| Score$\quad$ | Frequency$\quad$ | distributions$\quad$ | |
| (1) | (2) | (3) | |
| 2 | 0.3324 | 0.3312 | 0.3359 |
| 3 | 0.4996 | 0.5021 | 0.4982 |
| 4 | 0.1679 | 0.1665 |
0.1658
|
The theoretical probabilities would be $2/6 = 33.33 \%$, for
2, $3/6 = 50\%$ for 3, and $1/6 = 16.66\%$, which are very close to
the simulated values.
For a 2-spinner, there are equal probabilities of obtaining
the two numbers with which it is labelled. It corresponds to the
polynomial $x^m + x^n$. If $m$ and $n$ are equal, then the
probability ofobtaining that value is 1 ($100\%$).\par For a
3-spinner, I have 3 cases:
- All three numbers are different. There is a probability of 1/3 ($33.33\%$) of obtaining any of the three numbers. It corresponds to the polynomial $x^m + x^n + x^p$, with $m \neq n \neq p$
- Two of the three numbers are equal. There is a probability of 1/3 ($33.33\%$) of obtaining the unique number, and a probability of $66.67\%$ of obtaining the number which appears twice. It corresponds to the polynomial $x^m + 2x^n$, $m \neq n$.
- All 3 numbers are equal. The probability of obtaining this number is $100\%$.
Two ordinary dice are equivalent to 2 6-spinners. The
theoretical probabilities and the simulated ones are:
| Number | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Theoretical | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
| Frequency | 0.0281 | 0.0556 | 0.083 | 0.1123 | 0.1397 | 0.1674 | 0.1381 |
0.1093
|
0.0842
|
0.0539 | 0.028 |
This case corresponds to the polynomial $(x + x^2 + x^3 + x^4
+ x^5 + x^6)^2$. The expansion of this polynomial is: $$x^2 + 2x^3
+ 3x^4 + 4x^5 + 5x^6 + 6x^7 + 5x^8 + 4x^9 + 3x^{10} + 2x^{11} +
x^{12}.$$ This polynomial could be factorised as follows: $$(x +
x^2 + x^3 + x^4 + x^5 + x^6)^2 = x^2 (1+x)^2 (1+x^2+x^4)^2$$ and
from this decomposition different combinations could be obtained,
all of them producing the same frequency distribution.
$$(x + x^2 + x^3 + x^4)^2 = x^2 (1 + x)^2 (1+x^2)^2 =[x + 2x^2 +
x^3][x + 2x^3 + x^5].$$ corresponds to the re-labelling $1, 2, 2,
3$ and $1, 3, 3, 5$ .