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Spinners

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

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.