2020_II_12_ProbStats

2020_II_12_ProbStats

Part (i)

\begin{align} P(\text{same score twice}) &= \sum_{i=1}^nP(X=i)^2\\ &= \sum_{i=1}^n\left(\frac1n+\varepsilon_i\right)^2\\ &= \sum_{i=1}^n\left(\frac1{n^2}+\frac{2\varepsilon_i}n+\varepsilon_i^2\right)\\ &= \sum_{i=1}^n\frac1{n^2}+\sum_{i=1}^n\frac{2\varepsilon_i}n+\sum_{i=1}^n\varepsilon_i^2\\ &= \frac1n+0+\sum_{i=1}^n\varepsilon_i^2\\ &= \frac1n+\sum_{i=1}^n\varepsilon_i^2\\ \end{align}

Since the probability of a fair dice showing the same score on two successive rolls is \frac1n and \sum_{i=1}^n{\varepsilon_i}^2>0\!, it is more likely for the biased die to have the same score on two successive rolls.

Part (ii)
Consider a biased die where P(X=i)= \frac{x_i}{\sum x_i}\!.

\begin{align} \frac1n &\leqslant P(\text{same score twice})\\ \frac1n &\leqslant 1-P(\text{two different scores})\\ P(\text{two different scores}) &\leqslant \frac{n-1}n\\ \end{align}

Now consider the fact that to roll two different scores, either the first score is greater than the second or the second is greater than the first and these two cases have an equal likelihood of occurring, by symmetry. Therefore, P(\text{first score more than second})=\frac12P(\text{two different scores})\!.

\begin{align} P(\text{two different scores}) &\leqslant \frac{n-1}n\\ 2P(\text{first score more than second}) &\leqslant \frac{n-1}n\\ P(\text{first score more than second}) &\leqslant \frac{n-1}{2n}\\ \sum_{i=2}^n\sum_{j=1}^{i-1}\left(\frac{x_i}{\sum x_i}\right)\left(\frac{x_j}{\sum x_i}\right) &\leqslant \frac{n-1}{2n}\\ \left(\frac1{\sum x_i}\right)^2\sum_{i=2}^n\sum_{j=1}^{i-1}x_ix_j &\leqslant \frac{n-1}{2n}\\ \sum_{i=2}^n\sum_{j=1}^{i-1}x_ix_j &\leqslant \frac{n-1}{2n}\left(\sum_{i=1}^n x_i\right)^2\\ \end{align}

Part (iii)
For a fair die, P(\text{three same scores})=\frac1{n^2}\!. For the biased die defined in the question:

\begin{align} P(\text{three same scores}) &= \sum_{i=1}^nP(X=i)^3\\ &= \sum_{i=1}^n\left(\frac1n+\varepsilon_i\right)^3\\ &= \sum_{i=1}^n\left(\frac1{n^3}+\frac{3\varepsilon_i}{n^2}+\frac{3\varepsilon_i^2}n+\varepsilon_i^3\right)\\ &= \sum_{i=1}^n\frac1{n^3}+\sum_{i=1}^n\frac{3\varepsilon_i}{n^2}+\sum_{i=1}^n\frac{3\varepsilon_i^2}n+\varepsilon_i^3\\ &= \frac1{n^2}+0+\sum_{i=1}^n\varepsilon_i^2\left(\frac3n+\varepsilon_i\right)\\ &= \frac1{n^2}+\sum_{i=1}^n\varepsilon_i^2\left(\frac2n+P(x=i)\right)\\ &\geqslant\frac1{n^2} \end{align}

So it is more likely for a biased die to show the same score on three successive rolls.

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