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Let's say we have a set of discrete random variables $\{ X_1, X_2, X_3, ... X_n \}$ with known probability distributions, how to calculate the probability that $x_{max} = max(x_1, x_2, x_3 ... x_n)$ is equal to the realization of a specific random variable $x_i$? In other words, I want to calculate $Pr(x_{max}=x_i)$.

Example:

Let's say we have 4 dice $\{D_1, D_2, D_2, D_4\}$ representing 4 discrete random variables, what is the probability that $D_1$ will take the highest value after rolling the 4 dice? This probability can be denoted by $Pr(d_{max}=d_1)$

My current understanding:

First I am calculating the probability that $D_1$ is greater than $D_2, D_3, D_4$, that is, calculating $Pr(D_1 > D_2), Pr(D_1 > D_3),$ and $Pr(D_1 > D_4)$. Then $Pr(d_{max}=d_1)$ can be expressed as the product of these probabilities: $Pr(d_{max}=d_1) = Pr(D_1 > D_2) \times Pr(D_1 > D_3) \times Pr(D_1 > D_3)$

How correct is that ? My concern is that the sum of resulting probabilities is diffrent than 1, that is, $\sum\limits_{i = 1}^4 Pr(d_{max}=d_i) \ne 1$

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  • $\begingroup$ How do you deal with equality? If your 4 dices give $(6, 6, 6, 6)$, does it count in $Pr(d_{max} = d_1)$? If yes, then this event will be counted in every case, thus the sum will be greater than $1$. $\endgroup$
    – Alain Remillard
    Nov 18, 2022 at 15:18
  • $\begingroup$ @AlainRemillard Thank you for your reply. Yes, the aim is to count that case as well in $Pr(d_{max} = d_1)$. Would considering $Pr(D_1 \ge D_i)$ instead of $Pr(D_1 > D_i)$ solve the problem? $\endgroup$
    – Oussalvatore oussama
    Nov 18, 2022 at 15:32
  • $\begingroup$ Yes, it should be $Pr(D_1\ge D_2$. There is also the fact that those events aren't independant. If you know that $D_1\ge D_2$, there is a better chance that $D_1 \ge D_3$. Thus $Pr((D_1 \ge D_2) \text{ and }(D_1 \ge D_3)) \neq Pr(D_1\ge D_2) \times Pr(D_1\ge D_3)$. See Henry's answer for a way to compute your probabilities. $\endgroup$
    – Alain Remillard
    Nov 18, 2022 at 15:58
  • $\begingroup$ If we denote $E_i = \{X_{\max} = X_i\}$, the event that $X_i$ is the largest, then note that $\Pr(\cup_{i = 1}^n E_i) = 1$ and $\Pr(\cup_{i = 1}^n E_i) \leq \sum_{i = 1}^n \Pr(E_i)$ and equality holds only if $E_i's$ are all pairwise disjoint. $\endgroup$
    – sudeep5221
    Nov 18, 2022 at 16:02
  • $\begingroup$ @sudeep5221 Thank you for your comment. Yes, it makes sense, in my case, the events $E_i$ are not disjoint and can occur at the same time. My final aim is to calculate the probability distribution of the random variable $E_i$ but it seems that my approach is wrong. Any suggestions for calculating it? $\endgroup$
    – Oussalvatore oussama
    Nov 19, 2022 at 10:16

1 Answer 1

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Suppose

  • the random variables are iid
  • the possible values of the random variable are $a_1,a_2,\ldots$ (in the example of fair standard dice $1,2,3,4,5,6$)
  • $\Pr(X_i = a_j)=q_j$ (in the case of fair dice all $\frac16$ on the support)
  • $\Pr(X_i \le a_j)=Q_j$ (in the case of fair dice $\frac j6$ on the support)

Then since you want the values of the other $n-1$ random variable to be less than or equal to the value of the random variable you are interested in, you get $$\sum_j q_j Q_j^{n-1}$$ which in the case of your four dice is $$\frac16 \left(\frac{1}{6}\right)^3+ \frac16 \left(\frac{2}{6}\right)^3+\frac16 \left(\frac{3}{6}\right)^3+\frac16 \left(\frac{4}{6}\right)^3+\frac16 \left(\frac{5}{6}\right)^3+\frac16 \left(\frac{6}{6}\right)^3 = \frac{49}{144}\approx 0.34$$

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  • $\begingroup$ Thank you @Henry for the detailed reply and the clarification. But let's say that I want to use the resulting probabilities as a distribution for a new random variable $D_{max}$ that takes the value $\{D_1, D_2, D_3, D_3\}$ and that captures the identity of die with the highest value. This is not feasible with these probabilities as their sum differs from $1$ ($0.34 \times 4 = 1,36$). I was expecting that the dices have an equal probability of being maximal, which is equal to 0.25 ($0.25 \times 4 = 1$). Is there a way to capture that? $\endgroup$
    – Oussalvatore oussama
    Nov 19, 2022 at 10:02
  • $\begingroup$ @Oussalvatoreoussama Since there is a discrete distribution, there is a positive probability that two (or more) are both the highest value. What would you do if the values you saw were $5,4,1,5$? $\endgroup$
    – Henry
    Nov 19, 2022 at 12:23
  • $\begingroup$ In my real case scenario, choosing one of the highest value variables is enough. In your example, returning either $D_1$ or $D_4$. $\endgroup$
    – Oussalvatore oussama
    Nov 19, 2022 at 14:45
  • $\begingroup$ @Oussalvatoreoussama If you split ties uniformly at random then the probabilities of a particular die being chosen is indeed $0.25$ by symmetry/exchangeability. If you always choose the earlier die then the probabilities of being chosen are $\frac{49}{144} \approx 0.34$, $\frac{175}{648} \approx 0.27$, $\frac{35}{162}\approx 0.22$ and $\frac{25}{144}\approx 0.17$ $\endgroup$
    – Henry
    Nov 19, 2022 at 15:38
  • $\begingroup$ Thank you @Henry. So it is all about the ties-splitting policy. One last question, can you please elaborate on how you calculated the probabilities when always choosing the earlier die? thanks. $\endgroup$
    – Oussalvatore oussama
    Nov 20, 2022 at 11:25

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