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If I roll 3 dice with n sides, and then roll a 4th die of the same size, what are the odds of it being higher than at least one of the previous rolls?

I'm thinking it should be something like the odds of rolling a 1 in the first 3 dice times the odds of rolling a 2 or higher on the last die, giving me this equation.

$(1 - ((n-1)/n)^3)*((n-1)/n)$

Is this equation correct? I feel like I'm missing something.

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    $\begingroup$ I guess I misread the title at first glance, thinking you wanted the fourth (and last) roll to be higher than all the first three rolls. But as I reread everything, it seems more likely you ask only that the last roll exceeds at least one of the first three rolls. In that case you want the distribution of the smallest of three rolls, from which you can derive the chance that the fourth roll is above that. $\endgroup$ – hardmath Oct 23 '18 at 1:25
  • $\begingroup$ @hardmath exactly, I'm unsure of how to turn that into an equation though. $\endgroup$ – william porter Oct 23 '18 at 1:29
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There are $n^3$ sequences for the first three rolls, of which

  • 1 sequence has minimum roll $n$
  • $2^3-1^3=7$ sequences have minimum roll $n-1$. The expression as a difference of cubes can be seen by drawing a table of the minimum function
  • $3^3-2^3=19$ sequences have minimum roll $n-2$, etc.

Thus the probability that the fourth roll is less than or equal to the minimum of the first three rolls is $$\frac1{n^3}\sum_{k=1}^n((n-k+1)^3-(n-k)^3)\frac kn=\frac1{n^4}\sum_{k=1}^nk^3=\frac1{n^4}\left(\frac{n(n+1)}2\right)^2=\frac{(n+1)^2}{4n^2}$$ where the simplification of the sum comes first from the difference of cubes telescoping and then the formula for the sum of consecutive cubes. Therefore the probability that the fourth roll is higher than at least one of the previous three rolls is the complement of this, or $1-\frac{(n+1)^2}{4n^2}=\frac{(n-1)^2}{4n^2}$.

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What is the probability that the 4th roll is not the smallest?

One of the 4 dice must be the smallest. It is just as likely to be the 4th die rolled as one of the previous 3.

But there is a possibility that there is a tie.

As n gets to be very large, I would expect our probability to approach $\frac 14$

There are $n^4$ possibilities for the 4 dice to come up.

if you roll a $k$ on your last die there are $(n-k)$ ways to roll better than that, and $(n-k)^3$ to roll 3 dice better than that.

$\sum_\limits{k=1}^n (n-k)^3 = \sum_\limits{k=1}^{n-1} k^3 = \frac {n^2(n-1)^2}{4}$

$\frac {(n-1)^2}{4n^2}$

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