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A friend of mine recently created a game where your movements on the plateau are determined by series of dice throws. The rules are simple: you have exactly $6$ dice that you should throw simultaneously. However, by trading a specific resource (not discussed here), you can choose to throw again any dice you want (except the dice you previously chose to keep). Let's clarify this with an example.

Consider this: I want $6 \times 6$ (the maximal output).

Throw $1$, output $6, 6, 2, 1, 4, 5 \implies$ I keep the two $6$ and throw again the $4$ other dice.

Throw $2$, output $6, 5, 3, 4 \implies$ I keep the single $6$ and I guess you figured it out.

Now, for the question, I would like to compute the probability of successfully outputting $6$ times $6$ with at most n throws. The blunt answer is to considered all the favorable "paths" to my output, for example with $6$ times $6$ in $3$ throws: $0$ first throw, $0$ second throw, $6$ third throw, etc., which is $36$ paths I reckon.

Needless to say, the process of computing such probabilities for a high number of throws is quite the hassle. Obviously we can consider that outputting $6\times6$ in at most $3$ throws is equivalent to doing so in exactly $1$ throw OR in exactly $2$ throw OR in exactly $3$ throw which is convenient because you have an iterative process. However, there is still $21$ unique paths for successfully outputting $6\times6$ in exactly $3$ throws.

Is there a crafty way to compute said probability that I'm not envisioning?

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  • $\begingroup$ It's actually quite easy. Consider each dice separately. We want each dice to roll $6$ at least once in $n$ throws. So, for a given dice, the probability of getting a $6$ is $1-\left(\frac56\right)^n$. And, we want each dice to be a $6$, so our final answer is $$\left(1-\left(\frac56\right)^n\right)^6$$ $\endgroup$ – Don Thousand Mar 3 at 13:26
  • $\begingroup$ "Dice" is plural. the singular is "die". $\endgroup$ – saulspatz Mar 3 at 13:50
  • $\begingroup$ @DonThousand Slick. You should make that an answer. $\endgroup$ – saulspatz Mar 3 at 13:50
  • $\begingroup$ What do you mean by "at most" n throws"? $\endgroup$ – true blue anil Mar 3 at 14:15
  • $\begingroup$ @trueblueanil You might finish in fewer throws. For example, you could throw $6$ sixes on the first roll. $\endgroup$ – saulspatz Mar 3 at 15:02
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I'm making this an answer for completeness.

It's actually quite easy. Consider each die separately. We want each die to roll $6$ at least once in $n$ throws. So, for a given die, the probability of getting a $6$ is $1−\left(\frac56\right)^n$. And, we want each dice to be a $6$, so our final answer is $$\left(1−\left(\frac56\right)^n\right)^6$$

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  • $\begingroup$ Thank you for this answser. $\endgroup$ – kronenbouh Mar 3 at 14:40

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