# Overall probability of random events

This should be a simple problem of basic probabilities. Still, I can't wrap my head around it.

Problem description: You have a dice with 50 faces numbered from 1 to 50. What is the probability that rolling the dice 150 times, number 1 would appear at least once?

And (would be nice to have) a generalization: If you have a dice with N faces numbered from 1 to N, what is the probability that rolling the dice M times, the number 1 will appear at least once.

Thank you!

My effort (you can skip reading from now on): I tried to reduce the problem and try to generalize it afterwards. Say we have a dice with 3 faces and we roll it twice. All the possible outcomes would be {first_run,second_run}: {1,1},{1,2},{1,3},{2,1},{2,2},{2,3},{3,1},{3,2},{3,3}. The favorable cases are with bold. So the probability would be 5/9 here.

• Check out the Binomial Distribution. – InterstellarProbe Oct 16 '19 at 18:50
• HINT First figure out the probability that $A$ never appears on the screen – Bram28 Oct 16 '19 at 18:58
• As an aside, it is almost never a good idea to write down the full list of possible outcomes as it is easy to accidentally write one too many times or miss writing another. It can help build intuition, but it should otherwise be saved as a last resort. – JMoravitz Oct 16 '19 at 19:11
• Expanding on the previous hints, notice in your example of using 3faced dice and look at the pairings which aren't in bold. These can be neatly organized into a grid (or three-dimensional grid, or higher dimensional as you increase the number of dice), just as we could have arranged the total set of all outcomes in such a grid as well. For your example, have the rows correspond to the outcome of the first die and the columns correspond to the outcomes of the second die. This will lead you to discover the Rule of Product. – JMoravitz Oct 16 '19 at 19:15
• You should discover that there are $N^M$ possibilities for rolling an $N$ sided die $M$ times, and $(N-1)^M$ of those possibilities will be such that none of the dice showed a specific face. – JMoravitz Oct 16 '19 at 19:18

The chance that you get all Bs is $$.98^{150} = 0.048296$$. The chance you get at least one A is $$1 -.98^{150} = 0.951704.$$