Probability of specific result of dice rolling using inclusion-exclusion principle

Question

I have two dices and I roll them ten times. What's the probability that I will throw all pairs (i,i), where i=1,...,6 at least once?

I would appreciate any kind of help and hints!

Answer 1

Hint: for $i=1,\dots,6$ let $A_i$ be the event "$(i,i)$ never comes up". You want the probability of the complement of the event $(A_1\cup A_2\cup A_3\cup A_4\cup A_5\cup A_6)$. By the in-and-out principle, that probability is equal to $$1-\sum P(A_i)+\sum P(A_i\cap A_j)-\dots$$ Now $P(A_1)=(\frac{35}{36})^{10},\ P(A_1\cap A_2)=(\frac{34}{36})^{10}$, etc. That's $$1-6\left(\frac{35}{36}\right)^{10}+15\left(\frac{34}{36}\right)^{10}-20\left(\frac{33}{36}\right)^{10}+15\left(\frac{32}{36}\right)^{10}-6\left(\frac{31}{36}\right)^{10}+\left(\frac{30}{36}\right)^{10}.$$