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Two dice are thrown r times. I want to find the probability $p_r$ that each of the six combinations (1,1),...,(6,6)appears at least once.

My Answer:-$p_r=6*(1-(\frac{35}{36})^r)-15*(1-(\frac{34}{36})^r)+20*(1-(\frac{33}{36})^r)-15*(1-(\frac{32}{36})^r)+6*(1-(\frac{31}{36})^r)-(1-(\frac{30}{36})^r)$.

Author's solution:- If $A_k$is the event that (k,k)does not appear, then $1-p_r=6*(\frac{35}{36})^r-\binom{6}{2}(\frac{34}{36})^r+\binom{6}{3}(\frac{33}{36})^r-\binom{6}{4}(\frac{32}{36})^r+6*(\frac{31}{36})^r-(\frac{30}{36})^r$

The thereom used here is:- The probability $P_1$of the realisation of atleast one event among the events $A_1,A_2,...,A_N$ is given by

$P_1=S_1-S_2+S_3-S_4+-...\pm S_N.$ Where $S_1=\sum p_i,S_2=\sum p_{i,j},S_3=\sum p_{i,j,k}....$ Here $i<j<k<...\leq N$ and $S_r$ has $\binom{N}{r}$ terms. $p_i=P[A_i], p_{i,j}=[A_iA_j] ,p_{i,j,k}=[A_iA_jA_k]$

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    $\begingroup$ This is a coupon collector problem. The "coupons" are the ordered pairs of values. The twist here is that you don't want to collect all of them: you just want to collect six particular ones out of the 36. The basis of your answer is unclear because, as in your other questions, you haven't attempted to justify it. That it is incorrect is apparent by considering what happens as $r$ grows large: it decreases exponentially rather than approaching $1.$ $\endgroup$
    – whuber ♦
    Commented Sep 29, 2018 at 18:09
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    $\begingroup$ You seem to be fishing for a solution, because you haven't offered any reasons for the various formulas you have supplied. In the current case the same considerations as before will tell you whether it makes sense; namely, what happens to your $p_r$ as $r$ grows large? $\endgroup$
    – whuber ♦
    Commented Sep 30, 2018 at 15:08
  • $\begingroup$ @whuber $p_r$will reach 1 when r becomes larger and larger.Recall formula for union of two events and use it for N events. $\endgroup$
    – Win_odd Dhamnekar
    Commented Sep 30, 2018 at 15:23
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    $\begingroup$ You're right: I misread the formula because it is broken at a strange place across two lines. But, once again, I would like to suggest that if you would attempt an explanation of your formula you would be able to see for yourself whether it is correct or not. You might also be able to see that it is algebraically equivalent to the author's solution by using the fact that the limiting value of $p_r$ is indeed $1.$ $\endgroup$
    – whuber ♦
    Commented Sep 30, 2018 at 15:29

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