Example 1
Suppose 7 dice are rolled. What is the probability that at least 4 of the dice show the same face?
Example 2
Suppose that 6 job assignments are randomly assigned to 5 workers. What is the probability that at least 4 of the job assignments go to the same worker?
Example 2 is left as exercise.
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Discussion of Example 1
Fix a face (say 1). Finding the probability of that at least 4 of the dice show the face 1 is a binomial problem. Then multiplying this answer by 6 will give the desired answer.
Consider obtaining a 1 as a success. Let 
![\displaystyle \begin{aligned}(1) \ \ \ \ \ P(X \ge 4)&=1-P(X \le 3) \\&=1-P(X=0)-P(X=1) \\&- \ \ \ P(X=2)-P(X=3) \\&=1-\binom{7}{0} \biggl[\frac{1}{6} \biggr]^0 \biggr[\frac{5}{6} \biggr]^7 - \binom{7}{1} \biggl[\frac{1}{6} \biggr]^1 \biggr[\frac{5}{6} \biggr]^6 \\&- \ \ \ \binom{7}{2} \biggl[\frac{1}{6} \biggr]^2 \biggr[\frac{5}{6} \biggr]^5 - \binom{7}{3} \biggl[\frac{1}{6} \biggr]^3 \biggr[\frac{5}{6} \biggr]^4 \\&=\frac{4936}{279936} \end{aligned}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D%281%29+%5C+%5C+%5C+%5C+%5C+P%28X+%5Cge+4%29%26%3D1-P%28X+%5Cle+3%29+%5C%5C%26%3D1-P%28X%3D0%29-P%28X%3D1%29+%5C%5C%26-+%5C+%5C+%5C+P%28X%3D2%29-P%28X%3D3%29+%5C%5C%26%3D1-%5Cbinom%7B7%7D%7B0%7D+%5Cbiggl%5B%5Cfrac%7B1%7D%7B6%7D+%5Cbiggr%5D%5E0+%5Cbiggr%5B%5Cfrac%7B5%7D%7B6%7D+%5Cbiggr%5D%5E7+-+%5Cbinom%7B7%7D%7B1%7D+%5Cbiggl%5B%5Cfrac%7B1%7D%7B6%7D+%5Cbiggr%5D%5E1+%5Cbiggr%5B%5Cfrac%7B5%7D%7B6%7D+%5Cbiggr%5D%5E6+%5C%5C%26-+%5C+%5C+%5C+%5Cbinom%7B7%7D%7B2%7D+%5Cbiggl%5B%5Cfrac%7B1%7D%7B6%7D+%5Cbiggr%5D%5E2+%5Cbiggr%5B%5Cfrac%7B5%7D%7B6%7D+%5Cbiggr%5D%5E5+-+%5Cbinom%7B7%7D%7B3%7D+%5Cbiggl%5B%5Cfrac%7B1%7D%7B6%7D+%5Cbiggr%5D%5E3+%5Cbiggr%5B%5Cfrac%7B5%7D%7B6%7D+%5Cbiggr%5D%5E4+%5C%5C%26%3D%5Cfrac%7B4936%7D%7B279936%7D+%5Cend%7Baligned%7D&bg=ffffff&fg=111111&s=0 "\displaystyle \begin{aligned}(1) \ \ \ \ \ P(X \ge 4)&=1-P(X \le 3) \\&=1-P(X=0)-P(X=1) \\&- \ \ \ P(X=2)-P(X=3) \\&=1-\binom{7}{0} \biggl[\frac{1}{6} \biggr]^0 \biggr[\frac{5}{6} \biggr]^7 - \binom{7}{1} \biggl[\frac{1}{6} \biggr]^1 \biggr[\frac{5}{6} \biggr]^6 \\&- \ \ \ \binom{7}{2} \biggl[\frac{1}{6} \biggr]^2 \biggr[\frac{5}{6} \biggr]^5 - \binom{7}{3} \biggl[\frac{1}{6} \biggr]^3 \biggr[\frac{5}{6} \biggr]^4 \\&=\frac{4936}{279936} \end{aligned}")
Multiplying 
To give some perspective to this example, for each 
If we roll more dice, the event
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Answer to Example 2
