(a) What is the probability that if there are 4 dice thrown, there will be two same values gained and two different ones? Is this solution correct? P = 6 ⋅ 1 ⋅ 5 ⋅ 4 ⋅ 6 = 720.
My logic is: I have 6 possible values assigned to first die, than I must choose the same value on the second die, and the remaining dice may be chosen in $5\times4$ ways (they must differ from each other and also from the first two dice). I must count that the two same dice may be chosen from all 4 dice in 4C2 ways.
(b) If we throw 4 dices, we might gain 6^4 different results. These may be composed as:
we gain all the values different (6 ⋅ 5 ⋅ 4 ⋅ 3 = 360);
we gain all the values same (6);
we gain two same values and two different values (6 ⋅ 1 ⋅ 5 ⋅ 4 ⋅ 6 = 720);
we gain three same values and one differing (6 ⋅ 5 ⋅ 4 = 120). If counted, we would get this result: 1206 out of 1296? What possibilities are remaining?
2026-04-02 03:52:33.1775101953
Dice - throwing
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If we start by assuming the second dice matches the first and the third and fourth are both different the probability would be
$$\frac{1}{6} \cdot \frac{5}{6} \cdot \frac{4}{6} = \frac{5}{54}$$
But we don't care about the order. How many ways can we have two same and two different?
Lets list them SSDD, SDSD, SDDS, DSSD, DSDS, DDSS. So we can multiply this by 6 and the answer is:
$$ 6 \cdot \frac{5}{54} = \frac{5}{9} $$