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In the game Yahtzee, a player rolls five fair six-sided dice, and gets a Yahtzee if all five dice show the same number. After the initial roll, the player gets two chances to reroll some of the dice. What is the probability that, on the initial roll, at least two of the dice show the same number? Express your answer as a common fraction.

I tried complementary counting, and the only case is when all of the die rolled are different, implying that one of the numbers is not chosen. Thus, we have $\frac{6*6*5*4*3*2}{6^5}$ where $6*5*4*3*2$ represents the placement of the 5 numbers of the die. Reducing gives $\frac{5}{9}$. Is this solution correct? If so, what is wrong with it? Also, is there a better solution

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  • $\begingroup$ Why do you have two $6$'s in the numerator? Which are you trying to calculate... the probability that you do have all numbers different? Or the probability that you don't have all numbers different (and so have at least one number appearing at least twice)? What do you intend then as your final answer to the originally intended question? $\endgroup$
    – JMoravitz
    Commented Aug 30, 2020 at 22:00
  • $\begingroup$ As an aside... this is no different than the birthday problem... just with dice and numbers instead of people and birthdates. $\endgroup$
    – JMoravitz
    Commented Aug 30, 2020 at 22:01
  • $\begingroup$ Oops. Sorry, I intended $\frac{4}{9}$ to be solution. $\endgroup$
    – user771227
    Commented Aug 30, 2020 at 23:57

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Complementary counting is in my opinion the simplest way to go. As one of the comments suggests your numerator is wrong (you write 6 numbers for 5 dice). Also if you arrive at a probability of $ p_1 $, then that is the probability that all dice are different. The correct solution would be $ 1 - p_1 $.

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    $\begingroup$ The way you write this is confusing. You make it sound as though the correct solution is $\frac{4}{9}$ when it is not. You meant of course that if the earlier workings and arithmetic were correct then it was the complement that was intended, but as you and I pointed out, the earlier workings were not correct. $\endgroup$
    – JMoravitz
    Commented Aug 31, 2020 at 0:00
  • $\begingroup$ Thanks for the comment, I edited my answer to make it less confusing. $\endgroup$ Commented Aug 31, 2020 at 1:26

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