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Five die are tossed and the uppermost face on each die is observed. What is the probability that the uppermost faces of these five dice at least two of the uppermost faces show the same number?

I start by finding the total number of possibilities of rolling the five dice. which is 6*6*6*6*6 = 7776 total possibilities.

For the dice one of the dice has a probability of (6c1) , this is because it has 6 choices it can be but we want to choose one to set the other dice to be this number. The other dice that will have the same number has a probability of (1c1), because there is only 1 possibility for what it can be in order to match the other dice and we choose that possibility. The other three dice have 6 possibile options for what they can be.

Therefore the probability would be ((6c1)(1c1)(6)(6)(6)/(7776)) = 0.1666 or ~17%.

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Hint: The complement event is that none of the faces show the same number.

Find one minus the probability for having all distinct numbers.


 

 


PS: $\dfrac{{^6C_1}{^1C_1}6^3}{7776}$ is the probability that two specific dice have the same value.

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  • $\begingroup$ The probability they are all different is 0.09259. (6!)/7776 so, you are saying it would be 0.90741 for them to show to same faces? This seems high. But ok, thank you. $\endgroup$ – sawreals2 Oct 25 '16 at 2:10
  • $\begingroup$ what was the flaw in my logic? $\endgroup$ – sawreals2 Oct 25 '16 at 2:11
  • $\begingroup$ Also, why is this the complement? Shouldn't the complement be all dice being the same? The complement to this would be two dice being the different and the rest being the same? $\endgroup$ – sawreals2 Oct 25 '16 at 2:12
  • $\begingroup$ You can have: 5 singletons, a pair and 3 singlton, 2 pair and a singleton, a triplet and 2 singletons, a triplet and a pair, a quadruplet and singleton, or a quintuplet. Of these, all but the first have "at least two dice with the same value." @sawreals2 $\endgroup$ – Graham Kemp Oct 25 '16 at 2:18

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