I know that the probability should be over or exactly 1 since out 6 possible values the 7th dice will always be a duplicate.

My calculations are wrong though:

$\frac{{7\choose2} 5! 6*1}{6^7}$

Why? I'm guessing my error is in the denominator.

Edit: any one pair excluding triplets, full house etc

Edit 2: My way of thinking was: I need 6 distinct values on 6 of the dice and the 7th dice's value has to be equal to one of the other 6.

${7\choose2}$ = the different combination of dice pairs out of 7 dice

5! = the other combination of distinct values from the non pair dice

6 = how many numbers I get to choose my pair from

1 = ${6\choose6}$ choose value out of the distinct 6 values.

  • 1
    $\begingroup$ No probability is ever greater than $1$. $\endgroup$ – Ravi May 13 '16 at 21:58
  • $\begingroup$ The probability that two or more dice match is of course $1$, no computation needed. One could do an Inclusion/Exclusion calculation, but it would be tedious, and, if we did not make a mistake, we would end up at $1$. The probability that exactly two dice match is well under $1$. Is that the probability that you want to compute? If so, you have done it correctly. $\endgroup$ – André Nicolas May 13 '16 at 21:59
  • $\begingroup$ It can't be over 1. Yes, by the "pigeonhole principle" it must be 1. As for you calculation, I guess I would need some exposition on why you think that might be the answer, or how you chose each expression in that equation. $\endgroup$ – Doug M May 13 '16 at 21:59
  • $\begingroup$ Probabilities are never "over... 1". As you say, there cannot be seven distinct values (on rolling six-sided dice), so the probability is one. But it is unclear what your calculation is supposed to represent. You just give an expression, without explaining how it should be interpreted as a probability of an outcome. $\endgroup$ – hardmath May 13 '16 at 22:00
  • $\begingroup$ I just edited my question to make it clear how I thought about it. I understand now why it's indeed less than one now..but I need the probability and calculations for any one pair. $\endgroup$ – Del Wolf May 13 '16 at 22:23

You seem to calculate the probability of exactly a pair (i.e., no triplet, no two pairs, etc.) and that is indeed smaller than $1$.


The probability of rolling two $1´s$ at first is $\frac{1}{6}\cdot\frac{1}{6}$. The probability of rolling $ 2,3,4,5,6$ next is $\frac{1}{6}\cdot\frac{1}{6} \cdot\frac{1}{6} \cdot\frac{1}{6} \cdot \frac{1}{6} $.

The first two rolls can be also a pair of $ 2,3,4,5$ or $6$. Therefore an extra factor of $6$ is needed. And the outcome of the $7$ rolls can be arranged in $\frac{7!}{2!}$ ways.

In total the probability to get $\texttt{exactly}$ one pair is $ \frac{7!\cdot 6}{2!\cdot 6^7}=\frac{35}{648}\approx 5.4\%$


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