How can I calculate the probability of shaking $7$ of a kind using $10$ six-sided dice?


If you want exactly $7$ of a kind, you use binomial distribution to first find the probability that you roll $7$ of the same "good" number in $10$ trials. The binomial distribution shows that this probability is $$ \binom{10}{7}\left(\frac{1}{6}\right)^7\left(\frac{5}{6}\right)^3, $$ since the probability of success is $1/6$.

Then there are six numbers to choose one from which could be the "good" kind of which you want $7$. Hence you multiply the above expression by $\binom{6}{1}$: $$ \binom{6}{1}\binom{10}{7}\left(\frac{1}{6}\right)^7\left(\frac{5}{6}\right)^3. $$

Note that this gives the exact probability $\frac{625}{419904}\approx\frac{1}{672}$, consistent with the approximate value given by Wolframalpha.

  • $\begingroup$ Your last sentence is not correct. Notice what happens if you try to use this to determine the probability of 2 of a kind, for instance. $\endgroup$ – Matthew Conroy Jan 11 '11 at 20:44
  • $\begingroup$ @Matthew Conroy, thank you, I have removed it. Was the issue with the power of $3$? I was assuming that when changing the $7$, one would also change the $3$ to a new value, say $8$ in the case of $2$ of a kind. Is that the only issue, or am I missing something else? $\endgroup$ – yunone Jan 11 '11 at 20:49
  • $\begingroup$ The trouble is that the (5/6)^(10-n) bit allows the "other" dice to be anything, which could include pairs, or triples, etc. So when you consider the 2 of a kind case, for instance, you would not be considering the fact that one can have more than one two of a kind at a time. Hence, when multiplying everything by 6, you will have over-counted the ways to have a two of a kind. $\endgroup$ – Matthew Conroy Jan 11 '11 at 20:57
  • $\begingroup$ @Matthew, interesting, I hadn't thought about that. So for 5 of a kind of smaller, the problem becomes a little different? Thanks again for catching that error. $\endgroup$ – yunone Jan 11 '11 at 21:02

To be short: The result is $p\approx \frac{1}{672}$

You can use the second formula in the Wikipedia article and sum the probabilities for every face of 1-6: http://en.wikipedia.org/wiki/Liar%27s_dice#Basic_dice_odds

Note that Wolframalpha is also able to get this: http://www.wolframalpha.com/input/?i=10+dice (Press on more under probabilities)

  • 1
    $\begingroup$ Though the links are helpful, this answer is not correct: 1/672 is only an approximation. See yunone's correct answer. $\endgroup$ – Matthew Conroy Jan 11 '11 at 20:28
  • $\begingroup$ Thanks for the hint, I adjusted it and agree that yunones answer is better. Still the formula he mentions is basically the one that I linked on Wikipedia :-) Stupid me in trusting that wolframalpha calculates it accurately $\endgroup$ – Listing Jan 11 '11 at 20:33
  • $\begingroup$ Note all the "approximately equal" signs that WolframAlpha uses on that page. For instance, 0.2718 is clearly not 1/4. $\endgroup$ – Matthew Conroy Jan 11 '11 at 20:40
  • $\begingroup$ Yes, I thought the left is the approximation of the exact fraction on the right side not looking at the small examples. $\endgroup$ – Listing Jan 11 '11 at 20:45

Calculate the probability of exactly $7$ sixes.
Calculate the probability of exactly $8$ sixes.
Calculate the probability of exactly $9$ sixes.
Calculate the probability of exactly $10$ sixes.
Add them all up, and multiply by $6$.

  • $\begingroup$ Thanks for your answer, that is helpful. One more question though. I can calculate the probability of exactly 10 sixes but how do I calculate 9 out of 10 being a certain number? $\endgroup$ – Shaun Bowe Jan 11 '11 at 19:20
  • $\begingroup$ You need 9 dice to be 6 and 1 to be something else. There are 10 ways to select which one is the something else. So the probability is $\left(\frac{1}{6}\right)^9\frac{5}{6}10$ $\endgroup$ – Ross Millikan Jan 11 '11 at 20:58

Here's another way to get the same answer (and let's be explicit that we're looking for the probability of exactly 7 of a kind, not at least 7 of a kind):

Write the result as a sequence, e.g., the ten rolls might come out $(2,5,4,6,5,4,3,1,2,2)$. Obviously there are $6^{10}$ such sequences, each equally likely.

How many sequences have 7 of a kind? There are 6 possibilities for what we have 7 of, and there are $10\choose 7$ ways to pick the 7 places that are the same. Finally, there are $5^3$ ways to fill the other three places, since there are 5 choices for each place.

So the answer is $6\cdot{10\choose 7}\cdot 5^3 / 6^{10} = 625/419904$.


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