How can I calculate the probability of shaking $7$ of a kind using $10$ six-sided dice?
Chance of $7$ of a kind with $10$ dice
4 Answers
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.
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0@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. – 2011-01-11
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 $6$.
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0You 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 $\left(\frac{1}{6}\right)^9\frac{5}{6}10$ – 2011-01-11
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)
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0Yes, I thought the left is the approximation of the exact fraction on the right side not looking at the small examples. – 2011-01-11
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 $6\cdot{10\choose 7}\cdot 5^3 / 6^{10} = 625/419904$.