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If we had a 4 sided dice (numbers 1,2,3,4 on the faces) and rolled it 5 times and recorded the results. What would be the probability that we rolled the same number exactly 3 times? What about 3 or more times?

For finding out the probability of getting 3 matching numbers in our sequence of rolls, I came up with the following. $${5 \choose 3}×(4/4×1/4×1/4×3/4×3/4) = .3516$$ It does not intuitively seem correct at first, but I am trying to model that the first roll can be anything (4/4) and the next roll must match whatever it got, thus (1/4), and same thing for the next roll (1/4). The last two must not match the first, so I am finding (3/4) for both of those rolls. I am multiplying the whole thing by the different ways that those matching numbers could have come up in the sequence (5 choose 3). The result seems kind of high to me, but I understand that probability is often not intuitive for most people.

Given that my logic above is correct, could this be expanded in some way to account for the second part of my question about 3 or more matches?

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  • $\begingroup$ can't you just calculate same thing for rolling same number exactly 4 times and 5 times then add them up? This shouldn't be hard, since numbers are not that big or complicated $\endgroup$ – user160738 Mar 4 '15 at 18:05
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Your logic is correct. The probability of rolling exactly three $4$'s in five rolls of the die is $$\binom{5}{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2$$ where $(1/4)^3$ is the probability of rolling a $4$ three times, $(3/4)^2$ is the probability of rolling a number other than $4$ twice, and $\binom{5}{3}$ represents the number of ways that three of the five rolls could produce a $4$. The same calculation applies to rolling a $1$, $2$, or $3$ exactly three times in five rolls, so the probability that the same number is rolled exactly three times is $$4\binom{5}{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2$$ which agrees with your result.

The probability of rolling the same number at least three times in five rolls is found by adding the probabilities of rolling the same number exactly three times, exactly four times, and exactly five times, which is $$4\left[\binom{5}{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2 + \binom{5}{4}\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right) + \left(\frac{1}{4}\right)^5\right]$$

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  • $\begingroup$ @n-f-taussig If I were to expand my example to rolling the 4-sided dice 6 times instead of 5, with the same basic question 3 or more matches, would the pattern be basically the same? Or would there be additional work because you could have two groups of 3 (ex {1,1,1,3,3,3}? $\endgroup$ – user213351 Mar 4 '15 at 18:43
  • $\begingroup$ There would be additional work since you could have two groups of $3$. $\endgroup$ – N. F. Taussig Mar 4 '15 at 19:03
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As the probability for rolling the same number exactly 3 times is:

${5\choose3}\times(4/4\times1/4\times1/4\times3/4\times3/4)={5\choose3} (1/4)^2 (3/4)^2$

The probability for rolling the same number exactly 4 and 5 times can be found in a similar way:

Probability for rolling the same number exactly 4 times is:

${5\choose4}\times(4/4\times1/4\times1/4\times1/4\times3/4)={5\choose4} (1/4)^3 (3/4)$

Probability for rolling the same number exactly 4 times is:

${5\choose5}\times(4/4\times1/4\times1/4\times1/4\times1/4)={5\choose5} (1/4)^4$

So, to find the probability for rolling the same number 3 times or more, just sum the three probabilities (as they are mutually exclusive):

$\sum\limits_{i=3}^5 ({5\choose i}\times(1/4)^{i-1}\times(3/4)^{5-i})$

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