# Probability of rolling 5 dice?

Say you picked $\{4 5 2 6 5\}$ to show up in the dice rolls. You win if you have 3, 4, or 5 numbers in the correct sequence.

• 3 numbers in correct sequence would be; $\{4 5 2\}$ OR $\{5 2 6\}$ OR $\{2 6 5\}$
• 4 numbers in the correct sequence would be; $\{4 5 2 6\}$ OR $\{5 2 6 5\}$
• 5 numbers in the correct sequence would be; $\{4 5 2 6 5\}$

What would be the probability of winning? How would I calculate it?

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• First three dice match: $\frac 1{6^3}$.
• First is wrong, the next three match: $\frac56\cdot \frac 1{6^3}$.
• Second is wrong, the next three match: $\frac56\cdot \frac 1{6^3}$.

In total: $(1+\frac 56+\frac56)\frac 1{6^3}=\frac1{81}$.

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3 numbers correct: $(\frac{5}{6})^2\cdot\frac{1}{6^3}\cdot\binom{5}{2}$

4 numbers correct: $\frac{5}{6}\cdot\frac{1}{6^4}\cdot\binom{5}{1}$

5 numbers correct: $\frac{1}{6^5}$

total: $\frac{1}{6^5}\bigg(1+5\cdot\binom{5}{1}+5^2\cdot\binom{5}{2}\bigg)=\frac{276}{7776}=\frac{23}{648}$

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The probability of getting exactly $3$ correct numbers in sequence is the sum of $\left(\frac16\right)^3\cdot\frac56\cdot1$ (the $3$ are the first $3$), $\frac56\left(\frac16\right)^3\frac56$ (the $3$ are the middle $3$), and $1\cdot\frac56\cdot\left(\frac16\right)^3$ (the $3$ are the last $3$). –  Brian M. Scott Oct 20 '12 at 19:34

3 numbers in correct sequence

• for {452} = 2*(1/6)^3 *(5/6) + (1/6)^3

• for {526 }= 2*(1/6)^3 *(5/6) + (5/6)^2 *(1/6)^3

• for {265} = 2*(1/6)^3 *(5/6) + (1/6)^3

4 numbers in correct sequence

• for {4526} = (1/6)^4 *(5/6) + (1/6)^4
• for {5265} = (5/6)*(1/6)^4 + (1/6)^4

5 numbers in correct sequence

• that is {45265} = (1/6)^5