An unbiased die having the numbers 1,2,3,4,5,6 is rolled 4 times. What is the probability that the minimum face value is 2?

According to my reasoning the answer should be $\frac{5^4}{6^4}$ because we have the options 2,3,4,5 and 6 . Which can be chosen for four trials in $5^4$ ways .

However the correct answer seems to be $\frac{5^4-4^4}{6^4}$ and I can’t reason it out. Please help me where I’m missing out

Thank you !

  • 1
    $(5/6)^4$ is the probability that the minimum face value is $\ge 2$. – ncmathsadist Oct 19 at 17:34
  • @ncmathsadist so do you think that my answer is correct ? Please help – Aditi Oct 19 at 17:35
up vote 5 down vote accepted

There are $5^4$ roll sequences formed from the numbers 2, 3, 4, 5, 6 (and thus having minimum digit at least 2). Of these, $4^4$ roll sequences are formed from 3, 4, 5, 6, so cannot have minimum digit 2. The remaining $5^4-4^4$ sequences thus have minimum digit exactly 2, as required, and dividing by the $6^4$ rolls in total yields the correct probability.

  • Ohhh I didn’t know that it was strict to have 2 in the answer . They never mentioned that $2$ must be one of the outcomes though so I was a bit confused. How do you deduce that $2$ must be one of the outcomes ? – Aditi Oct 19 at 17:38
  • 3
    @Aditi The minimum face value is stipulated to be 2. This means that at least one die actually shows 2, to realise the minimum. – Parcly Taxel Oct 19 at 17:38
  • Ohh now I get it ! Thank you :) – Aditi Oct 19 at 17:39


Your answer would be good if the question were "What is the probability that the minimum face is not $1$"?

  • Thank you , but how do we deduce that 2 must be one of the outcomes ? – Aditi Oct 19 at 17:38
  • You can't. Every roll could be fives and sixes (for example). There it is the point. – ajotatxe Oct 19 at 17:39
  • Thank you for the hint ! – Aditi Oct 19 at 17:43

You can also calculate the numerator directly: ${4 \choose 1}*4^3$ sequences have a single 2; ${4 \choose 2}*4^2$ have two 2's; ${4 \choose 3}*4^1$sequences have 3 2's; 1 sequence of 2222 (all sequences containing no 1's). If you add these up, you get $369=5^4-4^4$.

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