The probability of rolling 2 dice and getting a 6 on either one of the die or both is : 11/36 or about 0.305.

Also I calculate the probability of rolling 4 dice and getting a 6 on either one, two, three or all four dice is : 421/1296 or about 0.32.

Is that correct? I am just surprised to find both the probabilities so close together.

Thank you.

  • 3
    $\begingroup$ The probability is 1 less the probability of no one six in the throw, i.e. $\Pr[X\ge 1]=1-\Pr[X=0]=1-\left(\frac56\right)^{4}$ where $X$ is the random variable that count the number of 6's in the throw. $\endgroup$
    – Masacroso
    May 23 '16 at 13:13

The probability of getting at least $1$ six when rolling four dice is

$$ 1-(5/6)^{4}=1-625/1296=671/1296\approx 0.518 $$

  • $\begingroup$ Hello and thank you so much for the quick reply! $\endgroup$
    – Oleo One
    May 23 '16 at 13:24

Method #$1$ - split it into disjoint events, and add up their probabilities:


Method #$2$ - calculate $1$ minus the probability of the complementary event:


The result in both cases is: $$\frac{671}{1296}$$

  • $\begingroup$ Hello and thank you so much for detailed reply! But I rather new to math and I am having alittle trouble finding where I am making an error : I am calculating the probability for getting six when rolling 4 dice like this : (1/1296) if all dice roll 6 + ( 20/1296) if 3 dice roll 6 + (100/1296) is 2 dice roll 6 + (300/1296) if 1 die rolls 6. Which gives me a total of (421/1296) what am I doing wrong? $\endgroup$
    – Oleo One
    May 23 '16 at 13:32
  • 1
    $\begingroup$ @OleoOne: Two errors in ($20/1296$) if 3 dice roll "6". First, you are not taking into consideration the fact that you have $\binom43$ different groups of $3$ out of $4$ dice. Second, you are not taking into consideration the fact that the $4$th die might also show "6", in which case, you are "double-counting" with the case of 4 dice roll "6". Of course, these two types of errors also apply for the rest of your cases. $\endgroup$ May 23 '16 at 14:08
  • $\begingroup$ Thank you! That eplains it beautifully. I finally understand the several errors I was making - I have the correct answer now. Thanks again for your help. $\endgroup$
    – Oleo One
    May 24 '16 at 17:41
  • $\begingroup$ @OleoOne: You're welcome. Feel free to accept it as the correct answer if you consider it the most appropriate one (since only one can be accepted), by clicking on the V next to it... :) $\endgroup$ May 24 '16 at 19:18

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