If you roll two fair six-sided dice, what is the probability that the sum is $4$ or higher?

The answer is $\frac{33}{36}$ or $\frac{11}{12}$. I understand how to arrive at this answer. What I don't understand is why the answer isn't $\frac{9}{11}$? When summing the results after rolling two fair six sided dice, there are $11$ equally possible outcomes: $2$ through $12$. Two of these outcomes are below four, meaning $9$ are greater than or equal to four which is how I arrived at $\frac{9}{11}$. Can someone help explain why that is wrong?

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    $\begingroup$ There are two outcomes when you play the lottery. You either win or you lose. The probability of winning however is usually not $\frac{1}{2}$. $\endgroup$
    – JMoravitz
    Mar 9 '18 at 4:15
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    $\begingroup$ @JMoravitz: That was one my father's jokes. "What's the probability that an elephant sits in front of our porch?". "Either there's an elephant, or there isn't one. The probability must be 50% then!" $\endgroup$ Mar 9 '18 at 8:16
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    $\begingroup$ @user1551 The problem is not ill-posed, calling a die fair implies that the outcomes of different rolls are independent. $\endgroup$
    – JiK
    Mar 9 '18 at 9:40
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    $\begingroup$ @user1551 One is going to have a lot of trouble interpreting probability questions by thinking that a fair die roll is allowed to depend on external events. $\endgroup$
    – JiK
    Mar 9 '18 at 9:47
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    $\begingroup$ Great, now I'm trying to figure out how to map quantum entanglement to fair dice $\endgroup$
    – Foon
    Mar 9 '18 at 14:39

It is wrong because it is not $11$ equally possible outcome.

There is exactly $1$ way to get the sum to be $2$. ($1+1=2$)

but there is more than one way to get $3$. ($1+2=3, 2+1=3$)

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    $\begingroup$ To add to this, there are 36 equally possible outcomes. 6 on one die and 6 on the other, 6 x 6 = 36. 1 of those outcomes is a 2, 2 of those outcomes is a 3 so (36 - 3)/36 outcomes 4 or higher. $\endgroup$
    – user370160
    Mar 10 '18 at 3:41

As pointed out by others, the possible sums of the dice don't all have an equal probability of showing up. To see this, you can write it all out:

Die 1 | Die 2 | result
  1       1       2
  1       2       3
  1       3       4
  6       5      11
  6       6      12

When you look at the resulting table, there are 36 combinations. 1 of those is a 2 (ie you have a 1 in 36 chance of getting a 2 from D1 + D2), 2 of those are '3' etc.

Now it's easy to see how to get the chance of getting a 4 or higher.

  • $\begingroup$ I like this because it's trivial to do, but it makes everything suddenly obvious. $\endgroup$
    – Beska
    Mar 9 '18 at 13:16

$\frac9{11}$ is wrong precisely because it assumes the probabilities of getting each sum are equal. Here they are not: there's only one way to roll a sum of two (the snake eyes of gambling jargon) but six ways to roll a seven.


if the correct answer is $ \frac {33}{36}$ it means there are 36 possible results.. You should try to write them down..It will be more clear. You just used equiprobability in a case where there isn't equiprobability..$P(1)=0 \quad P(2) =1/36,\quad P(3)=2/36 ,\quad P(4)=3/36,$

$ P(5)=4/36 \quad P(6)=5/36 \quad P(7)=6/36 \quad P(8)=5/36 \quad $

$ P(9)=4/36 \quad P(10)=3/36 \quad P(11)=2/36 \quad P(12)=1/36$

And the answer is just: $$P=1-P(1)-P(2)-P(3)=\frac {33}{36}$$

  • $\begingroup$ Nitpick: if the answer is $\frac{x}{36}$ then there are at least 36 possible results, but if you don't know it's dice, there could b $36*n$possibilities, with $x*n$ desired outcomes $\endgroup$ Mar 9 '18 at 16:19
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    $\begingroup$ "If the correct answer is $\frac{33}{36}$ it means there are 36 possible results." Definitely not. For one: I can build a weighted coin that comes up heads 33 times out of 36, despite there being only two possible outcomes. For another: $\frac{33}{36}=\frac{3300}{3600}$ and it would be silly to claim that this therefore means there are 3600 possible results. $\endgroup$ Mar 9 '18 at 16:39
  • $\begingroup$ yes @DanielWagner you are right $\endgroup$
    – MtGlasser
    Mar 9 '18 at 16:50

Your probabilities are wrong. For each value of die 1, there are 6 possible values of die 2, the available scores for a roll of two dice are: (7 is the most likely score)







So, out of the 36 total possible outcomes, 33 of them have a value of 4 or higher, or 33:36, or 11:12


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