What"s the probability of getting a total of \$7\$ or \$11\$ when a pair of fair dice is tossed?

I already looked it up on the internet and my answer matched the same answer on a site. However, though I am confident that my solution is right, I am curious if there"s a method in which I could compute this faster since the photo below shows how time consuming that kind of approach would be. Thanks in advance.

You are watching: Probability of rolling two dice and getting a sum of 7 or 11   For \$7\$, see that the first roll doesn"t matter. Why? If we roll anything from \$1\$ to \$6\$, then the second roll can always get a sum of \$7\$. The second dice has probability \$frac16\$ that it matches with the first roll.

Then, for \$11\$, I like to think of it as the probability of rolling a \$3\$. It"s much easier. Why? Try inverting all the numbers in your die table you had in the image. Instead of \$1, 2, 3, 4, 5, 6\$, go \$6, 5, 4, 3, 2, 1\$. You should see that \$11\$ and \$3\$ overlap. From here, just calculate that there are \$2\$ ways to roll a \$3\$: either \$1, 2\$ or \$2, 1\$. So it"s \$frac236 = frac118\$.

Key takeaways:

\$7\$ is always \$frac16\$ probabilityWhen asked to find probability of a larger number (like \$11\$), find the smaller counterpart (in this case, \$3\$).
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answered Aug 14 "20 at 2:33 FruDeFruDe
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To calculate the chance of rolling a \$7\$, roll the dice one at a time. Notice that it doesn"t matter what the first roll is. Whatever it is, there"s one possible roll of the second die that gives you a \$7\$. So the chance of rolling a \$7\$ has to be \$frac 16\$.

To calculate the chance of rolling an \$11\$, roll the dice one at a time. If the first roll is \$4\$ or less, you have no chance. The first roll will be \$5\$ or more, keeping you in the ball game, with probability \$frac 13\$. If you"re still in the ball game, your chance of getting the second roll you need for an \$11\$ is again \$frac 16\$, so the total chance that you will roll an \$11\$ is \$frac 13 cdot frac 16 = frac118\$.

Adding these two independent probabilities, the chance of rolling either a \$7\$ or \$11\$ is \$frac 16+ frac118=frac 29\$.

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answered Aug 14 "20 at 2:21 Robert ShoreRobert Shore
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Gotta love stars and bars method.

The number of positive integer solutions to \$a_1+a_2=7\$ is \$inom7-12-1=6\$. Therefore the probability of getting \$7\$ from two dice is \$frac636=frac16\$.

For \$11\$ or any number higher than \$7\$, we cannot proceed exactly like this, since \$1+10=11\$ is also a solution for example, and we know that each roll cannot produce higher number than \$6\$. So we modify the equation a little to be \$7-a_1+7-a_2=11\$ where each \$a\$ is less than 7. This is equivalent to finding the number of positive integers solution to \$a_1+a_2=3\$, which is \$inom3-12-1=2\$. Therefore, the probability of getting \$11\$ from two dice is \$frac236=frac118\$

Try to experiment with different numbers, calculate manually and using other methods, then compare the result.

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answered Aug 14 "20 at 2:33 \$endgroup\$
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Welcome to the les-grizzlys-catalans.org Stack Exchange.

There sure is a quicker way; you just have to quickly enumerate the possibilities for each by treating the roll of each die as independent events.

There are six possible ways to get 7 - one for each outcome of the first die - and two possible ways to get 11 - one each in the event that the first die is 5 or 6 - meaning you have eight total possibilities . There are \$6^2=36\$ possibilities for how the two dice could roll, so you have a \$frac836=frac29\$ chance of rolling either one.

See more: What Are Things That Come In A Dozen Cool Things In Our Online Collection For 12

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answered Aug 14 "20 at 2:26
Stephen GoreeStephen Goree