One of the discussions we looked at critical time involved rolling 3 dice and also getting at least one six. I didn’t walk into detail on the calculate there; yet I found another place wherein we questioned it at length. We’ll look at at the here.

You are watching: Probability of rolling a 6 with 3 dice

## A wrong way and a ideal way

The question came in 2013:

At the very least One "6" in three Dice Rolls, summed SimplyI recognize this is wrong, but not why.If you have actually 3 fair six-sided dice, and you great to find the probability of rolling at least one 6, why can"t you add the separation, personal, instance probabilities of obtaining a 6?In other words, if 1/6 is the probability of gaining a 6 in one throw, why is 1/6 + 1/6 + 1/6 no the probability of obtaining a 6 in 3 throws?I understand it is dorn to perform so, however why, exactly? where is the logical error in reasoning when adding the probabilities?In situation you wonder even if it is or not I know just how to actually fix such a basic problem, the actual systems is P(at the very least one 6 out of three throws) = 1 - P(no six) = 1 - (5/6)^3And I understand that calculating the probability for 6 tosses by adding them would give a probability that 1, i beg your pardon is impossible. Yet again, whereby is the logical error in including them?Eway has currently made 2 comments ns would have actually made if that hadn’t!

First, we have the right to tell that adding probabilities can’t it is in right, due to the fact that no issue what the probability of an event happening on one trial, if you add up enough of them, the total will exceed 1, and also no probability can be higher than 1.

Second, the quickest means to discover an “at the very least one” probability is to an initial find the probability that “none”. These room complementary events: If the result is no “at least one”, climate it is “none”, and also if that is not “none”, then it is “at least one”. Therefore the probability we want is 1 minus the probability of none, which is easily found by multiplication if the individual trials room independent.

## Why the not correct is wrong

I replied, starting with what friend can’t do:

You have the right to only add the probabilities of support exclusive occasions (unless you make adjustments, together I"ll show). This is since if you include events the "overlap," you room counting some possible outcomes more than once.I’ll have more to say around this below. But I can use a Venn diagram to highlight the issue. Quite than placed the variety of items in each region of the diagram, together we normally do, I will put an example of a roll; for example, (1, 2, 3) will mean we roll very first a 1, climate a 2, then a 3, for this reason that we rolled no sixes.

Because the the overlaps, adding the number (or probability) of rolls with the first, the second, and the 3rd being 6 will significantly overcount the rolls. Because that example, the role (6,6,3) would be counting twice, for the very first roll and for the 2nd roll.

## More appropriate ways

But the Venn diagram suggests several other means to perform it:

There are, however, alternate ways that perform involve adding. One way to rest up the occasion "at least one 6" right into mutually exclusive events is: P(exactly one 6) + P(exactly two 6"s) + P(exactly 3 6"s)This offers C(3,1)*(1/6)^1*(5/6)^2 + C(3,2)*(1/6)^2*(5/6)^1 + C(3,3)*(1/6)^3*(5/6)^0 = 3*25/216 + 3*5/216 + 1*1/216 = 91/216What does this mean? The probability that us roll specifically one six is 3 (the variety of ways to choose which one will be a six — the number of one-set areas in the diagram), times the probability the that roll will be a 6 (1/6), time the probability the the various other two rolls will not be 6 (5/6*5/6). Likewise for the various other two probabilities.

This takes some work, but is really straightforward.

This agrees through your calculation: 1 - (5/6)^3 = 216/216 - 125/216 = 91/216This is what Eway had actually done, i m sorry is the technique we commonly teach. The probability that *none* the the 3 rolls is a 6 is 5/6 (the probability that an individual roll is *not* a six), raised to the 3rd power. The probability the this walk *not* take place (that is, the at the very least one role *is* a six) is 1 minus that.

So currently we have actually two means to calculation the same answer.

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You asked about calculating this probability this way: P(first is 6) + P(second is 6) + P(third is 6)This would certainly count roll with an ext than one 6 more than once; e.g., the role 3, 6, 6 is count as having actually the 2nd roll it is in 6 and also having the third roll it is in 6.In order come correct because that this, friend would need to use the inclusion-exclusion principle, subtracting the end the outcomes count twice, however then adding back in the outcome that was added three times and then subtracted out three times: P(1st is 6) + P(2nd is 6) + P(3rd is 6) - P(1st and second are 6) - P(2nd and third are 6) - P(1st and third are 6) + P(all three space 6) = 1/6 + 1/6 + 1/6 - 1/36 - 1/36 - 1/36 + 1/216 = 36/216 + 36/216 + 36/216 - 6/216 - 6/216 - 6/216 + 1/216 = 91/216So there room three valid means to execute it. Which do you prefer?Looking earlier at ours Venn diagram, us are adding the 3 circles together, i beg your pardon counts every of the two-circle areas twice, and counts the middle region three times. Then us subtract each of the intersections of two circles to compensate because that the overlaps; yet this subtracts the middle region three times, totally removing that from our count (as it had been counted 3 times initially). So we have to include it back in.

This is an extension of the formula for the union of two sets:

(P(Acup B) = P(A)+P(B)-P(Acap B))

For 3 sets, the is

(P(Acup Bcup C) = P(A)+P(B)+P(C)-P(Acap B)-P(Bcap C)-P(Ccap A)+P(Acup Bcap C))