Probability of rolling at least one 6 in 3 tries.

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Hoodleehoo

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Can someone explain this to me? I'm studying for the GRE and am stuck on this concept.

"If you roll a dice three times, what is the probability of rolling a 6 at least once?"

The correct answer is 91/216

I can figure it out by calculating the chance of not rolling any 6's and then subtracting it from 1,but I can't figure out how to calculate it directly.

I tried 1/6 + 1/6 + 1/6 but that = 1/2. What am I missing?? I thought I understood the concept, but obviously I don't. The book says for "or" questions to use addition (assuming mutually exclusive). This one really tripped me up.

Thanks so much in advance!!
 
Hoodleehoo said:
Can someone explain this to me? I'm studying for the GRE and am stuck on this concept.
"If you roll a dice three times, what is the probability of rolling a 6 at least once?"
The correct answer is 91/216
Click to expand...
There are \(\displaystyle 6\cdot 6\cdot 6\) ways to roll three dice.
In those \(\displaystyle 216\) ways there are \(\displaystyle 5\cdot 5\cdot 5=125\) ways not to roll a six.

The opposite of none is at least one.
 
pka said:
There are \(\displaystyle 6\cdot 6\cdot 6\) ways to roll three dice.
In those \(\displaystyle 216\) ways there are \(\displaystyle 5\cdot 5\cdot 5=125\) ways not to roll a six.

The opposite of none is at least one.
Click to expand...

Right I can do it that way. But why doesn't 1/6 + 1/6 + 1/6 work? Can't it be restated "what is the probability of rolling a 6 on the first roll OR a 6 on the second roll OR a 6 on the third roll"? In which case you'd add the three fractions together.
 
Hoodleehoo said:
Right I can do it that way. But why doesn't 1/6 + 1/6 + 1/6 work? Can't it be restated "what is the probability of rolling a 6 on the first roll OR a 6 on the second roll OR a 6 on the third roll"? In which case you'd add the three fractions together.
Click to expand...

Your reasoning here is only somewhat correct. You're right that adding up all the possible ways of rolling at least one 6 will give you the desired answer. But, where you went wrong is that your calculation only accounts for the probability of rolling exactly one 6. In other words, you have to account for the probability of rolling two 6's and three 6's. If we let N be the number of sixes rolled, then the notation would look something like this:

P(N > 1) = P("6 on only first dice") + P("6 on only second dice") + P("6 on only third dice") + P("6 on first and second dice) + P("6 on first and third dice) + ... + P("6 on all dice")

Pka's method is far easier than working with that nasty expression:

P(N > 1) = 1 - P(N = 0)
 
Hoodleehoo said:
Right I can do it that way. But why doesn't 1/6 + 1/6 + 1/6 work? Can't it be restated "what is the probability of rolling a 6 on the first roll OR a 6 on the second roll OR a 6 on the third roll"? In which case you'd add the three fractions together.
Click to expand...
You can also do this: \(\displaystyle \mathcal{P}(N\ge 1)=1-\left(\frac{5}{6}\right)^3=\frac{91}{216}\).
That is subtracting three non-sixes, gives at least one six.
 
Last edited:
ksdhart2 said:
Your reasoning here is only somewhat correct. You're right that adding up all the possible ways of rolling at least one 6 will give you the desired answer. But, where you went wrong is that your calculation only accounts for the probability of rolling exactly one 6. In other words, you have to account for the probability of rolling two 6's and three 6's. If we let N be the number of sixes rolled, then the notation would look something like this:

P(N > 1) = P("6 on only first dice") + P("6 on only second dice") + P("6 on only third dice") + P("6 on first and second dice) + P("6 on first and third dice) + ... + P("6 on all dice")

Pka's method is far easier than working with that nasty expression:

P(N > 1) = 1 - P(N = 0)
Click to expand...

Thank you so much! This is excellent. ☺️
 
Hoodleehoo said:
Right I can do it that way. But why doesn't 1/6 + 1/6 + 1/6 work? Can't it be restated "what is the probability of rolling a 6 on the first roll OR a 6 on the second roll OR a 6 on the third roll"? In which case you'd add the three fractions together.
Click to expand...
1/6
1/6 + 1/6
1/6 + 1/6 +1/6
1/6 + 1/6 + 1/6 + 1/6
1/6 + 1/6 + 1/6 + 1/6 +1/6
1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6
....

These sums are getting bigger. Do you really think that the chance of rolling 6's back to back is greater than rolling a die and getting a 6?

How about rolling three 6s in a row. Is that going to happen more often than rolling 2 sixes in a row?
Of course not.

Using your method, the chances of rolling six 6's in a roll is 1. That means it will always happen.
How about rolling seven 6's in a row, that probability will be more than 1??

Even *if* what you did is corect you only considered the cases when you get one six. The problem asks for the probability of getting at least one 3.

P(getting at least one 6) = p(getting exactly one six OR getting exactly two 6's OR getting exactly 3three 6's) = p(getting exactly one six) +p(getting exactly two 6's) + p(getting exactly 3three 6's).

There are three probabilities to calculate vs only one using complements.

Let ~6 mean you rolled something other than a 6.

p(getting exactly one six) p({6,~6, ~6} OR {~6, 6, ~6} OR {~6, ~6, 6}) = 3*p({6,~6, ~6})= 3*p(6 AND THEN ~6 AND THEN ~6) 3*(1/6)(5/6)(5/6)=75/216

p(getting exactly two six) p({6, 6, ~6} OR {6, ~6, 6} OR {~6, 6, 6}) = 3*p(6, 6, ~6) = 3*(1/6)(1/6)(5/6) = 15/216

p(getting exactly three six) p(6, 6, 6) = (1/6)(1/6(1/6) = 1/216

Adding these up will give you p(at least one 6) =91/216
 
Jomo said:
1/6
1/6 + 1/6
1/6 + 1/6 +1/6
1/6 + 1/6 + 1/6 + 1/6
1/6 + 1/6 + 1/6 + 1/6 +1/6
1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6
....



These sums are getting bigger. Do you really think that the chance of rolling 6's back to back is greater than rolling a die and getting a 6?

How about rolling three 6s in a row. Is that going to happen more often than rolling 2 sixes in a row?
Of course not.

Using your method, the chances of rolling six 6's in a roll is 1. That means it will always happen.
How about rolling seven 6's in a row, that probability will be more than 1??

Even *if* what you did is corect you only considered the cases when you get one six. The problem asks for the probability of getting at least one 3.

P(getting at least one 6) = p(getting exactly one six OR getting exactly two 6's OR getting exactly 3three 6's) = p(getting exactly one six) +p(getting exactly two 6's) + p(getting exactly 3three 6's).

There are three probabilities to calculate vs only one using complements.

Let ~6 mean you rolled something other than a 6.

p(getting exactly one six) p({6,~6, ~6} OR {~6, 6, ~6} OR {~6, ~6, 6}) = 3*p({6,~6, ~6})= 3*p(6 AND THEN ~6 AND THEN ~6) 3*(1/6)(5/6)(5/6)=75/216

p(getting exactly two six) p({6, 6, ~6} OR {6, ~6, 6} OR {~6, 6, 6}) = 3*p(6, 6, ~6) = 3*(1/6)(1/6)(5/6) = 15/216

p(getting exactly three six) p(6, 6, 6) = (1/6)(1/6(1/6) = 1/216

Adding these up will give you p(at least one 6) =91/216
Click to expand...

Read the question again.
 
Hoodleehoo said:
Read the question again.
Click to expand...
OK, I read the question again.
So using your method, if you roll a die 7 seven then the probability of getting at least one 6 is 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 7/6 =1 1/6.
 
if you roll a die 7 seven then the probability of getting at least one 6
Click to expand...

ANSWER: \(\displaystyle 1-\left(\frac{5}{6}\right)^7\)
 
Jomo said:
OK, I read the question again.
So using your method, if you roll a die 7 seven then the probability of getting at least one 6 is 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 7/6 =1 1/6.
Click to expand...

? I'm pretty sure that's impossible
 
Hoodleehoo said:
? I'm pretty sure that's impossible
Click to expand...
Exactly. That is the point I am making. If your method works, ie p(at least one 6 in three rolls) = 1/6 + 1/6 + 1/6, then it would follow that p(at least one 6 in seven rolls) = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1 1/6.
Are you really having a hard time following that logic? Now since p(any event) is between 0 and 1 inclusive, it follows that what is above is wrong.
 
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