Help with a Dice problem (need an expert although the problem seems simple)

Count the number of times a 6 doesn't occur, and subtract it from the number of possible outcomes.

There are 6^3 = 216 different ways to roll three dice.

There are 5^3 = 125 different combinations without 6's.

The probability of getting at least one six is 1 - (125/216) = 42%.

- Warren

 

Did the guess of 50% come from multiplying the probability of rolling a 6 on one roll by the number of rolls? This is most definitely not the correct answer.

 

Well, then you get the fun of telling your friend he is wrong.

- Warren

 

Perhaps you should point out to your friend that his theory gives a probablity of 117% of rolling a 6 in seven rolls. Not only is a probability greater than 100% meaningless, it isn't even true. It's clearly possible to roll a die 7 times and to not roll a 6 on each one.

 

Let's see:

Total number of combinations is 6*6*6=216.

Number of combinations with exactly one 6 is 25*3=75.

Number of combinations with exactly two 6's is 5*3=15.

Number of combinations with exactly three 6's is one.

Thus the total number of combinations with at least one 6 is 91. The probability of rolling at least one 6 on three die is 91/216=42%.

That's my guess.

Doug

 

I explained the answer. I also believe several of us explained the answer quite simply. You have even indicated that you understand the answer. We also explained why your answer cannot be correct.

If you don't believe us, brute force it. Write down every single combination possible (it won't take THAT long, there are only 216 ways) and cross off all the ones without any sixes. You'll see for yourself.

42%.

Yes, it's the exact same question. All you're asking is "given three independent dice rolls, what is the probability that there will be at least one six?" I don't understand why you would think this new question is any different. Why would the person rolling the dice make any difference?

- Warren

 

Chroot gave a simpler explanation than the others. I see no reason for you to insult him.

 

I don't really know how to be any more clear than my first post, but I'll try.

You do agree, I assume, that there are 216 possible combinations. In case you don't, you can calculate them as such: there are six possible outcomes upon rolling the first die. For each of those six, there are six outcomes upon rolling the second. For each of those 36, there are six outcomes upon rolling the third. Thus, there are 6*6*6 = 216 different outcomes.

Now, if you want to find the probability of "at least one six," you must recognize that it is exactly the opposite of the probability of "no sixes at all."

It turns out to be much easier to calculate the number of combinations with no sixes than it is to calculate the number of combinations with 1, 2, or 3 sixes (though Dough showed that method, also).

How many outcomes have no sixes? You roll the first die, and there are five outcomes that are not six. For each of those five outcomes, there are five outcomes upon rolling the second die that are not six. And for each of those 25 outcomes, there are five outcomes upon rolling the third die that are not six. That's 5*5*5 = 125 possible outcomes with no sixes.

The probability of getting no sixes at all is thus 125 out of 216, or ~58%. The answer to the "opposite" question, "what is the probability of at least one six," can be found by simply subtracting this probability from one.

Thus the probability is 100% - 58% = 42%.

The reason you can't treat the dice rolls as independent is simple: they aren't. Your question is phrased as "what is the probability that there is at least one six?" This means that the answer to the question depends not on a single die, but on all three simultaneously. It doesn't matter what the second and third dice do, if the first die is a six -- thus the rolls are dependent.

A somewhat different question, "What is the probability that there will be exactly one six?" results in independent probability, with which you seem to be familiar. The probability of exactly one six is the sum of the three ways it can happen:

[tex]
\left( \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \right)
+ \left( \frac{5}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} \right)
+ \left( \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} \right) \approx 34 \%
[/tex]

The question "what is the probability of exactly two sixes?" can be found by summing the three ways it can happen:

[tex]
\left( \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} \right)
+ \left( \frac{5}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \right)
+ \left( \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} \right) \approx 7 \%
[/tex]

The question "what is the probability of exactly three sixes?" is, as Doug said, 1 out of 216:

[tex]\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \approx 0.5 \%[/tex]

Add them together -- it's about 42%.

Hopefully you understand now how to bridge the gap between your view of the dice rolls as being independent, and the view of them actually being dependent. Note that Doug said all of this already, albeit with less gusto.

- Warren

 

Oh, and Bryan, I might as well point out that you did not "specifically ask [me] for clairty beyond [my] first post" at all. In fact, you just reiterated your incorrect view of things, and went so far as to declare "There is no reason to combine the dice rolls that I can see, as each roll does not affect the others." As such, it didn't quite sound like you wanted help.

As you can hopefully see by now, the dice rolls are dependent.

- Warren

 

Shouldn't this be:


The question "what is the probability of exactly three sixes?" is, as Doug said, 1 out of 216:

[tex]\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \approx 0.5 \%[/tex]

 

I was a bit confused at first as to why you asked the same question twice and got different results. I think I figured it out. You meant
The question "what is the probability of exactly three sixes?" is, as Doug said, 1 out of 216:

[tex]\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \approx 0.5 \%[/tex]

Is this correct?

 

Whoooops, thanks master_coda and Stephen, you guys are quick!

- Warren

 

The horse may be cold but there's another way to beat it

The probability of getting a 6 on the first roll is:

[tex]\frac{1}{6}[/tex]

The probability of getting a 6 on the second roll given that we didn't get a 6 on the first is:

[tex]\frac{5}{6}\cdot\frac{1}{6}[/tex]

The probability of getting a 6 on the third roll given that we didn't get a 6 on the first two is:

[tex]\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}[/tex]

Since we don't care which roll we get the 6 we take the sum:

[tex]\frac{1}{6}+\frac{5}{6}\cdot\frac{1}{6}+\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}
=\frac{1}{6}+\frac{5}{36}+\frac{25}{216}
=\frac{36}{216}+\frac{30}{216}+\frac{25}{216}
=\frac{91}{216}\approx 42 \%[/tex]

Doug

 

chroot: It seems we may of had a miscommunication, anyway thanks for your reply. I understand your way of thinking perfectly now.

doug: I do have one question regarding a small piece of this.

"The probability of getting a 6 on the second roll given that we didn't get a 6 on the first is: 1/6*5/6"

Why is it given that we diddn't get a 6 on the first dice when we are rolling them at once and only grouping the results? Or another way of looking at it: Wouldn't the fraction be 6/6 * 1/6 if if were in fact given that the first roll succeded? I assume given is equal to 100% I think this is the part that made be begin to look at the problem as stange in the first place

Got my own account now,
-Bryan