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

Mazuz · Dec 16, 2003

Ok some friends and I have been arguing over this for hours!

If we are rolling 3 six sided dice, one at a time, what are the odds that one of the three dice will land on a 6 (lets say). Each dice clearly has a 1/6 chance of landing on the 6 each. We are not interested in more than one result of 6 in the series of dice we are rolling. So if it matters (I don't think it does) the rolling stops after a 6 is attained.

Help from someone who definitely knows how this works would be greatly appreciated. Our answers are 42% or 50%. But we cannot agree or prove decisively to each other which is correct.

chroot · Dec 16, 2003

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

master_coda · Dec 16, 2003

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.

Mazuz · Dec 16, 2003

Ok, my friend is still disagreeing with me. This is his description of his position. If you roll one dice the odds of getting a 6 are 1/6. Then *if* a second roll occurs the chance of getting a 6 on the second roll is also 1/6. Therefore in two instances of rolling the odds are 1/6+1/6= 1/3 or 2 in 6. *If* you get a third roll, the chance for rolling a 6 on it is still 1/6. There is no reason to combine the dice rolls that I can see, as each roll does not affect the others. Oh, and it doesnt help unless you can explain your position simply. Thanks.

chroot · Dec 16, 2003

Originally posted by Mazuz
Ok, my friend is still disagreeing with me. This is his description of his position. If you roll one dice the odds of getting a 6 are 1/6. Then *if* a second roll occurs the chance of getting a 6 on the second roll is also 1/6. Therefore in two instances of rolling the odds are 1/6+1/6= 1/3 or 2 in 6. *If* you get a third roll, the chance for rolling a 6 on it is still 1/6. There is no reason to combine the dice rolls that I can see, as each roll does not affect the others. Oh, and it doesnt help unless you can explain your position simply. Thanks.

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

- Warren

master_coda · Dec 16, 2003

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.

Doug · Dec 16, 2003

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

Mazuz · Dec 16, 2003

Thank you guys immensly for your help, I may be able to enjoy lord of the rings tonight now without having to argue with him :P I understood 42 percent to be right but couldn't prove it to his satisfaction, he is quite an annoying arguer, because he is very good at it even when he is wrong. I think you guys convinced him somehow. Think i'll go take some advil now ;)

Mazuz · Dec 16, 2003

This is the friend

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

- Warren

"Oh, and it doesnt help unless you can explain your position simply. Thanks"

Thanks for nothing.

-Bryan

None of you explained this in a simple way. Let me ask a simple question. What are the odds that if three different people each roll one dice, that at least one of them will roll al least one 6. Then explain to me whether or not this is logically the same question as the origional one or not and why. I really am trying to grasp this, I understand how and why your way works, but cant yet see how mine does not. No more wortless or condensending answers like the ego stroker quoted above, thanks a lot. Does anyone even understand what I am asking?

-Bryan

chroot · Dec 17, 2003

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.

What are the odds that if three different people each roll one dice, that at least one of them will roll al least one 6.

42%.

Then explain to me whether or not this is logically the same question as the origional one or not and why.

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

HallsofIvy · Dec 17, 2003

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

Mazuz · Dec 17, 2003

Well, hrm. So no, no one is able to explain the difference to me or answer my question (which is what problem was I origonally doing, and how is it is logically inconsistant with the other one), I can do the simple math myself, dont need it shown to me, what I need is a logical, rational, and progressive answer explaining to me that how, by grouping the individual dice rolls, makes it that your odds of getting a 6 in any given roll come out to anyting but 1 in 6 and why it is mathematically incorrect to combine and divide instances to get an average in this case. If some people have a one in 6 average of dying on tuesdays, out of any 3 of those people on any given tuesday, an average of 1 (50%) will be dead, (not 42% right?)It's ok. I wasn't really expecting much. Thanks anyhow. Please dont bother to answer unless you aree really willing to help, and can explain things in more than one superficial way. Im not asking just what the should be but why it is the way it is in this case *and* not the way my mind intuitivly grasps it. So can someone give me an explaination, and not a few basic problems I am already familer with. That would be appreciated.

-Bryan

Mazuz · Dec 17, 2003

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

No, all he ever said to me is "tell your friend he is wrong" after being specifically asked for clairty beyond his first post. This was not in any way helpful or a response to my post and I consider this to be ego stroking. Im sure he capible enough to respond for himself, and I'm not attacking him, so I dont really see why you bothered to post that. But its cool, would have been more useful if you had helped with the answer, but no worries.

-Bryan

chroot · Dec 17, 2003

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%.

what I need is a logical, rational, and progressive answer

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

chroot · Dec 17, 2003

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

master_coda · Dec 17, 2003

Originally posted by chroot
The question "what is the probability of exactly one six?" is, as Doug said, 1 out of 216:

[tex]\frac{1}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} \approx 0.5 \%[/tex]

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]

StephenPrivitera · Dec 17, 2003

Originally posted by chroot

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

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

chroot · Dec 17, 2003

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

- Warren

Doug · Dec 17, 2003

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

Magnus_Grey · Dec 18, 2003

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