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

1. Dec 16, 2003

### Mazuz

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.

2. Dec 16, 2003

### chroot

Staff Emeritus
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

3. Dec 16, 2003

### master_coda

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.

4. Dec 16, 2003

### 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.

5. Dec 16, 2003

### chroot

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

- Warren

6. Dec 16, 2003

### master_coda

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.

7. Dec 16, 2003

### Doug

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

8. Dec 16, 2003

### Mazuz

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 ;)

9. Dec 16, 2003

### Mazuz

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

10. Dec 17, 2003

### chroot

Staff Emeritus
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

11. Dec 17, 2003

### HallsofIvy

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

12. Dec 17, 2003

### Mazuz

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

13. Dec 17, 2003

### Mazuz

>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

14. Dec 17, 2003

### chroot

Staff Emeritus
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:

$$\left( \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \right) + \left( \frac{5}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} \right) + \left( \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} \right) \approx 34 \%$$

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

$$\left( \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} \right) + \left( \frac{5}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \right) + \left( \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} \right) \approx 7 \%$$

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

$$\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \approx 0.5 \%$$

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

Last edited: Dec 17, 2003
15. Dec 17, 2003

### chroot

Staff Emeritus
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

16. Dec 17, 2003

### master_coda

Shouldn't this be:

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

$$\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \approx 0.5 \%$$

17. Dec 17, 2003

### StephenPrivitera

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:

$$\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \approx 0.5 \%$$

Is this correct?

18. Dec 17, 2003

### chroot

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

- Warren

19. Dec 17, 2003

### Doug

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

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

$$\frac{1}{6}$$

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

$$\frac{5}{6}\cdot\frac{1}{6}$$

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

$$\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}$$

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

$$\frac{1}{6}+\frac{5}{6}\cdot\frac{1}{6}+\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6} =\frac{1}{6}+\frac{5}{36}+\frac{25}{216} =\frac{36}{216}+\frac{30}{216}+\frac{25}{216} =\frac{91}{216}\approx 42 \%$$

Doug

20. Dec 18, 2003

### Magnus_Grey

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

Last edited: Dec 18, 2003
21. Dec 18, 2003

### master_coda

I can see where that might be kind of confusing. A better way of saying it would be "The probability of rolling a 6 on the second roll and not rolling a 6 on the first roll is 5/6 * 1/6".

22. Dec 19, 2003

### Magnus_Grey

"I can see where that might be kind of confusing. A better way of saying it would be "

"The probability of rolling a 6 on the second roll and not rolling a 6 on the first roll is 5/6 * 1/6".

Thanks. But it still sounds like there is an assumption made that no 6 was rolled on the first dice. Why is this assumption made? By the way, if you roll one dice, then the other, then the other, but only roll the second and third ones if the first/second is safe, does it change the problem in any way?

-Bryan

Last edited: Dec 19, 2003
23. Dec 19, 2003

### master_coda

What we're doing is trying to break the situation down into different ways we can get a six, without having any of the cases overlap.

The first case we use is if a six is rolled on the first roll. We don't care what the other rolls are in that case, they could be sixes, you might not even roll them at all.

The second case is if we roll a six on the second roll, but not on the first. We have to add the "not on the first" because we already counted the case where we roll a six first.

The third roll is the same...we already counted the cases where a six was rolled on the first or second roll, so we only want to count the cases where a six is not rolled on either of the first two.

Note that the way we are counting the rolls allows us to ignore what happens after we roll a six. If we roll a six on the first roll, it counts as a success whether or not we roll the other two dice, whether or not they come up as sixes.

24. Dec 21, 2003

### Hessam

wait... couldnt this be a binomial distribution? only 2 outcomes are needed, either the dice lands on 6, or it doesnt... using the binomial distribution formula you should find the "Expected" value

25. Dec 21, 2003

### krab

Yes... Binomial probability is

$$\left(\begin{array}{cc}n\\x\end{array}\right)t^x(1-t)^{n-x}$$

This is the probability of x successes in n trials. You need to evaluate this for t=1/6, n=3, and add together for x=1,2,3. You still get 42%. In fact you get exactly the 3 terms Warren gave 10 posts back. Warren obtained the factor

$$\left(\begin{array}{cc}n\\x\end{array}\right)$$

for each of the terms simply by counting.

This problem and the way in which it is non-intuitive to those who do not often calculate such things, reminds me of another.

For a random group of 23 people, what is the probability that there are 2 of them that have the same birthday?

Intuitively, one thinks that since 23 is quite small compared with 365, the probability must be quite small. Such turns out not to be the case!