Probability of getting 4,5 or 6 in 4 tosses

[ISTER_REG]

Hi,

I want to find out what the probability is, for getting a 4,5 or a 6 in 4 tosses of a die. To calculate this I would use the multinomial distribution. But my problem here is that I do not realy know what events I can expect... Does the question indicate that I can have events like 4,5,6,X or only single occurrences of 4, 5 and 6?
I would be very grateful for any hints and approaches!

 

[Cubist]

Let's say that "p" is the probability of getting a 4,5 or a 6 in 4 tosses of a die. What would 1-p represent?

 

[Dr.Peterson]

Hi,

I want to find out what the probability is, for getting a 4,5 or a 6 in 4 tosses of a die. To calculate this I would use the multinomial distribution. But my problem here is that I do not realy know what events I can expect... Does the question indicate that I can have events like 4,5,6,X or only single occurrences of 4, 5 and 6?
I would be very grateful for any hints and approaches!

You're asking how to interpret the question. That's appropriate, because it is a little ambiguous.

I would take it to mean that at least one of the 4 rolls is a 4, 5, or 6; that is, you roll four times, and call it a success if you ever see a number greater than 3. Perhaps it could also be taken to mean that you get a 4, 5, or 6 on exactly one roll, but I don't see it that way.

I suppose it could even be taken as meaning that each of the 4 rolls is a 4, 5, or 6 (so that you never see anything else), but that would surely be written differently.

What you really need to do is to ask the author of the problem. Did you quote it exactly? Is there any other information (such as a supposedly correct answer)?

 

[Cubist]

Let's say that "p" is the probability of getting a 4,5 or a 6 in 4 tosses of a die. What would 1-p represent?

...this is as a hint (you were also asking for a hint) that assumes the following interpretation of the question (which is very probably the intended meaning)...

I would take it to mean that at least one of the 4 rolls is a 4, 5, or 6

 

[ISTER_REG]

Hey folks, thanks for your hints and answers!

I want to point out here three things, first the original question, second my approach, third a question related to multinomial usage in my approach.

1.) The original question comes from Schaums, I quote it here: "Find the probability of not getting a 1,2 or 3 in 4 tosses of a fair die". A solution without a way is also given with 1/16. The headline is "multinomial distribution", so it should be solved with this...

2.) My approach is concerning, that "not getting a 1,2 or 3 in 4 tosses" means, I can get 4,5,6. But the question is however which pattern this means e.g. 4XXX, 44XXX and so on. I was assuming that the following patterns are allowed, for which I also calculated the probability:
- Pattern 4444, 5555, 6666 (4 times same number) --> [MATH]3\left(\frac{1}{6}\right)^4[/MATH]- Pattern 4455, 4466 , ... (each to numbers are equal) --> [MATH]3\cdot \frac{4!}{2!2!}\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^2[/MATH]- Pattern 4456, 5546, ... (two numbers equal, and two different) --> [MATH]3\left(\frac{4!}{2!1!1!}\right)\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^1\left(\frac{1}{6}\right)^1[/MATH]- Pattern 444(5|6), ... (three equal numbers and one of two different) --> [MATH]6\left(\frac{4!}{3!1!}\right)\left(\frac{1}{6}\right)^3\left(\frac{1}{6}\right)^1[/MATH]
Putting it all together:

[MATH]3\left(\frac{1}{6}\right)^4+ 3\cdot \frac{4!}{2!2!}\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^2 + 3\left(\frac{4!}{2!1!1!}\right)\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^1\left(\frac{1}{6}\right)^1 + 6\left(\frac{4!}{3!1!}\right)\left(\frac{1}{6}\right)^3\left(\frac{1}{6}\right)^1 = \frac{1}{16}[/MATH]
3.) Now as I see my calculation it "looks" like multinomial distribution, but I'am not sure if so, therefore I would be very thankful, if someone can confirm if this is the right approach

 

[Cubist]

..."Find the probability of not getting a 1,2 or 3 in 4 tosses of a fair die"

Other readers please note that this question is different to the post#1 question

3.) Now as I see my calculation it "looks" like multinomial distribution, but I'am not sure if so, therefore I would be very thankful, if someone can confirm if this is the right approach

Your approach uses the multinomial distribution, and you have the correct answer. Therefore it looks OK to me. BUT, there's a way of doing this with a single multinomial calculation. Can you figure it out without opening the hint?

Spoiler: hint

what's the probability of getting 0 fours, 0 fives, 0 sixes, and 4 lots of <something?>

 

[Dr.Peterson]

1.) The original question comes from Schaums, I quote it here: "Find the probability of not getting a 1,2 or 3 in 4 tosses of a fair die". A solution without a way is also given with 1/16. The headline is "multinomial distribution", so it should be solved with this...

By not quoting the actual problem and its context, you totally violated the guidelines. If you had said this, we could have gone straight to the answer.

The question you asked, "the probability ... for getting a 4,5 or a 6 in 4 tosses of a die", absolutely does not mean the same as the problem you quote now. If you don't get any of 1, 2, or 3, then you get ONLY 4, 5, and 6. Thus, what you want is the third option I mentioned, "I suppose it could even be taken as meaning that each of the 4 rolls is a 4, 5, or 6 (so that you never see anything else), but that would surely be written differently."

And telling us the topic of the section tells us what method is appropriate. You mentioned the multinomial distribution, but as if it were just your preference.

But the problem can be solved far more easily. On each toss, the probability of not getting 1, 2, or 3 is 1/2. On four tosses, it's (1/2)^4 = 1/16.

 

[Cubist]

But the problem can be solved far more easily. On each toss, the probability of not getting 1, 2, or 3 is 1/2. On four tosses, it's (1/2)^4 = 1/16.

SPECULATION: Maybe this problem was set in order to illustrate that the multinomial distribution is a generalisation of the binomial distribution. Which, in turn is a generalisation of the probability of n repeated outcomes.

Spoiler: answer

[math] \frac{4!}{0! \times 0! \times 0! \times 4!} \left(\frac{1}{6}\right)^0 \left(\frac{1}{6}\right)^0 \left(\frac{1}{6}\right)^0 \left(\frac{1}{2}\right)^4 [/math]
[math]= \binom{4}{4}\left(p\right)^4 \left(1-p\right)^0 [/math] where \( p=\frac{1}{2} \), binomial

[math]= \left(\frac{1}{2}\right)^4[/math], n repeated outcomes

[math]= \frac{1}{16}[/math]

 

[ISTER_REG]

Hey,

thanks for the answer(s) and sorry for not quoting the original question first. In my understanding asking for "not getting 1, 2, or 3 " is related to asking for "getting 4, 5, or 6" (so I solved for myselfe).

SPECULATION: Maybe this problem was set in order to illustrate that the multinomial distribution is a generalisation of the binomial distribution. Which, in turn is a generalisation of the probability of n repeated outcomes.

Yes, within my post #5, I wanted to say this. So the task was "Find the probability of not getting a 1,2 or 3 in 4 tosses of a fair die" and above there was the headline "multinomial distribution", which indicates that this problem has to be solved by using multinomial distribution.

Back to the math

@Cubist in your answer of post #8 (the spoiler) I have a short question regarding that. Why is there a 4! in the denominator, I have at the moment no intuitive understanding of the reasoning behind that (the rest of the approach is understanable for me ). In #6 you explained this by 4 slots of something, I try to imagine that but for my understanding there is a times 3 (I would have expect this) missing (4444, 5555, 6666)

 

[Cubist]

@Cubist in your answer of post #8 (the spoiler) I have a short question regarding that. Why is there a 4! in the denominator, I have at the moment no intuitive understanding of the reasoning behind that (the rest of the approach is understanable for me ). In #6 you explained this by 4 slots of something, I try to imagine that but for my understanding there is a times 3 missing (4444, 5555, 6666)

Sorry, my hint in post#6 was wrong. I meant to say...

Calculate the probability of getting 0 ones, 0 twos, 0 threes, and 4 outcomes of {four,five or six}. The probability of getting a {4,5, or 6} is...

[MATH]p_4=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}[/MATH]
Please see this page on the multinomial distribution. My method corresponds to \(n=4, x_1= x_2= x_3=0, x_4=4\). And the probabilities are \(p_1=p_2=p_3=\frac{1}{6}, p_4=\frac{1}{2} \). We don't have to multiply this by anything since the 4 rolls only happen once.

Using a regexp notation
- Pattern [456][456][456][456] which matches 4 characters and all of them must be a 4,5, or 6

EDIT: Obviously you could miss out \(x_1,x_2,\) and \(x_3\) but I included them to show that they don't influence the end result.