# Throwing Dice, Probability

• June 4th 2011, 11:10 AM
qspeechc
Throwing Dice, Probability
Hi everyone.

Here's the question: How many dice do you need to throw so that the probability of one of each face showing (i.e. getting at least one dice to show '1', and at least one to show '2', and... etc.) is at least 0.5?

My answer was 44, which is seriously wrong because the correct answer is 17. I would like to know how to arrive at the correct answer. Anyway, here's how I got my answer.

The possible number of configurations our n dice can have is (n+5) choose 5, i.e. C(n+5, 5) is the notation I will use. I get this by saying: use five dividers to partition the dice into six groups, corresponding to the numbers 1, 2, 3, 4, 5, 6, the partitions telling us what the dice will show on its face.

The number of ways to get at least one of each number showing is to take 6 dice and assume they have the numbers 1,2,...,6 showing. Then configure the remaining n-6 as you please. There are C(n-1, 5) ways to do this.

Therefore for n dice the probability of getting one of each face is C(n-1, 5)/C(n+5, 5). To get this greater than or equal to 0.5 we need n = 44.

As I said, my answer is seriously wrong. I would like to know what is the right way to arrive at the answer, and also, if possible, where I went wrong. Thank you for any help.
• June 4th 2011, 02:57 PM
Plato
QUESTION?
How do you know that the correct answer is 17?
$\sum\limits_{k = 0}^5 {\left( { - 1} \right)^k \binom{6}{k} \left( {\frac{{6 -k}}{6}} \right)^N }$ is then probability of getting at least one of each in N throws.

That sum agrees with $N=6$ which is $\frac{6!}{6^6}$
• June 5th 2011, 10:37 AM
qspeechc
Quote:

Originally Posted by Plato
QUESTION?
How do you know that the correct answer is 17?

That's what the book I am using says. Must be a typo.

Quote:

Originally Posted by Plato
$\sum\limits_{k = 0}^5 {\left( { - 1} \right)^k \binom{6}{k} \left( {\frac{{6 -k}}{6}} \right)^N }$ is then probability of getting at least one of each in N throws.

That sum agrees with $N=6$ which is $\frac{6!}{6^6}$

Please explain how you arrived at that. Why sum from 0 to 5? You mean 0 to N?
• June 5th 2011, 12:04 PM
Plato
What textbook are you using?
This is not an easy problem to grasp.
I will give you a brief version. Say we toss a die ten times.
What is the probability of not getting a 1? It is $\left(\frac{5}{6}\right)^{10}$
That is the same for not getting a 2.
But what is the probability of not getting a 1 or 2?
Is it not $2\left(\frac{5}{6}\right)^{10}-\left(\frac{4}{6}\right)^{10}$.
Thus by subtracting that number from 1, we have the probability of of getting at least one 1 and at least one 2.
That is using inclusion/exclusion.

So the sum I gave in reply #2 is the result of expanding that to all six numbers.
• June 6th 2011, 04:24 AM
qspeechc
Oh, I see. It is difficult to explain. Basically here's how I understand it.

Call $A_i$ the probability of not getting an i=1,2,...,or 6 in N throws. Then we want $1-P(A_1\cup A_2\cup \ldots \cup A_6)$. So first we find $P(A_1\cup A_2\cup \ldots \cup A_6)$ by finding $P(A_1\cup A_2)$ then $P(A_1\cup A_2\cup A_3)$ and so on.

But everything simplifies because
$P(A_i) = P(A_j)$
$P(A_i\cap A_j) = P(A_k\cap A_m)$
and similarly for intersections of 3, 4, 5 sets.

Thank you Plato.

Btw, I'm using some old course notes from a friend, I'm trying to learn this stuff on my own. I have never taken an probability/statistics course myself. Is there any book you can recommend?
• June 6th 2011, 12:31 PM
Plato
Quote:

Originally Posted by qspeechc
Btw, I'm using some old course notes from a friend, I'm trying to learn this stuff on my own. I have never taken an probability/statistics course myself. Is there any book you can recommend?

I just now saw your P.S.
For readability I very much like Jim Pitman's book.
• June 9th 2011, 02:29 PM
bryangoodrich
Reviewing probability theory from Casella and Berger "Statistical Inference," I saw an example that reminded me of this (see page 24). I used the same approach and got the same answer Plato did (i.e., 13 throws). The book wanted to know the probability for the event that a gambler could throw at least 1 six in 4 rolls of a die. We get

$P(\text{at least 1 six in 4 rolls}) = 1 - P(\text{no six in 4 rolls})$
$= 1 - \prod_{i = 1}^4 P(\text{no six on roll } i)$

The last equality follows by independence of rolls. Since the probability of not getting a given numbered face is 5/6, we have

$P(\text{at least 1 six in 4 rolls}) = 1 - (5/6)^4 = 0.518$

The reason this seemed analogous here was because we're asking the same thing, but for all six of the die, and we want the end result to be an inequality such that P(.) > 0.5. If we instead replace "4" by "k" in the above example, then we want

$(1 - (5/6)^k)^6 \geq 0.5$

For k = 12 and k = 13, I got 0.49 and 0.56, respectively. This approach seemed a lot simpler than Plato's above and required no subtraction, which threw me off in his explanation. Thoughts?

On reflection, I wonder if this is correct, though. It could be bad intuition, but I want to say there is a lack of independence between the rolls and we have to discount something (which might raise k to 17, unless we are concluding the given solution was in error). I say this because while this formula works for independent events for a given number to appear in k rolls, each roll has a chance to reveal one of the 6. So I may not, say, roll a six on the first roll, but I an guaranteed to roll something on the first roll. I'm probably wrong, but that kept me from thinking my approach was valid. However, when I got the same answer Plato did, I thought I would comment.
• June 9th 2011, 02:54 PM
Plato
Quote:

Originally Posted by bryangoodrich
Reviewing probability theory from Casella and Berger "Statistical Inference," I saw an example that reminded me of this (see page 24). I used the same approach and got the same answer Plato did (i.e., 13 throws). The book wanted to know the probability for the event that a gambler could throw at least 1 six in 4 rolls of a die. We get
On reflection, I wonder if this is correct, though. It could be bad intuition, but I want to say there is a lack of independence between the rolls and we have to discount something (which might raise k to 17, unless we are concluding the given solution was in error). I say this because while this formula works for independent events for a given number to appear in k rolls, each roll has a chance to reveal one of the 6. So I may not, say, roll a six on the first roll, but I an guaranteed to roll something on the first roll. I'm probably wrong, but that kept me from thinking my approach was valid. However, when I got the same answer Plato did, I thought I would comment.

For that reason, I did ask for the exact source of that suggested answer of 17. I suspect that answer is the result of using some statistical model to approximating the process. Whereas, mine is a pure counting argument.