Questions about probability

[kenny1999]

Hi. What is the probability of throwing three dice at the same time while we are getting at least one of the particular number?

I mean, for example, the probability of getting at least one number 2 among the three dice thrown

Is it 1/6+1/6+1/6 = 3/6 = 1/2 ?

How to explain to my daughter that it is 1/6+1/6+1/6 but isn't 1/6 x 1/6 x 1/6 ?

 

[Svein]

I'm not a statistician, but I seem to remember that the "AND" operation is a multiplication. So - the probability of not getting a "2" in three throws would be 5/6*5/6*5/6 = 125/216. Thus, the probability of getting at least one "2" would be 1 - 125/216 = 91/216.

 

[andrewkirk]

The answer is neither 1/6 x 1/6 x 1/6 nor 1/6 + 1/6 + 1/6. It is in between the two numbers.
It is:

probability of three twos + probability of two twos + probability of one two

The first item is 1/6 x 1/6 x 1/6

The second item has three cases - according to whether the die that doesn't give 2 is die 1, die 2 or die 3. For each case, the probability is 5/6 x 5/6 x 1/6,
as 5/6 is the probability of not getting a 2. Times the answer by three to get the total prob of two twos.

The third item has three cases - according to whether the die that gives 2 is die 1, die 2 or die 3. For each case, the probability is 5/6 x 1/6 x 1/6. Times the answer by three to get the total prob of two twos.

Add the three items up to get the answer.

EDIT: just saw @Svein 's answer. His approach is more elegant than mine. Fortunately, they both give the same answer.

 

[StoneTemplePython]

Hi. What is the probability of throwing three dice at the same time while we are getting at least one of the particular number?

I mean, for example, the probability of getting at least one number 2 among the three dice thrown

Is it 1/6+1/6+1/6 = 3/6 = 1/2 ?

How to explain to my daughter that it is 1/6+1/6+1/6 but isn't 1/6 x 1/6 x 1/6 ?

One more approach:
first come up with an arbitrary labeling of the dice -- so you have dice numbered 1, 2 and 3.

let $A_k$ be the event that after a single roll, dice number $k$ is a $2$. These events are independent so you can multiply associated probabilities, e.g. $P\big(A_1\cap A_2\big) = P\big(A_1\big)P\big(A_2\big)$

Now, what you've actually done is
$P\big(A_1 \cup A_2 \cup A_3\big) \leq P\big(A_1\big) + P\big(A_2\big) + P\big(A_3\big)= 3 \cdot \frac{1}{6}$
this is an upper bound called Boole's Inequality or a union bound. This inequality becomes an equality iff the events are mutually exclusive -- but clearly they aren't mutually exclusive because you can e.g. roll a 2 with both dice 1 and with dice 2. So the problem is you've confused an inequality with an equality.

Enter Bonferonni's inequality:
dealing with some of these dependencies we can bound this below by considering the pairwise intersections/ interactions
$(A_1 \cap A_2)$, $(A_2 \cap A_3)$, and $(A_1 \cap A_3)$

$3 \cdot \frac{1}{6} - 3 \cdot \frac{1}{6^2} = \Big(P\big(A_1\big) + P\big(A_2\big) + P\big(A_3\big)\Big)- \Big(P(A_1 \cap A_2)+ P(A_2 \cap A_3) + P(A_1 \cap A_3)\Big) \leq P\big(A_1 \cup A_2 \cup A_3\big)$

and finally, if we iterate on this inequality to completion by considering the 3 way interactions of events i.e. $\big(A_1 \cap A_2 \cap A_3\big)$, we recover inclusion-exclusion and get

$P\big(A_1 \cup A_2 \cup A_3\big)$
$=\Big(P\big(A_1\big) + P\big(A_2\big) + P\big(A_3\big)\Big)- \Big(P(A_1 \cap A_2)+ P(A_2 \cap A_3) + P(A_1 \cap A_3)\Big) +\Big(P\big(A_1 \cap A_2 \cap A_3\big)\Big)$
$=\big(3 \cdot \frac{1}{6}\big) - \big(3 \cdot \frac{1}{6^2}\big) + \big(\frac{1} {6^3}\big)$
$= \frac{91}{216}$

- - - -
post number 2 is the expedient route which makes use of the identity
$P\big(A_1 \cup A_2 \cup A_3\big) = P\big((A_1^C \cap A_2^C \cap A_3^C)^C\big)=1 - P\big(A_1^C \cap A_2^C \cap A_3^C\big)$

and $P\big(A_1^C \cap A_2^C \cap A_3^C\big) = P\big(A_1^C\big)P\big(A_2^C\big)P\big(A_3^C\big)= \big(\frac{5}{6}\big)^3$
by stochastic independence.
- - - -
in either case, drawing a picture and looking at the venn diagram with 3 interlocking circles and 3 different shadings... is how I'd finish this off to explain it to someone.

- - - -
edit:
a nice way of deriving Bonferonni's Inequalities and Inclusion-Exclusion is to take

$P\big(A_1 \cup A_2 \cup ... \cup A_n\big) = 1 - P\big(A_1^C \cap A_2^C \cap ... \cap A_n^C\big) = E\Big[1 - \big(1 - \mathbb I_{A_1}\big)\big(1 - \mathbb I_{A_2}\big)...\big(1 - \mathbb I_{A_n}\big)\Big]$
where the right hand side is shown in factored form... if you multiply out the factored polynomial you can expand this into a standard polynomial and then take expectations from there -- this recovers inclusion-exclusion with characteristic alternating sign formula and elementary symmetric function of roots (indicators r.v.'s in this case), and if you truncate your evaluation of this factored polynomial by just evaluating lower order terms, the truncated result is an upper or lower bound depending the sign of the next highest order term (a basic taylor polynomial result). There's a little more to it with nesting properties, but this is the core of Bonferonni's Inequalities.

 

[PeroK]

Hi. What is the probability of throwing three dice at the same time while we are getting at least one of the particular number?

I mean, for example, the probability of getting at least one number 2 among the three dice thrown

Is it 1/6+1/6+1/6 = 3/6 = 1/2 ?

How to explain to my daughter that it is 1/6+1/6+1/6 but isn't 1/6 x 1/6 x 1/6 ?

The simplest and best approach is to do an experiment. Get the three dice, roll them repeatedly and count how many times you get at least one 2.

If the answer were 1/6 x 1/6 x 1/6 = 1/216, you would almost never get a 2. Somehow, by throwing three dice at the same time you prevent any die coming up 2, except rarely. Also, in the this case, the chance of getting at least one 1 would be 1/236, and the same for 3, 4, 5 and 6. So, most of the time the three dice would have to show blanks or something, as you'd rarely get any of the numbers 1-6.

So, what does have probablity 1/216?

 

[mathman]

Easy answer: Probability of getting none is $(5/6)^3$, so probability of getting at least one is$1-(5/6)^3=0.421296296296296$.