# Thread: More dice trick questions

1. ## More dice trick questions

I hope these aren't too hard for the basic statistics forum, but some friends and I are disagreeing on the answers to these two questions:

1) Show that if two dice are loaded with the same probability distribution, then the probability of doubles is at least 1/6.
2) If P2 is the probability of a 2 showing and we are told that at least one 2 has shown, what is the probability that upon looking at the pair of dice the sum will be 4?

Now for each question:
1) My initial solution to this was to show that the probability of doubles on die with even distributions are 1/6, so any variation in the distribution leads to a smaller number. But I can't quite prove this algebraically and I'm stuck. Is my approach wrong?

2) The simple answer (which some are sticking to) is that it's simple: The answer is P2. But I don't believe it could be that easy, and here's the answer I came up with:
Let A = the event where the sum is 4
B = The event where at least one 2 shows up
A is equal to the probability of getting a (1,3), (3,1), or (2,2). Since we know one die is a two, the probability of 1,3 or 3,1 is zero and P[A] = P[(2,2)].
As for B:
$P[B] = 1 - P[not B] = 1 - (1-P_2 )(1-P_2 ) = P_2 (2 - P_2 )$

Now, we want to know P[A|B]:

$P[A|B] = \frac{P[A \cap B]}{P[B]} = \frac{P[(2,2)]}{P[B]} = \frac{P_2^2}{P_2 (2 - P_2 )} = \frac{P_2}{2-P_2}$
I can't explain it...it just feels wrong. Like I did too much. Is the simple answer really correct or is this one right?

2. 1) You have to show that if $0 \leq p_i \leq 1$ for $i = 1, 2, \dots , 6$
and $p_1 + p_2 + \dots + p_6 = 1$,
then $p_1^2 + p_2^2 + \dots + p_6^2 \geq 1/6$.

Hint: You may find the Cauchy-Schwartz Inequality useful.

2) You are correct.

3. Originally Posted by RespeckKnuckles
I hope these aren't too hard for the basic statistics forum, but some friends and I are disagreeing on the answers to these two questions:
1) Show that if two dice are loaded with the same probability distribution, then the probability of doubles is at least 1/6.
2) If P2 is the probability of a 2 showing and we are told that at least one 2 has shown, what is the probability that upon looking at the pair of dice the sum will be 4?

Now for each question:1) My initial solution to this was to show that the probability of doubles on die with even distributions are 1/6, so any variation in the distribution leads to a smaller number. But I can't quite prove this algebraically and I'm stuck. Is my approach wrong?
I will try to help with #1. In may give you ideas for #2.

Set up the probabilities. Let $P(X=k)=p_k,~k=1,2,3,4,5,6$.
Then it must be the case that $\sum\limits_{k = 1}^6 {p_k } = 1$.
Let $D$ be the event that doubles are tossed. Then $P(D) = \sum\limits_{k = 1}^6 {\left( {p_k } \right)^2 }$.

We are going to use the basic inequality: $2ab\le a^2+b^2.$
Now $1 = \left( {\sum\limits_{k = 1}^6 {p_k } } \right)^2 = \sum\limits_{k = 1}^6 {p_k ^2 } + \sum\limits_{1 = k < j}^6 {2p_k p_j } \leqslant 6\sum\limits_{k = 1}^6 {p_k ^2 } = 6P(D)$

4. I'm still confused about something, though. In problem 2,
B = The event where at least one 2 shows up

Why can't we then say:
$P[B] = P[D_1 = 2 (OR) D_2=2] = P[D_1 = 2] + P[D_2 = 2] = 2P_2$

which is clearly different from the "correct" answer. Why doesn't it yield the same thing?

this time you have counted (2,2) twice.

6. Originally Posted by Plato
this time you have counted (2,2) twice.
But since the roll of dice 1 and dice 2 are two disjoint events, shouldn't the probability of the events together be the sum of their independent probabilities?

7. There are only eleven pairs that contain at least one two.
Do you understand that?

8. Yes. I am trying to figure out why some of these solutions don't work algebraically. I already have the correct answer, but I'm trying to figure out why some of these other approaches I might have used do not work.

I realize now that my assertion that P(D1 = 2 OR D2 = 2) = P(D1=2) + P(D2=2) doesn't work because the two events are not mutually exclusive. For anyone wondering, the formula should then be:
$P(D_1 = 2 \cup D_2 = 2) = P(D_1 = 2) + P(D_2 = 2) - P(D_1 = 2 \cap D_2
= 2) = P_2 + P_2 - P_2^2 = P_2(2 - P_2)$

Which is correct. But now I have a different question. If this were seen as a bernoulli trial, we would then use the formula for a binomial distribution:
$\binom{2}{1} P_2^1 (1 - P_2)^{(2-1)} = 2P_2 (1 - P_2) = P_2 (2 - 2P_2) \neq P_2 (2-P_2)$

So does the binomial distribution not apply here?

9. $P(\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\})$
$=p_2 p_1 + p_2 p_2 + p_2 p_3 + p_2 p_4 + p_2 p_5 + p_2 p_6=p_2$.
Factor out $p_2$ the other factor sums to 1.

The same is true for pairs $(x,2)$.

But we have counted $(2,2)$ twice.

So $P(D_1\cup D_2)=p_2+p_2-(p_2)^2$.

10. thanks, that definitely makes sense. But any idea why the binomial distribution formula does not work?

11. Originally Posted by RespeckKnuckles
thanks, that definitely makes sense. But any idea why the binomial distribution formula does not work?
Your formula is for the probability of exactly one 2. Add the probability of two 2's to your result and you get the answer that you expect.