# Thread: Dice Probability (FUN QUESTION!)

1. ## Dice Probability (FUN QUESTION!)

Person x rolls a die three times. What is the probability that her second and third rolls are both larger than her first roll?
(ex 5:6:6) (not ex:5:5:6)
also
the result of her second roll is greater than that of her first roll and the result of her third roll is greater than the second?
(ex: 4:5:6) (not ex: 5:6:6

part a:
first roll she can roll 1-5, second roll she can roll 2-6, third roll she can roll 2-6.
thats all i've got.

2. ## Re: Dice Probability (FUN QUESTION!)

Hey lhurlbert.

Hint: Consider that each dice roll is independent and then consider all the cases for what the first value takes.

Then calculate P(Y,Z>X|X=1) + P(Y,Z>X|X=2) + ... + P(Y,Z>X|X=6)

3. ## Re: Dice Probability (FUN QUESTION!)

Person x rolls a die three times. What is the probability that her second and third rolls are both larger than her first roll?
(ex 5:6:6) (not ex:5:5:6)
You have 3 independent events: First Roll, Second Roll, Third Roll. Remember, too, that ORDER MATTERS here, so (1,2,3) is NOT the same as (2,1,3) or (3,2,1). The first order of business is to compute the total number possible outcomes. Since we are presumably working with a fair die, all possible outcomes are equiprobable. Having equiprobable outcomes is important because the solution to this problem can then be written as an easy-to-understand fraction:

$\displaystyle P(\text{condition}) = \frac{\text{number of outcomes that satisfy condition}}{\text{total number of possible outcomes}}$

Since Roll 1 has 6 possible outcomes, Roll 2 has 6 possible outcomes and Roll 3 has 6 possible outcomes (all rolls independent and thus holding the "with replacement" property), there are $\displaystyle 6\cdot 6\cdot 6 = 6^3 = 216$ total possible (and equiprobable) outcomes.

Now the challenge, how many of the 216 possible outcomes satisfy our condition?. Our condition is satisfied if the values of the 2nd and 3rd rolls are both strictly greater than the value obtained from the first roll.

Consider the 1st Roll is a 1. Then, our condition is satisfied IFF the second and third rolls are both strictly greater than 1. That is, Rolls 2 and 3 must both be between 2 and 6 for our condition to be satisfied. Thus, there are 5 possible outcomes for Roll 2 and 5 possible outcomes for Roll 3 that do not violate our condition. 5 times 5 is 25, so there are 25 outcomes that satisfy our condition given that the first roll is a 1.

Next, consider the 1st Roll is a 2. There are then 4 acceptable outcomes (3,4,5,6) for Roll 2 and 4 acceptable outcomes for Roll 3. So, if the first roll is a 2, then there are $\displaystyle 4 \cdot 4 = 4^2 = 16$ possible outcomes that satisfy our condition.

Repeat this process for the 1st Roll being 3, then 4, then 5, and you will see what happens finally at 6.

When the first roll is 3, then the second and third rolls both must choose from 4, 5 or 6 to satisfy our condition. That is, the second and third rolls both choose among 3 acceptable values. $\displaystyle 3\cdot 3 = 3^2 = 9$ possible outcomes, given that the first roll is 3, satisfy our condition.

When the first roll is 4, the second and third rolls both must choose either 5 or 6. So, if the first roll is a 4, then there are only $\displaystyle 2\cdot2=2^2=4$ outcomes possible that satisfy our condition.

When the first roll is 5, the second and third rolls must both be 6. Thus, when the first roll is 5, there is only one outcome, namely (5,6,6), that satisfies our condition.

When the first roll is 6, our condition is automatically violated by virtue of no higher number available.

Alas, we tally up the total number of outcomes that satisfy our condition. 55 of the 216 possible outcomes (or permutations) satisfy the condition that the second and third rolls are both strictly larger than the first roll

4. ## Re: Dice Probability (FUN QUESTION!)

Here is another approach. Let's say the rolls are (a,b,c). There are 6^3 possible triples, each of which we assume is equally likely. We want to count the number of triples where a < b and a < c.

First, let's consider the cases where a, b, and c are distinct. There are $\displaystyle \binom{6}{3}$ triples where a < b < c, and another $\displaystyle \binom{6}{3}$ triples where a < c < b. So there are $\displaystyle 2 \binom{6}{3}$ triples (a,b,c) where a, b, and c are distinct.

Now consider the cases where a < b = c. There are $\displaystyle \binom{6}{2}$ of these.

So all together there are
$\displaystyle 2 \binom{6}{3} + \binom{6}{2} = 55$
triples (a,b,c) where a < b and a < c.