A Question for the Statistics Experts

Southpaw

Southpaw

Well-Known Member
This question does not pertain to the game of blackjack, so the mods may move this thread at their own discretion. It may seem to pertain to the game of craps because the problem involves two dice--thus making it seem as if it belongs in the Other Games Section--but the reason I ask this is to improve my understanding of statistics.

The question reads as follows:

"A certain board game is played by rolling a pair of fair six-sided dice and then moving one's piece forward the number of spaces indicated by the sum showing on the dice. A player is "frozen" if her opponent's piece comes to rest in the space already occupied by her piece. If player A is about to roll and is currently six spaces behind player B, what is the probability that player B will be frozen after player A rolls?"

After you understand the problem, you realize that what they are asking simplifies to the following:

"What is the probability that the sum of two dice rolled will be 6?"

You find that there are 36 possible outcomes total by knowing that there are six possibilities for each dice and the result of one die is independent of the other. (6*6) = 36

You realize that to get a sum of six you could roll (1) a one and a five, (2) a two and a four or (3) a pair of threes.

You see that there are two ways to get the first two options. You could either roll (1,5) or (5,1) to satisfy the first. Similarly, you could roll (2,4) or (4,2) to satisfy the second.

The answer to the problem is 5/36 being that there are five combinations of the thirty-six that satisfy the problem: (1,5), (5,1), (2,4), (4,2) or (3,3).

When solving the problem on my own, I came up with the answer 6/36 or 1/6 because, intuitively, I wanted to say that there are two ways to get (3,3) in the same way that there are two ways to roll a five and a one: either (1,5) or (5,1).

Can anyone explain to me why there is, in fact, only one way to roll two threes, while there is two ways to roll a five and a one?

Suppose that you rolled the dice individually and you wanted to compare the probability of rolling a 1 and then a 5 to rolling a pair of threes:

Prob (1,5) vs. (3,3)

It just seems that it should be twice as likely that you'd get a pair of threes, since it seems that there should be two ways to do it.

Someone please explain this to me. I don't doubt that my intuition is incorrect. I just can't exactly see why it is incorrect.

AssumeR? MangoJ? BJA? SleightofHand?

Spaw
 
K

k_c

Well-Known Member
Southpaw said:
This question does not pertain to the game of blackjack, so the mods may move this thread at their own discretion. It may seem to pertain to the game of craps because the problem involves two dice--thus making it seem as if it belongs in the Other Games Section--but the reason I ask this is to improve my understanding of statistics.

The question reads as follows:

"A certain board game is played by rolling a pair of fair six-sided dice and then moving one's piece forward the number of spaces indicated by the sum showing on the dice. A player is "frozen" if her opponent's piece comes to rest in the space already occupied by her piece. If player A is about to roll and is currently six spaces behind player B, what is the probability that player B will be frozen after player A rolls?"

After you understand the problem, you realize that what they are asking simplifies to the following:

"What is the probability that the sum of two dice rolled will be 6?"

You find that there are 36 possible outcomes total by knowing that there are six possibilities for each dice and the result of one die is independent of the other. (6*6) = 36

You realize that to get a sum of six you could roll (1) a one and a five, (2) a two and a four or (3) a pair of threes.

You see that there are two ways to get the first two options. You could either roll (1,5) or (5,1) to satisfy the first. Similarly, you could roll (2,4) or (4,2) to satisfy the second.

The answer to the problem is 5/36 being that there are five combinations of the thirty-six that satisfy the problem: (1,5), (5,1), (2,4), (4,2) or (3,3).

When solving the problem on my own, I came up with the answer 6/36 or 1/6 because, intuitively, I wanted to say that there are two ways to get (3,3) in the same way that there are two ways to roll a five and a one: either (1,5) or (5,1).

Can anyone explain to me why there is, in fact, only one way to roll two threes, while there is two ways to roll a five and a one?

Suppose that you rolled the dice individually and you wanted to compare the probability of rolling a 1 and then a 5 to rolling a pair of threes:

Prob (1,5) vs. (3,3)

It just seems that it should be twice as likely that you'd get a pair of threes, since it seems that there should be two ways to do it.

Someone please explain this to me. I don't doubt that my intuition is incorrect. I just can't exactly see why it is incorrect.

AssumeR? MangoJ? BJA? SleightofHand?

Spaw
Click to expand...
5-1 or 1-5
1st dice can be 5 or 1, prob = 1/3
2nd dice must be either 1 or 5, depending on 1st dice, prob = 1/6
prob(5-1) or (1-5) = 1/3 * 1/6 = 1/18

4-2 or 2-4
1st dice can be 4 or 2, prob = 1/3
2nd dice must be either 2 or 4, depending on 1st dice, prob = 1/6
prob(4-2) or (2-4) = 1/3 * 1/6 = 1/18

3-3
1st dice must be 3, prob = 1/6
2nd dice must be 3, prob = 1/6
prob(3-3) = 1/6 * 1/6 = 1/36
 
Sonny

Sonny

Well-Known Member
Southpaw said:
Suppose that you rolled the dice individually and you wanted to compare the probability of rolling a 1 and then a 5 to rolling a pair of threes:

Prob (1,5) vs. (3,3)

It just seems that it should be twice as likely that you'd get a pair of threes, since it seems that there should be two ways to do it.
Click to expand...
The two dice are independent. Instead of thinking about rolling two dice, think about rolling one die twice. What is the probability of rolling a 1 then a 5?

p(5) * p(1) = (1/6) * (1/6) = 1/36

Now what is the probability of rolling a 3 twice in a row? It's the same because there is only one way to roll a 3 twice in a row. To put it another way, after rolling a 3, there is only one way to roll another 3.

I don't know if that example made it clearer or not so let me try one more thing. If you are rolling two dice (we'll make one red and one green) then it is true that there is more than one way to roll 2 threes. You could roll {R3,G3} or {G3,R3} as you stated. But now you are talking about permutations instead of combinations. You can use permutations if you want, but you have to include the fact that there are now more ways to roll a {1,5} and {5,1}:

{R1,G5}
{G1,R5}

{R5,G1}
{G5,R1}

If you use combinations then half of those will cancel out, leaving you with the answer from your original post.

-Sonny-

 
T

tthree

Banned
If you play craps all the combinations should be second nature.
6 ways: 7
5 ways each: 6 and 8
4 ways each: 5 and 9
3 ways each: 4 and 10
2 ways each: 3 and 11
1 way each: 2 and 12

A total of 36 ways. Convert each number of ways into a fraction of the total for each number.
5/36 for a 6.
 
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