Probability for Dice GAME!!?

Question

This game involved with the host and the player with only one round. Each the player and the host is to throw two dice.

For the player to win, one of the followings must occur when the player throws the dice:
1. the sum is the same for both players
2. the sum is 7

Please help me with the probability for the player to win for each situation.

Answer 1

Let's do part B first, since it is easier:

PART B:

When throwing dice, the possible outcomes of the sums are:
2 = 1/36
3 = 2/36
4 = 3/36
5 = 4/36
6 = 5/36
7 = 6/36
8 = 5/36
9 = 4/36
10 = 3/36
11 = 2/36
12 = 1/36

You can confirm this by figuring out the ways to make each sum.
2 = 1,1
3 = 1,2 ; 2,1
4 = 1,3; 2,2; 3,1
etc.
10 = 4,6; 5,5; 6,4
11 = 5,6 ; 6,5
12 = 6,6

P(roll has a sum of 7)
= 6/36
= 1/6
≈ 16.67%

That gives us the answer to PART B.

PART A:

The chance that they both roll 2 is:
1/36 x 1/36 = 1/1296
The chance that they both roll 3 is:
2/36 x 2/36 = 4/1296
....
The chance that they both roll 7 is:
6/36 x 6/36 = 36/1296
....
The chance that they both roll 11 is:
2/36 x 2/36 = 4/1296
The chance that they both roll 12 is:
1/36 x 1/36 = 1/1296

Adding them up we have:
(1 + 4 + 9 + 16 + 25 + 36 + 25 + 16 + 9 + 4 + 1) / 1296
= 146 / 1296
= 73 / 648
≈ 11.27%

That's the answer to PART A.

PART C:

If you need the chance of either of these events happening:
P(A or B) = P(A) + P(B) - P(A and B).

P(A) = 73/648
P(B) = 1/6

Probability of both events is the same as the player and host rolling seven = 1/6 x 1/6 = 1/36

Putting it all together, the probability the player wins is:
73/648 + 1/6 - 1/36
= 73/648 + 108/648 - 18/648
= (181 - 18) / 648
= 163 / 648
≈ 25.15%

Answer 2

I can help a little but i don't know if it's what your looking for.

1. wouldn't the probability be 1/12? the most the two dice can add up to is twelve. so it goes to figure that the host has a sum between there and the player has to get one of those twelve sums... 12 chances? or did i miss something...

2. 1+6, 2+5, 3+4. Three ways to get 7. But then you have to take into account the probability of getting those numbers for each die then adding the pairs together.

I know... it's kinda confusing.... It's College Math... at least thats what i'm learning right now and sorry that's the best i can explain it. Hope it helps a little.

Answer 3

Who is Homer? Each throw of the dice is an independent event and the probability of 5 or 6 on each occasion is 1/6 + 1/6 = 1/3.

Answer 4

x = sum
x: -----2-----3-----4-----5------6------7-----8------9-----10----11----12
p(x)-1/36-2/36-3/36-4/36-5/36-6/36-5/36-4/36-3/36-2/36-1/36
1- how many players ???? i assume there are 2 players, i guess it's about .1126
2- look at the table