question below.?

Question

In a dice game, two tice are thrown. If the sum of the dots is under seven, the player recives 8 points. If the sum of the dots is seven or greater, the player loses 6 points. How many points can a player expect to win or lose if 100 games are played? (to the nearest point and indicate whether it is a gain or loss)

Please give me a good explaination, for I am not that good at math and i'm only in middle school. THANKS!!

Answer 1

you have two dice, each die has 6 numbers
you have 36 possible outcomes (6*6)
if its easier, you can list them out

you can roll a

1 and 1=2
1 and 2=3
1 and 3=4
1 and 4=5
1 and 5=6
1 and 6=7
2 and 1=3
2 and 2=4
2 and 3=5
2 and 4=6
2 and 5=7
2 and 6=8
3 and 1=4
3 and 2=5
3 and 3=6
3 and 4=7
3 and 5=8
3 and 6=9
4 and 1=5
4 and 2=6
4 and 3=7
4 and 4=8
4 and 5=9
4 and 6=10
5 and 1=6
5 and 2=7
5 and 3=8
5 and 4=9
5 and 5=10
5 and 6=11
6 and 1=7
6 and 2=8
6 and 3=9
6 and 4=10
6 and 5=11
6 and 6=12

there are 15/36 (5/12) combinations listed above with a sum of less than 7 dots

so 5/12 of 100 games the player will receive 8 pts.
5/12 * 100 = 41.67 games that player will receive 8 pts
41.67*8 = 333.33 points

there are 21/36 (7/12) combinations listed above with a sum of 7 or more dots

so 7/12 of 100 games the player will lose 6 points
7/12 * 100 = 58.33 games that player will lose 6 point
58.33*6 = 350 points that a player will lose

if a player wins 333 pts and loses 350 pts, he nets (17 pts)

hope this is correct

Answer 2

Okay, what you want is to find ou the probability that the sum of the dice is less than 7 and the prob that the sum is greater or equal to 7.

To figure this out, draw up a table with numbers 1 to 6 on the top and 1 to 6 on the side. in each square (eg (3,4)) you have the prob that you get eg 3 on one die and 4 on the other. This is just 1/36 for all of the squares. Now you just add up the numbers eg 3+4=7 and hey presto thats the prob that the sum is 7. All the probs on the diagonal should be the prob that the sum is 7. So all you have to do is add up the probs from sum=2 to sum=6, and then sum=7 to sum=12, to get the probs. You then multiply these by the points( 8 for winning and -6 for losing). Then you multiply by 100 and hey presto theres your answer. I won't dignify you with the actual answer because I want you to understand how to do it.

Answer 3

Figure that there are 36 (6x6) possible combinations of points in a two dice throw. Of these:

1,1 2,1 3,1 4,1 5,1
1,2 2,2 3,2 4,2
1,3 2,3 3,3
1,4 2,4
1,5

Are less than 7 (15)

1,6
2,5
3,4
4,3
5,2
6,1

Are equal to 7 (6)

So that means that 15 (36 - 15 - 6) are greater than 7.

The probability that any throw will be <7 is 15 / 36 and the probability that any throw will >7 is 15 / 36. In one hundred throws, a player can expect to have 100 * 15 / 36 = 41.67 throws above 7 and 41.67 throws below 7, so 41.67 * 8 = $333.33 won and 41.67 * 6 = $250.02 lost. 333.33 - 250.02 = $83.31 profit each 100 games.

Answer 4

The question is, how much is the probability of getting a sum under 7. It is the number of outcomes under sum 7 divided by the number of all outcomes.
The number of all outcomes is 6*6 = 36.
The outcomes below sum 7 are:
1,1
1,2 2,1
1,3 2,2 3,1
1,4 2,3 3,2 4,1
1,5 2,4 3,3 4,2 5,1

/1,6 2,5 3,4 4,3 5,2 6,1
2,6 3,5 4,4 5,3 6,2
3,6 4,5 5,4 6,3
4,6 5,5 6,4
5,6 6,5
6,6/

alltogether 15.

(I put all the combinations over sum 7 in /-s to check if I get all the 36)

So the probability is P = 15/36 = 5/12 = 41.66 %

the expected number of points in one flip is: P*8 + (1-P)*(-6) = 2P-6 = -5.166

100 times:

-516.66