Simple discrete math dice problem  TOPIC_SOLVED

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Simple discrete math dice problem

Postby liftingmath on Tue Nov 19, 2013 8:31 pm

Okay, the problem states I have to find which situation gives me a better probability with coming out with a sum of 8. Throwing 2 dice or 3 three dice.

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Since a dice has 6 sides the amount of sets for 2 dice is 6^2 or 36.
Now we do the same for three dice 6^3 or 216.

Now I need to find how many combos out of 36 and 216 equal 8. I drew out every combo for the two dice since it was only 36 and came out with 5 different combos equaling 8. So 5/36. My teacher told us there is an easier way using the combo formula
C(n,r). She said since no set of 1's can equal 8 we only have 5 possible n's.

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She showed us in class that to figure out for 2 dice we do:

My question is why does R = 1? Would it not be 2 since it's two dice?

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And for 3 dice she showed
C(6,1) + C(5,1) + C(4,1) + C(3,1) + C(2,1) + C(1,1)

I'm flat out not understanding the formula for the 3 dice. Can anyone make any sense of this for me? Thanks everyone. Forum rocks. I should be spending my time on these sites instead of bodybuilding forums lol.
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Re: Simple discrete math dice problem  TOPIC_SOLVED

Postby stapel_eliz on Thu Nov 21, 2013 6:35 pm

It really would have helped if she'd provided her reasoning. As it is, her "work" is quite cryptic. My guess is that she's not showing a bunch of her reasoning!

For the two-dice case, I think that she's assuming that the first die has been thrown, so she knows the first value. To get a sum of eight, the first dice has to show more than 1 (because the highest value for the second die is 6, and 1 + 6 doesn't equal 8). So it'll be 2 through 6. For the second die, the value has to be 8 - (whatever the first value was) to equal 8, so she's needing a 2 through a 6. The "C(5, 1)" may be her way of saying that she's counting the number of ways to get one of those five values. Maybe...?

Then maybe for the second situation, she's saying that she's thrown the first die, and gotten something between 1 and 6. (Since she's throwing three dice, she can use 1's this time: 1 + 1 + 6 = 8.) Suppose the first die rolls a 1. Then she's looking at the number of ways of getting 7 with the remaining two dice. Say she rolls an "x" on the second die; then she's needing 7 - x for the third die. Since 1 < x < 6 for the second die, then the target value for the third die is 6, 5, 4, 3, 2, or 1. There are six ways to pick one of these values.

Now suppose the first die rolls a 2. Then she's looking at the number of ways of getting 6 with the remaining two dice. Using "x" again for the value of the roll of the second die, she'll be needing 6 - x for the third die. The target value will be 5, 4, 3, 2, or 1. There are five ways to pick one of these values.

And so forth.

But that's just my guess. 8-)
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