Sorry! for posting in the wrong forum. Should have posted in PS.

manik11 wrote:Brian plays a game in which two fair, six-sided dice are rolled simultaneously. For each die, an even number means that he wins that amount of money and an odd number means that he loses that amount of money. What is the probability that Brian makes money on his first roll?

11/12

7/12

1/2

5/12

1/3

OA : B
Source : Veritas Prep

So here's how I solved it and would like to know from the experts whether my approach is legit.
I would also like to see any alternate approach which could save me some time on such questions.

Since even on any dice = win
D1 : 2,4,6 (irrespective of the number on D2)
D2 : 2,4,6 (irrespective of the number on D1)

The above info gives us 3 unique favorable outcomes

Now for combination of dice we need an even number for (D1 + D2)

D1 D2
1 1
1 3
1 5
------
2 2
2 4
2 6
.
.
.
Here I can spot a pattern that will give me 18 favorable outcomes and rolling two dice together accounts for 36 possible outcomes

So, Probability of a win = 18+3/36 = 21/36 = 7/12

Thanks,
Manik

I think you may have misunderstood the premise here. We're not interested in the sum of the two dice, but in the net effect. Say, for example, that I rolled a (1, 1) In this case, because I lose money on each individual die that comes up odd, each of those rolls causes me to lose $1. So, a (1,1) is a $2 loss: (-1) + (-1) = -2.

However, if I roll a (2,1), I'd be +2 for my even roll and -1 on my odd roll, so this combination grants me a net of +1, and it's a gain for me.

So it isn't the sum of the dice that determines the outcome, it's the combined effect of each individual die.

First, we can calculate the total number of possible outcomes. 6*6 = 36. So that's our denominator.

Then we can find our desired outcomes. If we use brute force, we'd go roll by roll.

If I roll a '1' on the first, I only win if I get a 2, 4, 6, on the second toss. (It's a -1 for the first roll, so the +2, +4, or +6, will leave me with a net positive.) So 3 winning outcomes

If I roll a '2' on the first roll, I win money if I get a 1, 2, 4, or 6 on the second roll. 4 positive outcomes

If I roll a '3' on the first, I win if I get a 4 or a 6 on the second roll; 2 positive outcomes

If I roll a '4 on the first, I win if I get a 1, 2, 3, 4, or 6 on the second roll; 5 positive outcomes

If I roll a '5' on the first, I only win if I roll a 6 the second; 1 positive outcome

If I roll a '6' on the first, I win no matter what I get on the second roll, so 6 positive outcomes.

Add up the positive outcomes: 3 + 4 + 2 + 5 + 1 + 6 = 21

21/36 = 7/12

Alternatively, if you were pressed for time, you can start by looking at the answer choices. 11/12 seems awfully high. So as soon as I can prove that the odds are better than 1/2, I'm done. And logically, the odds should be better than 1/2 that I come out with a net positive, as the odd numbers (1, 3, 5,) are, on average, smaller than the even numbers (2, 4, 6)

Thanks a lot Dave...I completely misunderstood this one

manik11 wrote:Thanks a lot Dave...I completely misunderstood this one

This is a tough one - don't feel bad!