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Mary and Joe are to throw three dice each. The score is the [#permalink]
14 Jul 2008, 22:23

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Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?

Yes you are correct my friend as total no. of the desired outcomes is 8(11,12,13,14,15,16,17,18) and total no. of outcomes are 16 (3-18) so the probability is 8/16 ie 1/2

durgesh79 wrote:

i'm not sure if this is the right way to do it. I actually made an excel table of all 216 options to prove it

suppose there in one dice score can range from 1 to 6, probablity that score will be greater than 3.5 (average of 1 and 6) = 3/6 = 1/2

suppose there are two dice score can range from 2 to 12, probability that score will be greater than 7 (average of 2 and 12) = 1/2

similarly if there are 3 dice, score can range from 3 to 18, avarage = 10.5, probability = 1/2 = 32/64

The logic : the probablity will be evenly distributed on both sides of avegare sum.

Yes you are correct my friend as total no. of the desired outcomes is 8(11,12,13,14,15,16,17,18) and total no. of outcomes are 16 (3-18) so the probability is 8/16 ie 1/2

it will only work when the sum we are looking for starts from middle of the range

For example if will not work in below case

what is the probablility that the sum is more than 17.

in this case the right answer is 1/216 ..... using your logic of desired outcome (only one .... 18) and total no of outcome 16 (3-18) one can make a mistake of calculating 1/16 which is wrong ....

Yes you are right +1 to you for correcting me.....

Is there any other method to calculate P(6)= (1(2,2,2)+6(3,2,1)+3(4,1,1))/ 216 or is there any other method to solve this as in this there is any chance to forget some cases..............

I first started thinking about the different results for each die, which is 6 * 6 * 6 = 216. But this was pointless. We need to focus on the results. The result of her throw is 10. The way she goes about getting it doesn't matter. If she throws 2-2-6, 3-1-6, 4-4-2, it doesn't matter. There are not 216 possibilities...there are 16. She can roll anything from 1-1-1 up to 6-6-6. That's a total of 3 up to 18. so 18-3 = 15, but include the 3, so 16 total chances.

We have to figure out the probability that Joe scores an 11 or higher. Simply count them. 8. 8 out of 16 = 1/2. While there isn't a 1/2 in the answer choices, there is a 32/64...I choose that as my answer. B

I'm not sure how sound this approach is. Will it work for other situations?

aaron22197 wrote:

Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?

So what would be a good, simple way to approach this problem? Lets say same scenario, but we want to know the probability of scoring greater than 13 (i.e, 14 - 18)

___ ___ ___

We all agree, there are 216 total possibilities.

First can be anything, so 6 Second can be 2 through 6, but this result will change options available with #3 Third is dependent upon second roll:

Second Roll | Possible 3rd rolls to Sum >= 14 1 {0} Not possible because 6 + 1 = 7 leaving 7 left and there isn't a 7 on the die 2 {6} Only 1 possible. 6 + 2 = 8, 14 - 8 = 6 (must be a 6 if second roll = 2) 3 {6,5} 4 {6,5,4} 5 {6,5,4,3} 6 {6,5,4,3,2}

So there are 30 possibilities for the first 2 rolls (6*5). Then....I get lost from this point. Can someone help me out. I think i've got it pretty organized, but i'm not sure how to go about computing the 3rd roll being dependent upon the second roll.

HELP!! lol _________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

So what would be a good, simple way to approach this problem? Lets say same scenario, but we want to know the probability of scoring greater than 13 (i.e, 14 - 18)

___ ___ ___

We all agree, there are 216 total possibilities.

First can be anything, so 6 Second can be 2 through 6, but this result will change options available with #3 Third is dependent upon second roll:

Second Roll | Possible 3rd rolls to Sum >= 14 1 {0} Not possible because 6 + 1 = 7 leaving 7 left and there isn't a 7 on the die 2 {6} Only 1 possible. 6 + 2 = 8, 14 - 8 = 6 (must be a 6 if second roll = 2) 3 {6,5} 4 {6,5,4} 5 {6,5,4,3} 6 {6,5,4,3,2}

So there are 30 possibilities for the first 2 rolls (6*5). Then....I get lost from this point. Can someone help me out. I think i've got it pretty organized, but i'm not sure how to go about computing the 3rd roll being dependent upon the second roll.

HELP!! lol

Your early approach doesn't work, durgesh is right: it actually only works if the events are equiprobable and it is NOT the case with sum of dices.

For your second problem, I don't see a better solution than counting the favorable outcomes:

First dice is 6: - second dice is 6 : 6 possibilities for dice 3 - second dice is 5 : 5 possibilities for dice 3 - second dice is 4 : 4 possibilities for dice 3 - second dice is 3 : 3 possibilities for dice 3 - second dice is 2 : 2 possibilities for dice 3 - second dice is 1 : 1 possibilities for dice 3 => 21 possibilities

First dice is 5: - second dice is 6 : 5 possibilities for dice 3 - second dice is 5 : 4 possibilities for dice 3 - second dice is 4 : 3 possibilities for dice 3 - second dice is 3 : 2 possibilities for dice 3 - second dice is 2 : 1 possibilities for dice 3 => 15 possibilities

First dice is 4: - second dice is 6 : 4 possibilities for dice 3 - second dice is 5 : 3 possibilities for dice 3 - second dice is 4 : 2 possibilities for dice 3 - second dice is 3 : 1 possibilities for dice 3 => 10 possibilities

First dice is 3: - second dice is 6 : 3 possibilities for dice 3 - second dice is 5 : 2 possibilities for dice 3 - second dice is 4 : 1 possibilities for dice 3 => 6 possibilities

First dice is 2: - second dice is 6 : 2 possibilities for dice 3 - second dice is 5 : 1 possibilities for dice 3 => 3 possibilities

First dice is 1: - second dice is 6 : 1 possibilities for dice 3 => 1 possibility

Total: 21+15+10+6+3+1 = 56 possibilities out of the 6^3 = 216 total possiblities

Answer is 56/216 = 23/108

Edit: I can formalize it a posteriori

If first dice = k and second dice = i, then there are k+i-6 possibilities for the third dice if k+i > 6, 0 otherwise

# possibilities is therefore \sum_{k=1}^6 \left( \sum_{i=7-k}^6 (k+i-6) \right) (that I will let you compute if you want as an exercise )

Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?

I agree with Oski that there is no faster way to solve these problems than to enumerate the possibilities. In more advanced probability, you learn to solve these questions using 'generating functions'. Generating functions are pretty interesting, and aren't too difficult to understand, so I'll explain them, but first I want to make absolutely clear: this will not be at all helpful for the GMAT. For interest only!

We have a die, numbered from 1 to 6. Consider this polynomial:

t^1 + t^2 + t^3 + t^4 + t^5 + t^6

For each face on the die, I've added a term to the polynomial, and the exponents in the polynomial are equal to the numbers on the die. The idea here is to think of the polynomial as representing the number of ways you can get each result: the t^1 term means 'there is one way to get a 1', the t^2 term means 'there is one way to get a 2', and so on.

Now let's see what happens if we square this polynomial:

Notice, when you calculate the '4t^5' term, for example, you arrive at it because you add (t^1)*(t^4) + (t^2)*(t^3) + (t^3)*(t^2) + (t^4)*(t^1). If you think of t^1 to mean 'rolling a 1', t^2 to mean 'rolling a 2', and so on, you can see that when we square our original polynomial, the '4t^5' term adds up all of the different ways we might roll a 5 on two dice: we can roll (1,4), (2,3), (3,2) or (4,1). That is, this polynomial:

tells us exactly how many ways we can get any sum on two dice: we can get a 2 in one way, a 3 in two ways, a 4 in three ways, and so on- the exponents represent the sums we can get, and the coefficients (the numbers in front of the t terms) tell us how many ways we can get that sum. So if you want to know how many ways you can get a sum of 13 on three dice, you can work out:

(t^1 + t^2 + t^3 + t^4 + t^5 + t^6)^3

and look at the number in front of t^13.

That's still a lot of work, but one of the fun things about generating functions is that you can answer any kind of dice question- and a lot of other questions as well. Say you have one six-sided die with the numbers 4, 4, 4, 5, 5, 6 on it. Then we can write down the generating function:

That tells us there's three ways to get a 4, two ways to get a 5, and one way to get a 6. Say we have another die with the numbers 2, 2, 3, 3, 4, 4 on it. We have the generating function:

2t^2 + 2t^3 + 2t^4

How many ways can we get a sum greater than 8 if we roll these dice together? Just multiply the two generating functions:

(3t^4 + 2t^5 + t^6)*(2t^2 + 2t^3 + 2t^4)

= 6t^6 + 10t^7 + 12t^8 + 6t^9 + 2t^10

So there are six ways to get a sum of 9, and two ways to get a sum of 10. If you need to know the probability of getting a sum greater than 8, just divide by 6*6 = 36, the total number of possible outcomes (which is also, of course, the sum of the coefficients: 6 + 10 + 12 + 6 + 2 = 36).

You can even use generating functions to answer the question "How many ways can you get exactly three heads if you flip a coin five times?" If we flip a coin once, we can get either 1 head or 0 heads. So we can use the generating function t^0 + t^1 = 1 + t, which just means 'there is one way to get zero heads, and one way to get one head'. The exponent counts the number of heads we can get. If we flip a coin five times, we can work out

(1 + t)^5 = 1 + 5t^1 + 10t^2 + 10t^3 + 5t^4 + t^5

So there are 10 ways to get exactly three heads. This is one of many ways to demonstrate the relationship between coin-tossing questions and the binomial theorem.

On the GMAT, there will *always* be a faster way to solve a problem than using generating functions, but in more advanced counting/probability, this can be very useful. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

The best thing to do here is take into account the symmetry. Sums of 3 and 18 are equally likely, as are 4 and 17 and 5 and 16 ... and 10 and 11. Thus the probability that the sum is greater than 10 is 1/2

I agree with Oski that there is no faster way to solve these problems than to enumerate the possibilities. In more advanced probability, you learn to solve these questions using 'generating functions'. Generating functions are pretty interesting, and aren't too difficult to understand, so I'll explain them, but first I want to make absolutely clear: this will not be at all helpful for the GMAT. For interest only!

Thanks ! It was indeed very interesting (I never thought of that ).