2 Dice and the GMAT. How Does It Apply?
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Bret is a GMAT Instructor for Kaplan. Click here to read more articles from Kaplan and to learn more about Kaplan's GMAT classes. |
While my students certainly accept that they have to take the GMAT, I often hear complaints about the applicability of GMAT topics – especially math – to real world situations. My usual response to this observation is that the GMAT is using math questions to test critical thinking skills that business schools consider essential.
However, some GMAT problems have surprising real world applications outside the realm of business. The sample problem below is an example of such a question. Specifically, it can help you win at craps.
If you want to try the problem on your own, skip down to it now and then return here – part of this discussion will give you clues about the correct answer.
For those of you not familiar with craps, the basic play is fairly straightforward. You roll two dice and if the sum of the result on each die is 7 or 11 you win. If the sum of the results are 2, 3 or 12 you lose. If you roll any other result the game progress to a second step that we do not need to concern ourselves with for purposes of this discussion.
Someone who is not familiar with probability might look at those rules and think, “this game is not fair – I have three ways to lose, but only two ways to win.” But, by understanding probability we can see that craps is actually the most fair game in the casino.
The key is that every result does not have an equal probability of occurring. To role a 2, the first die must roll a 1 and the second die must also role a 1. Likewise, to roll a 12, both dice must come up with a 6. A three is slightly more likely – the first die can be a 1 and the second a 2 or the first die can be a 2 and the second a 1. This gives us four ways to lose in our initial roll.
Rolling an 11 is just as likely as rolling a 3. You can roll a 5 on the first die and a 6 on the second or a 6 on the first and a 5 on the second. However, we can roll a 7 in many ways. We can roll a 1 first and a 6 second, a 2 first and a 5 second, a 3 first and a 4 second, a 4 first and a 3 second, a 5 first and a six second, or a 6 first and a 1 second. Thus, we have six ways to make a seven when we roll our dice.
This means that we have eight ways to win and only four ways to lose. While the rest of the game pushes the odds back into the casino’s favor, on that initial roll we are twice as likely to win as lose.
This concept of desired versus undesired outcomes will show up on many probability questions on the GMAT – give the problem below a try to see how.
Problem:
What is the probability of rolling a total of 7 with a single roll of two fair six-sided dice, each with the distinct numbers 1 through 6 on each side?
(A) 1/12
(B) 1/6
(C) 2/7
(D) 1/3
(E) 1/2
Solution:
When confronted with a probability question, always remember two points. First, probability is just desired outcomes divided by possible outcomes, so you will need to find this ratio. Second, if more than one event occurs, “and” means multiply and “or” means add.
In this problem, we want to roll a combined 7. We can accomplish this in a number of ways. We can roll a 1 with our first die and a 6 with our second, a 2 first and a 5 second, a 3 first and a 4 second, a 4 first and a 3 second, a 5 first and a 2 second, or a 6 first and a 1 second.
This gives us a total of 6 desired outcomes. Next, we will need to find our possible outcomes. Since we have 6 possible outcomes for the first die and 6 possible outcomes for the second die we should multiply 6 x 6 = 36 possible outcomes. Also note, that this is an “and” situation – we roll the first AND second die – so we want to multiply rather than add.
Finally, we put desired outcomes over possible outcomes, to get 6/36 as the probability of rolling a combined 7. 6/36 simplifies to 1/6, which is answer (B).
Written by Kaplan GMAT instructor Bret Ruber.
3 comments
Arunangsu on January 26th, 2012 at 9:19 am
(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)
6/36=1/6
Daniel Toma on January 28th, 2012 at 11:01 am
If you play the "Don't Pass" line, the expectations on the initial roll are reversed, and then switch back to the players' favor if there's at least a second roll.
Alyse on February 5th, 2012 at 10:32 pm
I solved this problem in a much easier way after reading the background. Since the first roll can be any of the numbers or 6/6=1, you can disregard it. But whatever the first roll is, there is only one second roll which will cause their sums to equal 7. The probability of the second roll being the "correct" one is 1/6 and the correct answer minus the extended multiplication!