<p>Is there a faster way of explaining today's SAT Question of the Day?</p>
<p>The number of ordered pairs (a,b) such that a<b, and a and b belong to {1,2,3,4,5,6},
is equal to the number of two digit combinations (order is not important) with no repeated digits from the set {1,2,3,4,5,6}:
6x5/2 = 15, or on your calculator 6C2=15.
The total number of ordered pairs (a,b) is 6x6=36, so the probability of a<b is
15/36 = 5/12.</p>
<p>How did you get 15?</p>
<p>kevaoe - Combinations. You have 6 all together, you choose 2, how many pairs is that?
6 Cr 2 =15</p>
<p>I just did it like so:
If you roll a 1, then the odds of the second roll being greater is 5/6.
If you roll a 2, then the odds of the second roll being greater is 4/6.
If you roll a 3, then the odds of the second roll being greater is 3/6.
If you roll a 4, then the odds of the second roll being greater is 2/6.
If you roll a 5, then the odds of the second roll being greater is 1/6.
If you roll a 6, then the odds of the second roll being greater is 0/6.</p>
<p>Add these results up, and take the average: 2.5/6 = 5/12.</p>
<p>^ smash, I am not sure why you are taking the average, but the first part of your approach makes sense.
(1/6)(5/6) + (1/6)(4/6) + (1/6)(3/6) + (1/6)(2/6) + (1/6)(1/6) = 5/12.</p>
<p>Another way - just direct listing of all favorable outcomes:
12,13,14,15,16,
23,24,25,26,
34,35,36,
45,46,
56;
altogether now:
5+4+3+2+1=15.
Total number of outcomes is 6x6=36, so the answer is
15/36 = 5/12.</p>
<p>=========================================
Edit:
In case somebody missed the QOTD for Monday, January 12, here it is:
A 6-sided number cube, with faces numbered 1 through 6, is to be rolled twice. What is the probability that the number that comes up on the first roll will be less than the number that comes up on the second roll?</p>
<p>Total no. of possibilities is 36
Total no. of possibilities in which the same numbers turn up in both the throws=6
remaining possibilities =30
since the possibility of a smaller no. turning up in the first throw=the possibility of a larger no. turning up in the first throw......................total no. of favourable outcomes is 15
therefore probability=15/36=5/12</p>
<h1>^cool!</h1>
<p>Just to collect what everyone has said together:</p>
<p>The way to think "out of the box" is probably ccprofile's reasoning. There are 36 different possible ordered pairs of rolls, where the first number in the pair comes from the first roll and the second number in the pair comes from the second roll. Of those pairs, 6 are two rolls of the same number and fail the condition specified. For each of the remaining 30 ordered pairs, if a is not equal to b and if (a,b) is a pair, then (b,a) is also a pair. Only one of them has the first number less than the second. In other words, exactly half of those 30 pairs, 15 pairs, have the first number less than the second.</p>
<p>This reasoning is another way of saying that you choose 2 different numbers from {1,2,3,4,5,6} (credit: gcf101). This is a combination, as Vertigo220h said: 6C2 = 6!/(2!4!) = 15.</p>
<p>Finally, the reason smash20's argument works is because of the following, which is a bit more complicated to explain but still easy to understand:</p>
<p>Think of the problem as two independent events. The first event is the first roll of the dice. The second event is the second roll of the dice. Clearly, the first does not affect the outcome of the second, so the events are independent.</p>
<p>Now, there is a 1/6 chance that the first roll is a 1. If so, there is a 5/6 chance that the second roll is greater. In total, there is a 1/6 * 5/6 = 5/36 chance that the first roll will be a 1 and that the second roll is greater than the first.</p>
<p>There is a 1/6 chance that the first roll is a 2. If so, there is a 4/6 chance that the second roll is greater. In total, there is a 1/6 * 4/6 = 4/36 chance that the first roll will be a 2 and that the second roll is greater than the first.</p>
<p>Repeat for all first rolls up to 6. Since all these events are mutually exclusive, we can add these probabilities up to get the final probability that two dice rolls will have the second greater than the first. Thus, we get</p>
<p>1/6<em>5/6 + 1/6</em>4/6 + 1/6<em>3/6 + 1/6</em>2/6 + 1/6<em>1/6 + 1/6</em>0/6 = 1/6(5/6 + 4/6 + 3/6 + 2/6 + 1/6) = 5/12</p>
<p>This is only an average because the probability of each number for the first roll is equal (it would be different in a weighted die, for example).</p>
<p>Note that this last approach is equivalent, in this case, to just listing out all the combinations, as gcf101 did. You will find that:</p>
<p>For a first roll of 1, 5 acceptable ordered pairs.
For a first roll of 2, 4 acceptable ordered pairs.</p>
<p>and so on, and in the end, you will get a probability of (5+4+3+2+1)/(6*6) = 5/12.</p>