1 00:00:00,000 --> 00:00:00,820 2 00:00:00,820 --> 00:00:01,640 Hi. 3 00:00:01,640 --> 00:00:04,500 Today we're going to do another fun problem that 4 00:00:04,500 --> 00:00:07,750 involves rolling two dice. 5 00:00:07,750 --> 00:00:11,210 So if you guys happen to frequent casinos, this problem 6 00:00:11,210 --> 00:00:13,750 might be really useful for you. 7 00:00:13,750 --> 00:00:14,680 I'm just kidding. 8 00:00:14,680 --> 00:00:18,310 But in all seriousness, this problem is a good problem, 9 00:00:18,310 --> 00:00:22,760 because it's going to remind us how and when to use the 10 00:00:22,760 --> 00:00:24,060 discrete uniform law. 11 00:00:24,060 --> 00:00:26,650 Don't worry, I'll review what that says. 12 00:00:26,650 --> 00:00:29,140 And it's also going to exercise your understanding of 13 00:00:29,140 --> 00:00:31,030 conditional probability. 14 00:00:31,030 --> 00:00:32,110 So quick recap. 15 00:00:32,110 --> 00:00:35,140 The discrete uniform law says that when your sample space is 16 00:00:35,140 --> 00:00:38,670 discrete, and when the outcomes in your sample space 17 00:00:38,670 --> 00:00:42,750 are equally likely, then to compute the probability of any 18 00:00:42,750 --> 00:00:47,855 event A, you can simply count the number of outcomes in A 19 00:00:47,855 --> 00:00:52,730 and divide it by the total number of possible outcomes. 20 00:00:52,730 --> 00:00:55,130 OK, so coming back to our problem. 21 00:00:55,130 --> 00:00:59,060 The problem statement tells us that we roll two fair 22 00:00:59,060 --> 00:01:01,150 six-sided die. 23 00:01:01,150 --> 00:01:05,069 And it also tells us that each one of the 36 possible 24 00:01:05,069 --> 00:01:08,940 outcomes is assumed to be equally likely. 25 00:01:08,940 --> 00:01:12,220 So you know alarm bell should be going off in your head. 26 00:01:12,220 --> 00:01:14,810 Our sample space is clearly discrete. 27 00:01:14,810 --> 00:01:16,930 And it says explicitly that all 28 00:01:16,930 --> 00:01:18,380 outcomes are equally likely. 29 00:01:18,380 --> 00:01:22,020 So clearly, we can use the discrete uniform law. 30 00:01:22,020 --> 00:01:25,850 And again, this is helpful because it reduces a problem 31 00:01:25,850 --> 00:01:30,090 of computing probabilities to a problem of counting. 32 00:01:30,090 --> 00:01:32,950 OK, and before we go any further, I just want to review 33 00:01:32,950 --> 00:01:34,850 what this graph is plotting. 34 00:01:34,850 --> 00:01:38,990 You've seen it a few times, but just to clarify, on one 35 00:01:38,990 --> 00:01:42,430 axis, we're plotting the outcome of the first die roll, 36 00:01:42,430 --> 00:01:44,960 and on the second axis, we're plotting the outcome of the 37 00:01:44,960 --> 00:01:46,810 second die roll. 38 00:01:46,810 --> 00:01:50,750 So if you got a 4 on your first die, and you get a 1 on 39 00:01:50,750 --> 00:01:54,590 your second die, that corresponds to this point over 40 00:01:54,590 --> 00:01:56,940 4 and up 1. 41 00:01:56,940 --> 00:02:01,060 OK, so part a asks us to find the probability 42 00:02:01,060 --> 00:02:04,500 that doubles our rolls. 43 00:02:04,500 --> 00:02:07,480 So let's use some shorthand. 44 00:02:07,480 --> 00:02:10,960 We're going to let D be the event that doubles are rolled. 45 00:02:10,960 --> 00:02:14,720 And we want to compute the probability of D. 46 00:02:14,720 --> 00:02:17,570 I argue before we can use the discrete uniform law. 47 00:02:17,570 --> 00:02:22,470 So if we apply that, we just get the number of outcomes 48 00:02:22,470 --> 00:02:26,050 that comprise the event "doubles rolled" divided by 49 00:02:26,050 --> 00:02:30,360 36, because there are 36 possible outcomes, which you 50 00:02:30,360 --> 00:02:34,380 can see just by counting the dots in this graph. 51 00:02:34,380 --> 00:02:37,710 Six possible outcomes for the first die, six possible 52 00:02:37,710 --> 00:02:38,970 outcomes for the second die. 53 00:02:38,970 --> 00:02:39,555 That's how you-- 54 00:02:39,555 --> 00:02:42,800 6 times 6 is 36. 55 00:02:42,800 --> 00:02:45,440 So I've been assuming this entire time that you know what 56 00:02:45,440 --> 00:02:46,250 doubles are. 57 00:02:46,250 --> 00:02:50,470 For those of you who don't know, doubles is essentially 58 00:02:50,470 --> 00:02:54,190 when that number on the first die matches the number on the 59 00:02:54,190 --> 00:02:56,120 second die. 60 00:02:56,120 --> 00:03:00,480 So this outcome here 1-1 is part of the event "doubles 61 00:03:00,480 --> 00:03:08,076 rolled." Similarly, 2-2, 3-3, 4-4, 5-5, and 6-6-- 62 00:03:08,076 --> 00:03:11,510 these six points comprise the event "doubles rolled." 63 00:03:11,510 --> 00:03:16,000 So we can go ahead and put 6 over 36, 64 00:03:16,000 --> 00:03:18,340 which is equal to 1/6. 65 00:03:18,340 --> 00:03:20,730 So we're done with part a. 66 00:03:20,730 --> 00:03:22,530 We haven't seen any conditioning yet. 67 00:03:22,530 --> 00:03:24,840 The conditioning comes in part b. 68 00:03:24,840 --> 00:03:28,060 So in part b we're still interested in the event D, in 69 00:03:28,060 --> 00:03:31,020 the event that "doubles are rolled." But now we want to 70 00:03:31,020 --> 00:03:35,710 compute this probability conditioned on the event that 71 00:03:35,710 --> 00:03:39,930 the sum of the results is less than or equal to 4. 72 00:03:39,930 --> 00:03:45,150 So I'm going to use this shorthand sum less than or 73 00:03:45,150 --> 00:03:50,180 equal to 4 to denote the event that the role results in the 74 00:03:50,180 --> 00:03:53,010 sum of 4 or smaller. 75 00:03:53,010 --> 00:03:55,570 So there's two ways we're going to go about 76 00:03:55,570 --> 00:03:57,660 solving part b. 77 00:03:57,660 --> 00:03:59,900 Let's just jump right into the first way. 78 00:03:59,900 --> 00:04:03,330 The first way is applying the definition of conditional 79 00:04:03,330 --> 00:04:04,430 probability. 80 00:04:04,430 --> 00:04:10,440 So hopefully you remember that this is just probability of D 81 00:04:10,440 --> 00:04:14,480 intersect sum less than or equal to 4, divided by 82 00:04:14,480 --> 00:04:19,870 probability of sum less than or equal to 4. 83 00:04:19,870 --> 00:04:23,980 Now, sum less than or equal to 4 and D intersect sum less 84 00:04:23,980 --> 00:04:26,950 than or equal to 4 are just two events. 85 00:04:26,950 --> 00:04:29,740 And so we can apply the discrete uniform law to 86 00:04:29,740 --> 00:04:33,170 calculate both the numerator and the denominator. 87 00:04:33,170 --> 00:04:36,050 So let's start with the denominator first because it 88 00:04:36,050 --> 00:04:38,070 seems a little bit easier. 89 00:04:38,070 --> 00:04:42,930 So sum less than or equal to 4, let's figure this out. 90 00:04:42,930 --> 00:04:45,970 Well, 1-1 gives us a sum of 2, that's less 91 00:04:45,970 --> 00:04:48,000 than or equal to 4. 92 00:04:48,000 --> 00:04:49,535 2-1 gives us 3. 93 00:04:49,535 --> 00:04:51,346 3-1 gives us 4. 94 00:04:51,346 --> 00:04:55,960 4-1 gives us 5, so we don't want to include this or this, 95 00:04:55,960 --> 00:04:57,160 or this point. 96 00:04:57,160 --> 00:04:59,380 And you can sort of convince yourself that the next point 97 00:04:59,380 --> 00:05:01,420 we want to include is this one. 98 00:05:01,420 --> 00:05:05,040 That corresponds to 2-2, which is 4, so it makes sense that 99 00:05:05,040 --> 00:05:09,410 these guys should form the boundary, because all dots 100 00:05:09,410 --> 00:05:12,540 sort of up and to the right will have a bigger sum. 101 00:05:12,540 --> 00:05:14,350 3-1 gives us 4. 102 00:05:14,350 --> 00:05:18,170 And 1-2 gives us 3. 103 00:05:18,170 --> 00:05:20,172 So these six points-- 104 00:05:20,172 --> 00:05:22,850 1, 2, 3, 4, 5, 6-- 105 00:05:22,850 --> 00:05:26,970 are the outcomes that comprise the event sum less than or 106 00:05:26,970 --> 00:05:28,910 equal to 4. 107 00:05:28,910 --> 00:05:34,480 So we can go ahead and write in the denominator, 6 over 36, 108 00:05:34,480 --> 00:05:38,310 because we just counted the outcomes in sum less than or 109 00:05:38,310 --> 00:05:40,620 equal to, 4 and divided it by the number 110 00:05:40,620 --> 00:05:43,170 of outcomes in omega. 111 00:05:43,170 --> 00:05:45,510 Now, let's compute the numerator. 112 00:05:45,510 --> 00:05:49,410 D intersect sum less than or equal to 4. 113 00:05:49,410 --> 00:05:51,920 So we already found the blue check marks. 114 00:05:51,920 --> 00:05:55,380 Those correspond to sum less than or equal to 4. 115 00:05:55,380 --> 00:05:58,350 Out of the points that have blue check marks, which one 116 00:05:58,350 --> 00:05:59,740 correspond to doubles? 117 00:05:59,740 --> 00:06:02,590 Well, they're actually already circled. 118 00:06:02,590 --> 00:06:04,370 It's just these two points. 119 00:06:04,370 --> 00:06:08,920 So we don't even need to circle those, so we get 2 over 120 00:06:08,920 --> 00:06:11,830 36, using the discrete uniform law. 121 00:06:11,830 --> 00:06:15,080 And you see that these two 36s cancel each other. 122 00:06:15,080 --> 00:06:19,980 So you just get 2/6 or 1/3. 123 00:06:19,980 --> 00:06:23,780 So that is one way of solving part b, but I want to take 124 00:06:23,780 --> 00:06:25,930 you, guys, through a different way, which I think is 125 00:06:25,930 --> 00:06:30,240 important, and that make sure you really understand what 126 00:06:30,240 --> 00:06:32,630 conditioning means. 127 00:06:32,630 --> 00:06:39,000 So another way that you can solve part b is to say, OK, we 128 00:06:39,000 --> 00:06:41,710 are now in the universe, we are in the conditional 129 00:06:41,710 --> 00:06:44,730 universe, where we know the sum of our 130 00:06:44,730 --> 00:06:49,090 results is 4 or smaller. 131 00:06:49,090 --> 00:06:55,600 And so that means our new sample space is really just 132 00:06:55,600 --> 00:06:58,125 this set of six points. 133 00:06:58,125 --> 00:07:01,430 134 00:07:01,430 --> 00:07:05,720 And one thing that it's worth noting is that conditioning 135 00:07:05,720 --> 00:07:09,780 never changes the relative frequencies or relative 136 00:07:09,780 --> 00:07:12,510 likelihoods of the different outcomes. 137 00:07:12,510 --> 00:07:16,530 So because all outcomes were equally likely in our original 138 00:07:16,530 --> 00:07:19,770 sample space omega, in the conditional worlds the 139 00:07:19,770 --> 00:07:22,280 outcomes are also equally likely. 140 00:07:22,280 --> 00:07:25,400 So using that argument, we could say that in our sort of 141 00:07:25,400 --> 00:07:27,330 blue conditional universe all of the 142 00:07:27,330 --> 00:07:29,570 outcomes are equally likely. 143 00:07:29,570 --> 00:07:32,910 And therefore, we can apply a conditional version of the 144 00:07:32,910 --> 00:07:34,540 discrete uniform law. 145 00:07:34,540 --> 00:07:37,800 So namely, to compute the probability of some event in 146 00:07:37,800 --> 00:07:39,160 that conditional world. 147 00:07:39,160 --> 00:07:42,760 So the conditional probability that "doubles are rolled", we 148 00:07:42,760 --> 00:07:46,440 need only count the number of outcomes in that event and 149 00:07:46,440 --> 00:07:49,430 divide it by the total number of outcomes. 150 00:07:49,430 --> 00:07:55,000 So in the conditional world, there's only two outcomes that 151 00:07:55,000 --> 00:07:58,670 comprise the event "doubles rolled." These are the only 152 00:07:58,670 --> 00:08:00,770 two circles in the blue region, right? 153 00:08:00,770 --> 00:08:04,290 So applying the conditional version number 154 00:08:04,290 --> 00:08:05,770 law, we have two. 155 00:08:05,770 --> 00:08:09,240 And then we need to divide by the size of omega. 156 00:08:09,240 --> 00:08:12,920 So our conditional universe, we've already said, has six 157 00:08:12,920 --> 00:08:14,320 possible dots. 158 00:08:14,320 --> 00:08:18,020 So we just divide by 6, and you see that we get the same 159 00:08:18,020 --> 00:08:20,420 answer of 1/3. 160 00:08:20,420 --> 00:08:23,010 And so again, we used two different strategies. 161 00:08:23,010 --> 00:08:26,040 I happen to prefer the second one, because it's slightly 162 00:08:26,040 --> 00:08:30,260 faster and it makes you think about what does conditioning 163 00:08:30,260 --> 00:08:31,480 really mean. 164 00:08:31,480 --> 00:08:34,840 Conditioning means you're now restricting your attention to 165 00:08:34,840 --> 00:08:36,179 a conditional universe. 166 00:08:36,179 --> 00:08:39,700 And given that you're in this conditional universe where the 167 00:08:39,700 --> 00:08:42,730 sum was less than or equal to 4, what is then the 168 00:08:42,730 --> 00:08:46,192 probability that doubles also happened? 169 00:08:46,192 --> 00:08:48,790 OK, hopefully you, guys, are following. 170 00:08:48,790 --> 00:08:51,610 Let's move on to part c. 171 00:08:51,610 --> 00:08:55,680 So part c asks for the probability that at least one 172 00:08:55,680 --> 00:08:58,510 die roll is a 6. 173 00:08:58,510 --> 00:09:03,190 So I'm going to use the letter S to denote this, the 174 00:09:03,190 --> 00:09:06,120 probability that at least one die roll is a 6. 175 00:09:06,120 --> 00:09:08,250 So let's go back to our picture and 176 00:09:08,250 --> 00:09:11,310 we'll use a green marker. 177 00:09:11,310 --> 00:09:17,120 So hopefully you agree that anything in this column 178 00:09:17,120 --> 00:09:20,520 corresponds to at least one 6. 179 00:09:20,520 --> 00:09:24,850 So this point, this point, this point, this point, this 180 00:09:24,850 --> 00:09:30,630 point, and this point your first die landed on a 6, so at 181 00:09:30,630 --> 00:09:32,720 least one 6 is satisfied. 182 00:09:32,720 --> 00:09:38,440 Similarly, if your second die has a 6, then we're also OK. 183 00:09:38,440 --> 00:09:41,960 So I claim we want to look at these 11 points. 184 00:09:41,960 --> 00:09:43,885 Let me just check that, yeah, 6 plus 5-- 185 00:09:43,885 --> 00:09:44,850 11. 186 00:09:44,850 --> 00:09:50,110 So using the discrete uniform law again, we get 187 00:09:50,110 --> 00:09:52,150 11 divided by 36. 188 00:09:52,150 --> 00:09:54,660 189 00:09:54,660 --> 00:09:58,210 OK, last problem, we're almost done. 190 00:09:58,210 --> 00:10:02,930 So again, we're interested in the event S again, so the 191 00:10:02,930 --> 00:10:05,810 event that at least one die roll is a 6. 192 00:10:05,810 --> 00:10:09,340 But now we want to compute the probability of that event in 193 00:10:09,340 --> 00:10:12,380 the conditional world where the two dice land 194 00:10:12,380 --> 00:10:13,630 on different numbers. 195 00:10:13,630 --> 00:10:16,020 196 00:10:16,020 --> 00:10:19,590 So I'm going to call this probability of S. Let's see, 197 00:10:19,590 --> 00:10:21,910 I'm running out of letters. 198 00:10:21,910 --> 00:10:26,350 Let's for lack of a better letter, my name is Katie, so 199 00:10:26,350 --> 00:10:29,640 we'll just use a K. We want to compute the probability of S 200 00:10:29,640 --> 00:10:33,970 given K. And instead of using the definition of conditional 201 00:10:33,970 --> 00:10:37,660 probability, like we did back in part b, we're going to use 202 00:10:37,660 --> 00:10:39,520 the faster route. 203 00:10:39,520 --> 00:10:48,820 So essentially, we're going to find the number of outcomes in 204 00:10:48,820 --> 00:10:51,715 the conditional world. 205 00:10:51,715 --> 00:10:54,720 206 00:10:54,720 --> 00:10:57,770 And then we're also going to compute the number of outcomes 207 00:10:57,770 --> 00:11:00,710 that comprise S in the conditional world. 208 00:11:00,710 --> 00:11:04,610 So let's take a look at this. 209 00:11:04,610 --> 00:11:10,160 We are conditioning on the event that the two dice land 210 00:11:10,160 --> 00:11:11,590 on different numbers. 211 00:11:11,590 --> 00:11:15,860 So hopefully you agree with me that every single dot that is 212 00:11:15,860 --> 00:11:18,760 not on the diagonal, so every single dot that doesn't 213 00:11:18,760 --> 00:11:23,160 correspond to doubles, is a dot that we care about. 214 00:11:23,160 --> 00:11:27,970 So our conditional universe of that the two dice land on 215 00:11:27,970 --> 00:11:34,066 "different numbers", that corresponds to these dots. 216 00:11:34,066 --> 00:11:37,550 217 00:11:37,550 --> 00:11:42,900 And it corresponds to these dots. 218 00:11:42,900 --> 00:11:46,740 I don't want to get this one. 219 00:11:46,740 --> 00:11:49,650 OK, that's good. 220 00:11:49,650 --> 00:11:54,100 So let's see, how many outcomes do we have in our 221 00:11:54,100 --> 00:11:56,660 conditional world? 222 00:11:56,660 --> 00:11:59,315 And I'm sorry I don't know why I didn't include this. 223 00:11:59,315 --> 00:12:00,940 This is absolutely included. 224 00:12:00,940 --> 00:12:03,460 I'm just testing to see if you, 225 00:12:03,460 --> 00:12:05,160 guys, are paying attention. 226 00:12:05,160 --> 00:12:08,720 So we counted before that there are six dots on the 227 00:12:08,720 --> 00:12:12,350 diagonal, and we know that there are 36 dots total. 228 00:12:12,350 --> 00:12:15,660 So the number of dots, or outcomes to use the proper 229 00:12:15,660 --> 00:12:20,670 word, in our conditional world is 36 minus 6, or 30. 230 00:12:20,670 --> 00:12:24,640 So we get a 30 on the denominator. 231 00:12:24,640 --> 00:12:28,260 And now we're sort of using a conditional version of our 232 00:12:28,260 --> 00:12:29,910 discrete uniform law, again. 233 00:12:29,910 --> 00:12:33,960 And the reason why we can do this is, as I argued before, 234 00:12:33,960 --> 00:12:36,870 that conditioning doesn't change the relative frequency 235 00:12:36,870 --> 00:12:38,130 of the outcomes. 236 00:12:38,130 --> 00:12:41,560 So in this conditional world, all of the outcomes are still 237 00:12:41,560 --> 00:12:44,740 equally likely, hence we can apply this law again. 238 00:12:44,740 --> 00:12:49,680 So now we need to count the number of outcomes that are in 239 00:12:49,680 --> 00:12:55,090 the orange conditional world, but that also satisfy at least 240 00:12:55,090 --> 00:12:57,760 one die roll is a 6. 241 00:12:57,760 --> 00:12:59,610 So you can see-- 242 00:12:59,610 --> 00:13:02,720 1-- we just need to count the green circles that are also in 243 00:13:02,720 --> 00:13:03,590 the orange. 244 00:13:03,590 --> 00:13:08,910 So that's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 245 00:13:08,910 --> 00:13:17,460 So we get a 10, so our answer is 10 over 30, or 1/3. 246 00:13:17,460 --> 00:13:19,340 So now we're done with this problem. 247 00:13:19,340 --> 00:13:23,360 As you see, hopefully, it wasn't too painful. 248 00:13:23,360 --> 00:13:24,920 And what are the important takeaways 249 00:13:24,920 --> 00:13:26,560 here for this problem? 250 00:13:26,560 --> 00:13:32,290 Well, one is that whenever you have a discrete sample space, 251 00:13:32,290 --> 00:13:35,540 in which all of outcomes are equally likely, you should 252 00:13:35,540 --> 00:13:39,570 think about using the discrete uniform law, because this law 253 00:13:39,570 --> 00:13:43,290 lets you reduce the problem from computing probabilities 254 00:13:43,290 --> 00:13:46,740 to just counting outcomes within events. 255 00:13:46,740 --> 00:13:50,860 And the second takeaway is the way we thought about 256 00:13:50,860 --> 00:13:51,970 conditioning. 257 00:13:51,970 --> 00:13:55,340 So we talked about one thing, which is that in your 258 00:13:55,340 --> 00:13:59,340 conditional world, when you condition, the relative 259 00:13:59,340 --> 00:14:02,310 likelihoods of the various outcomes don't change. 260 00:14:02,310 --> 00:14:05,640 So in our original universe, all of the outcomes were 261 00:14:05,640 --> 00:14:06,770 equally likely. 262 00:14:06,770 --> 00:14:09,380 So in our conditional universe, all of the outcomes 263 00:14:09,380 --> 00:14:10,840 are equally likely. 264 00:14:10,840 --> 00:14:14,900 And we saw it was much faster to apply a conditional version 265 00:14:14,900 --> 00:14:17,820 of the discrete uniform law. 266 00:14:17,820 --> 00:14:19,380 So that's it for today. 267 00:14:19,380 --> 00:14:21,750 And we'll do more problems next time. 268 00:14:21,750 --> 00:14:23,000