Statistics and probability
Sal determines if a few different methods of a magician choosing a volunteer are fair. Created by Sal Khan.
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- This may have been answered somewhere else, but how exactly does a random number generator work?(12 votes)
- This is a very complex and studied question.
True randomness is not achievable using mathematical methods, the best we humans can do is a pseudorandom number generator, in simple terms:
- you choose a seed (any number), most (pseudo)random number generators use the current time in seconds as the seed and then do some calculations with that value to output a (pseudo)random number.
There's more details in the Wikipedia article: http://en.wikipedia.org/wiki/Pseudorandom_number_generator(11 votes)
- Hi! I have a hard time solving for the percentage so i can find the weighted average. The phrase says: the final examination in a certain course counts three times as much as each of the three examinations. How will i interpret this?(4 votes)
- It's possible that there's a word missed out. If the sentence said "the final examination ... counts three times as much as each of the OTHER three examinations" then that would make sense -- there are four examinations, and the final one is worth three times as many marks as each of the first three. I think this is the most reasonable way to interpret the sentence. It doesn't make much sense as it is.(3 votes)
- how do you make a time stamp for a video?(2 votes)
- The number of windows is random but not evenly distributed cause there is no info or guarantee that the house has the same windows in every room. Did I correctly understand? If yes why it is mandatory to evenly distributed?(1 vote)
- Since the magician is trying to get each child to have an equal chance of being chosen, doesn't the fact that he starts from the birthday boy influence the entire process?(1 vote)
- How to calculate a probability of choose 3 off-suited and off-ranked cards out of 4 given cards from a standard 52-cards deck.
E.g. you're dealt 3 cards at a time.
P.S. By simulating the situation on a computer (1mln tryies) the answer must me like 54.5%)
P.P.S. The same question but to get 2 cards (off-suited and off-ranked)
P.P.S. The same question byt 1 card (off-suited and off-ranked) or what is the same 3 cards either same suit or same rank(1 vote)
- I am quite confused as to what you are actually asking here. You say you want 3 off-suit, off-rank cards out of 4 but then say you are dealt 3 cards at a time. Can you give an example of what would be a "success"?(1 vote)
- Regarding "Using probability to make fair decisions" exercise 1: Peter, Austin, Jason, and Matt each want a different type of pizza
The question does not specifiy whether cards are chosen without replacement or not. It should explain that first in the question.(1 vote)
- Is Method 3--which is unfair--even viable? What if there's 0 windows or >30 windows in the room?(1 vote)
- In that case, you can assume that they would have gone to another room and retried. Either way, the point is that it's not equally likely that a random room will have 2 windows or to have 10 or 30 or 22 windows. Usually, some outcomes would be much more likely than others. How many rooms do you know of that have 30 windows?(1 vote)
A magician performing at a birthday party stands inside a circle of 15 kids. He's going to choose a volunteer and he wants each kid to have the same chance of getting chosen, fair enough. Now we have three methods by which he can do it. Let's just think about whether each of these methods are fair, where each kid does have the same chance of getting chosen. If they don't lead to each kid having the same chance of getting chosen, think about why that is the case. So, method one, the magician starts with the birthday boy and moves clockwise, passing out 100 pieces of paper numbered one through 100. He cycles around the circle until all the pieces are distributed. He then uses a random number generator to pick an integer, one through 100, and chooses the volunteer with that number. So just think about what's happening. So there are 15 kids in a circle. So there's one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15. I planned that out amazingly well, I didn't think I would be able to fit exactly 15, but it worked out. So 15 kids in a circle, and then he's going to hand out pieces of paper. So he's going to give one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15. This person is going to get pieces of paper one and 16. Now this person is going to get two and 17. You're going to keep going round and round and round until all 100 pieces of paper are going to get distributed. Now, something to think about is whether every child here is going to get the same number of pieces of paper. I encourage you to pause this video and think about that. If we just keep cycling around all the way to 100, does each child get the same number of pieces of paper? Well just think about it. In order to get the same number of pieces of paper, 100 has to be divisible by 15, and we know 100 isn't divisible by 15. 15 goes into 100 six times. Six times 15 is 90, and you have a remainder of 10. So what's going to happen is all 15 kids are going to get six pieces of paper, and then another 10 of the 15 are going to get a seventh piece of paper. So they're not all getting an equal number of pieces of paper. So even though he's randomly picking an interger between one and 100 some of the students are going to have a higher chance than the other ones. The 10 that have seven pieces of paper are going to have a higher chance than the other five who only have six pieces of paper. So I would say method one is not fair. Let me write this down. Not fair. Sometimes life isn't fair, but in this case it's not fair where we define fair as the same chance of getting chosen. That's because they all have different numbers of pieces of paper. All of the students are not equally likely to get picked. Let's look at method two. The magician starts with the birthday boy and moves counterclockwise passing out 75 pieces of paper numbered one through 75. He cycles around the circle until all the pieces are distributed. He then uses a random number generator to pick an integer between one through 75 and chooses the volunteer with that number. So I encourage you to pause this video and think about whether this one, method two, whether that one is fair. Well method two is the same as method one except for instead of using 100 pieces of paper, we're using 75 pieces of paper. So think about, is 75 divisible by 15? And 75 is, five times 15 is 75. So in this situation, each student is going to get five pieces of paper. Each gets five pieces of paper. So they all have an equally likely chance of getting picked, and he's using a random number generator to pick them. So they all have an equally likely chance. I would say method two is, indeed, fair. They all have the same chance of getting chosen. Now think about method three. The magician starts with the birthday boy and moves clockwise, passing out 30 pieces of paper numbered one through 30. So they're all going to get the same number of pieces of paper. They're all going to get two pieces of paper each. 15 children getting two pieces of paper each would be 30 pieces of paper. So that looks reasonable so far. He cycles around the circle until all the pieces are distributed. Everyone gets two pieces. He gives number one to the birthday boy, number two to the next kid and so on, so that all seems reasonable. Kind of consistent with method two except instead of 75 it's 30, and obviously 75 was overkill. Even here this is overkill, he just really needs 15 pieces of paper. He then counts the number of windows in the room and chooses the volunteer with that number. So the question here is, is the number of windows in the room, is it random and is it evenly distributed? So maybe, maybe you can make a case depending on what building it's in, and someone's house, it's somewhat random on how many windows that house happens to have, the house that's happening to host the birthday party, but it's not going to be evenly distributed. I don't know, there's probably some, if you were to sit and plot all of the houses that had a birthday party, you'd probably see that it's more likely that they'd have 10 windows than one window, and definitely more likely that they have 10 windows than, let's say, 30 windows, or even maybe 15 windows even. So it's not going to be evenly distributed. Every house has a somewhat different number of windows, and the house that's happening to host the party seems to be somewhat random, but it's not going to be evenly distributed here. I would say it's not a kind of a really good random number generator because it's not evenly generated. So I would say method three is not fair. The number of windows is not a really good random number generator. A good random number generator, we would want, say, number one through 75 where any of these have an equally likely chance of getting picked. It's somewhat random, the number of windows that building has, but they're not all equally likely.