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BS: Mathematical Probability Query
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Subject: Mathematical Probability Query From: Fmaj7 Date: 09 Oct 00 - 05:38 PM I've tried looking elsewhere without succees, and felt sure that there might be someone here who could help. Hope you don't mind me asking this. I remember hearing / reading that in a group of (I think) 23 people, there is a more than 50% chance that two of the group will share the same birthday. Absurd as it sounds on first glance, the maths make it quite clear. Unfortunately, I've forgotten the maths. If anyone could explain or point me to a site that explains, I'd be most grateful. Thanks Fmaj7 |
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Subject: RE: BS: Mathematical Probability Query From: Liz the Squeak Date: 09 Oct 00 - 05:40 PM Well one way to check it is to see the birthday threads here! I share my birthday with another catter, if there are 21 others out there, then the odds are pretty good.... LTS |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Ed Pellow Date: 09 Oct 00 - 05:44 PM Try looking here All is explained. I love elegant maths Ed |
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Subject: RE: BS: Mathematical Probability Query From: Mrs.Duck Date: 10 Oct 00 - 04:35 PM I enjoyed that!! |
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Subject: RE: BS: Mathematical Probability Query From: Ed Pellow Date: 10 Oct 00 - 04:50 PM I love this problem, and the fact that with 55+ people, it's nigh on certain that at least 2 people will share the same birthday. Does anyone know any other probability problems which initially seem counter intuitive? Ed
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Subject: RE: BS: Mathematical Probability Query From: Mark Clark Date: 10 Oct 00 - 05:04 PM A racing team is preparing their car for a race. It's a one mile track and they want to observe it's performance in a test. They tell the driver he must drive around the track once and maintain an average speed of 60 MPH. During the first half of the lap the car has some problems and, when he is half way around the circuit, the driver realizes he's only averaged 30 MPH. What speed must the driver now average for the remainder of the track in order to average 60 MPH for the entire circuit? Have fun, - Mark |
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Subject: RE: BS: Mathematical Probability Query From: Bert Date: 10 Oct 00 - 05:13 PM Nice one Mark. I won't spoil it by telling the answer. |
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 10 Oct 00 - 06:08 PM By my reckoning, 1/2 a mile at 30 MPH would take him 1 minute but 1 mile at 60 MPH also takes 1 minute so he would have to do the second 1/2 mile in no time at all! Jon |
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Subject: RE: BS: Mathematical Probability Query From: Jim the Bart Date: 10 Oct 00 - 06:36 PM Sorry, Jon. He still has a half mile to make up the difference. I think I know how he'd do it, but I'll wait to see if I'm correct. |
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Subject: RE: BS: Mathematical Probability Query From: Jeri Date: 10 Oct 00 - 06:43 PM Bartholemew, Jon's disgusting ;-), but he's right. The guy has to drive the mile in one minute to average 60 MPH. His minute's up. |
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Subject: RE: BS: Mathematical Probability Query From: Ed Pellow Date: 10 Oct 00 - 06:57 PM OK, The driver need to cover 1 mile in one minute. Over the first 30 seconds, he's averaged 30 mph, (which works out at 0.5 miles per minute) Therefore he's covered 0.25 of a mile. He needs to cover 0.75 miles within the next 30 seconds. So he needs to travel at 1.5 miles per minute, which works out at 90 mph. For some reason, (whilst it seems logical) I've got a feeling that my answer is wrong. Ed |
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Subject: RE: BS: Mathematical Probability Query From: IvanB Date: 10 Oct 00 - 07:03 PM No, Ed, he's driven halfway around the circuit (1/2 mile) at 30 mph. Therefore he's used up his whole minute, and Jon's answer is correct. |
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 10 Oct 00 - 07:14 PM OK, here is an old one: If a block of cheese weighs 10 pounds plus 1/2 of its weight, how much does it weigh? Jon |
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Subject: RE: BS: Mathematical Probability Query From: Ed Pellow Date: 10 Oct 00 - 07:15 PM Ivan, You are of course right. I hate it when maths is put forward this way, as if to trick people. No wonder so many people don't like it, even though working through the probability of the original question is quite rewarding... Ed |
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Subject: RE: BS: Mathematical Probability Query From: Bradypus Date: 10 Oct 00 - 07:23 PM I remember reading an interesting study on the birthday problem. In a football (soccer) match, there are 23 people on the pitch at kick-off (11 on each side, + the referee). Someone surveyed all the Premier League games on a given Saturday (maybe more than just the one division) and found that in almost exactly half of them two of the people on the pitch shared a birthday ! Bradypus |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Murray MacLeod Date: 10 Oct 00 - 07:38 PM I remember programming my old Amstrad computer back in the early eighties to work out the birthday probabilities for all numbers from 2 to 367 (obviously if you have 367 people at least two of them MUST share the same birthday, whereas with 366 it is theoretically possible they could all have different birthdays. What was interesting was that at 80 people the computer returned the probability as certainty. Of course it was a Z-80 processor, maybe my Pentium would go much further, except you can't program these damned things, nowadays, it was much more fun back with the Apple 48k and the Sinclairs ......... Murray |
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Subject: RE: BS: Mathematical Probability Query From: IvanB Date: 10 Oct 00 - 07:42 PM Jon, it, of course, weighs 20 pounds. |
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Subject: RE: BS: Mathematical Probability Query From: catspaw49 Date: 10 Oct 00 - 07:43 PM Jon.........infinity. Spaw |
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Subject: RE: BS: Mathematical Probability Query From: Amos Date: 10 Oct 00 - 08:03 PM x = 10+(1/2 x) .5 x = 10 x = 20 Proof: 20 = 10 + (20/2) A |
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Subject: RE: BS: Mathematical Probability Query From: Marion Date: 10 Oct 00 - 09:27 PM Here's one where the obvious answer that everyone gives first is wrong, though not everybody accepts the right answer because it seems so counterintuitive. I once got into a long discussion on this on another board, and eventually won over those who argued against me, but let's see how it goes here. OK, you are shown three cups, one of which has a prize under it. You are asked to choose (and point out) one of the cups, but not lift it yet. After you have made your guess, the house lifts one of the cups that you didn't choose and shows you that it is empty. You are then offered the opportunity to choose again, this time for keeps. Should you stick with your original guess? Should you change to the other cup? Or does it not matter (i.e. your chances of winning are the same whether you change your guess or not)? Enjoy, Marion |
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Subject: RE: BS: Mathematical Probability Query From: Amos Date: 10 Oct 00 - 09:49 PM Statistically you should switch. Never could figure why though.
A |
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Subject: RE: BS: Mathematical Probability Query From: Troll Date: 10 Oct 00 - 09:52 PM Read this carefully. Which is heavier; a pound of gold or a pound of feathers? troll |
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Subject: RE: BS: Mathematical Probability Query From: catspaw49 Date: 10 Oct 00 - 09:53 PM uh Marion? Shell games and Monty have no statistical probability outside of the dealer's choice. Spaw |
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 10 Oct 00 - 10:00 PM Troll, that is mean but a pound of feathers weigh more (I think). Marion, I am temped to say 50/50 chance on the other 2 cups so sticking with the original is as good as anything but I have a horrible feeling that there is a cathc to this one. Jon |
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Subject: RE: BS: Mathematical Probability Query From: Amos Date: 10 Oct 00 - 10:04 PM They're measured on different scales, I believe. But a kilogram of feathers and a kilogram of gold mass the same. And they weigh the same at rest. But they would probably accelerate diffferently through atmosphere and have different terminal velocities. I think. A |
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Subject: RE: BS: Mathematical Probability Query From: Jeri Date: 10 Oct 00 - 10:11 PM Jon's right again - a pound can buy a lot more feathers weight-wise than gold. Marion, I'd say it wouldn't make a difference which of the two cups you chose. You have an equal chance of being correct no matter which cup you pick. You start off with a one-in-three chance, then go to one-in-two. You can switch cups - the probability is still 50%. |
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 10 Oct 00 - 10:16 PM Slight deviation but another old classic: 3 men go into a restaraunt for a meal. The bill came to £30 so they each pay £10 each. The manager saw that these people were loyal regulars and asked the waiter to knock £5 off the bill. The waiter was dishonest and decided to give them £1 back each and pocketed the remaining £2 for himself. Now the men each paid £10 and got £1 back so they each paid £9 and the waiter took £2 but £9 x £3 = £27 and £27 + the £2 the waiter took = £29. What happened to the missing £1? Jon |
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Subject: RE: BS: Mathematical Probability Query From: Troll Date: 10 Oct 00 - 10:22 PM Jon and Amos. The pound of feathers is HEAVIER.Gold is weighed on the troy scale which is 12 oz. to the pound.Feathers are weighed on the avoirdupois scale; 16 oz.to the pound. I used to win more beer with that one.***sigh*** troll |
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 10 Oct 00 - 10:35 PM Murray, I have never been that good and probability, as with much of maths, has never been something that I am good at but I can relate to what you are saying. I had a Commadore 64 and used to enjoy the occasional challenge.. As for the newer computers and programming, I still have an old dos version of Turbo Pascal which I still like and use ocassionally. Number crunching wise, I have a Fortran Compiler which I believe is good but I haven't got round to learning the language and I probably would be incapable of taking advantage of its power as my maths is not good enough. Jon |
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Subject: RE: BS: Mathematical Probability Query From: Peter T. Date: 10 Oct 00 - 10:35 PM True story: A student came into one of my classes this morning and said, Professor, the people in my house are complaining about you. And I said, oh why? and he said, you suggested that if we wanted to study exponential growth outside of a petri dish we should leave a piece of pizza lying around outside the fridge, and see what happens. I left it in my room for three days so no one would eat it before it started going bad, and then left it on the kitchen windowsill. I have been doing that for over a week, and when they come in and see it, I say that I am studying exponential growth, and blame you for how disgusting it looks. I confess to having been somewhat thrilled. Simple pleasures. yours, Peter T. |
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Subject: RE: BS: Mathematical Probability Query From: Mrrzy Date: 10 Oct 00 - 10:41 PM And about the other one, about the bill, you have to subtract, not add. The bill was 30, less 5, is 25 that the restaurant charged, but it took 27, if you count the unscrupulous waiter's 2. The guests each paid 9, which matches the 27 the restaurant, and its personnel, took. |
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Subject: RE: BS: Mathematical Probability Query From: Amos Date: 10 Oct 00 - 10:41 PM Fie on obfuscatory semantics, quotha! Th'art a knave an' thy trickster's tongue is quick with canards and darts of shrill vexation. Get the to a cobbler, an thee be not a heel, he mought mend thy sole, that thou hold thy tongue.... A |
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Subject: RE: BS: Mathematical Probability Query From: Mary in Kentucky Date: 10 Oct 00 - 10:43 PM So Marion, what is the correct answer? I'm inclined to say don't change your guess. Even though your original odds have changed, your remaining choices still have the same odds. This reminds me of the coin toss question. If you're tossing a coin, and you get 17 heads in a row, what are the chances that the next toss will be tails? |
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Subject: RE: BS: Mathematical Probability Query From: Marion Date: 10 Oct 00 - 11:13 PM Amos, you're right, and I'll try to explain below. Catspaw, sorry, but I don't understand you. Who or what is Monty? Jon and Jeri, you've given the most intuitive answer - that at the time of your second choice there are only two cups in the running, and you don't know which, so you think it's a 50-50 thing. But this is not so. In fact your best bet is to switch cups. If you switch, you have a 2/3 chance of winning, but if you don't switch, it's only 1/3. I know this is really counterintuitive, so I'll try two different ways of explaining it. One: remember that if you guessed right the first time, switching will definitely make you lose. If you guessed wrong the first time, switching will definitely make you win, since the other wrong cup has been eliminated for you. The probability of guessing right the first time is 1/3, so the probability that switching will make you lose is also 1/3. The probability of guessing wrong the first time is 2/3, so the probability that switching will make you win is also 2/3. Two: think of it this way: when you make your first guess and point out a cup, what you are really doing is dividing the cups into two groups: a small group with one cup, and a big group with two cups. When you are given a second chance to guess, what you are really doing is saying whether you think it's more likely that the prize will be in the big group or the small group. It's more reasonable to bet that the prize will be somewhere in the big group. When the house lifts an empty cup, that just shows that at least one of the big group is empty, but you knew that already, so that's not really relevant to the question of whether the big group or the small group is a better bet (although it is useful information because it tells which member of the big group would have the prize if one of them does). There's a 2/3 chance that the prize is in the big group, so your chances are better if you switch over to the big group. I know this sounds terribly convoluted compared to the beautiful simplicity of "two cups, the one you picked and one not touched yet, so 50-50 chances", so if you're still not convinced, try this mental experiment: Imagine you sat down with a ridiculously patient friend to play this game 1000 times. The plan is for your friend to switch every time, then you'll see if the number of times he wins is closer to 500 or 667. Work through in your mind what would happen. He would pick an empty cup the first time approximately 667 times, assuming that his guessing and your prize-hiding were random. So he would win approximately 667 times. |
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Subject: RE: BS: Mathematical Probability Query From: Marion Date: 10 Oct 00 - 11:33 PM Mary, there's always a 50% chance that the next coin toss will be tails. But I don't think that that puzzle has much bearing on my cups game. Each coin toss in the series is an independent event. But in the cups game, your two chances to guess are related to each other; the chance that your second decision will be successful is directly dependent on the the success of your first guess. The cups game could be considered a series of independent random events if you forgot which cup you had chosen first and simply guessed randomly between the two cups that hadn't been lifted by the house. In that case, you would win about 50% of the time; it would be a question of choosing between two cups, not of choosing between sticking with a first guess that was probably wrong or abandoning a first guess that was probably wrong. But in 2/3 of the games you win, your winning cup would happen to be one that you didn't pick on your trial run guess. Here's another little puzzle, for free: Suppose you have a basket that can hold ten apples. You take out three. How many apples do you have? Marion
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Subject: RE: BS: Mathematical Probability Query From: Mary in Kentucky Date: 10 Oct 00 - 11:39 PM You have three apples, but I'm still thinking about the cups. |
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 10 Oct 00 - 11:40 PM 3 Jon |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Murray MacLeod Date: 10 Oct 00 - 11:49 PM I am sorry Marion , but I cannot accept your reasoning on this, and wull have to ask you to refer me to an academic text which explains this and which agrees with your reasoning. I mean you didn't make this problem up yourself, did you? This problem seems to be related to one which was first aired in Scientific American in the 60's, and which stirred a fair amount of debate. There are three cards, one is red on both sides, one is white on both sides, and one is red on one side and white on the other. The dealer places them in a bag, shakes them then slides out one card onto the table so that only one face is visible. The face is red. What is the probability that the other face is red? Murray |
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Subject: RE: BS: Mathematical Probability Query From: Mary in Kentucky Date: 11 Oct 00 - 12:01 AM Marion, give me some time to think about this one...in the meantime, does anyone remember the monk walking up the mountain puzzle which illustrates the difference between right brain and left brain thinking? I'll see if I can reconstuct that one...later... |
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 11 Oct 00 - 12:03 AM |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Jim Dixon Date: 11 Oct 00 - 12:12 AM I agree with Marion's answer. Here's a variant of the puzzle which might make it more clear: Instead of 3 cups, suppose there are 1,000 cups. You choose number 483 at random. Then the house turns over all the cups EXCEPT numbers 483 and 722, showing that all 998 cups are empty. Now, would you keep number 483? Or switch to number 722? |
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Subject: RE: BS: Mathematical Probability Query From: Escamillo Date: 11 Oct 00 - 12:14 AM I'm so sorry! Please would you promise that my comments will be taken as another point of view, and that there is no intention to rain in anybody's parade ? That none will feel attacked in his/her beleifs and enjoyment ?? :)) Let's see the first problem. Dr.Math takes this FIRST PREMISE: "We'll start by figuring out the probability that two people have the same birthday. The first person can have any birthday. That gives him 365 possible birthdays out of 365 days, so the probability of the first person having the "right" birthday is 365/365, or 100%. The chance that the second person has the same birthday is 1/365. To find the probability that both people have this birthday, we have to multiply their separate probabilities. (365/365) * (1/365) = 1/365, or about 0.27%." This premise is absolutely false. He is taking for sure that the first person has a birthday that MATCHES everyone's birthday, so its probability is 1. In fact, the probability of any given couple of persons to have the same birthday is 1/365 multiplied by 1/365 and nothing else. Thus, the whole reasoning is false. Problem of the cups: sorry, the probability of winning when there are only two cups left, is 1/2, exactly 50%, no matter what have been the previous results.It is the same apparent paradox of the probability of 50% in a coin toss, after 6, or after 200 tails in a row: still 50%. A different thing is to calculate the probability of the whole sequence (not one toss): two tails in a row is 1/4, (will probably happen in 1 of every four sequences of two tosses), three in a raw is 1/2 * 1/2 * 1/2 = 1/8 = 12.5%,.. and so on. Interesting ! Un abrazo - Andrés (no Math degree..snif)
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Murray MacLeod Date: 11 Oct 00 - 12:25 AM Escamillo, I am sorry but your reasoning here is faulty. The question is, what is the probability that in a group of 23, AT LEAST two share the same birthday. Note that the question is not (or should not be) what is the probability that exactly two people share the same birthday? The only way to calculate this is to calculate the probability that everybody has a different birthday. Then you subtract that probability from 1 and you have the probability that evrybody in the group does NOT have a different birthday. That is another way of saying that at least two members of the group share ethe same birthday QED Murray |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Murray Macleod Date: 11 Oct 00 - 12:30 AM Ah, yes, Marion, I see it now, Jim's contrubution made the penny drop. Good problem ! Murray |
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Subject: RE: BS: Mathematical Probability Query From: Marion Date: 11 Oct 00 - 01:43 AM Mary: Damn it, I thought I might catch some people with the apples puzzle, because after the cups puzzle people wouldn't expect a simple solution... Murray: I heard the cups problem from my macroeconomics professor. I can't remember what connection he made between it and macroeconomics. The red/white thing does seem related at first blush, but I'd like to think about it some more before I comment. I'm intrigued that Jim's answer was so helpful to you; while it's nice to have someone agree with me, I couldn't understand how his explanation made it clearer. The thing that fascinates me about this problem is that the wrong answer is so simple and obvious whereas the right answer takes a lot of persuasion and there are a number of lengthy, counter-intuitive ways to explain. There's probably some lesson about life here. Andres: suppose you played this game several times, and suppose that you believed it didn't matter whether you kept your first guess or switched... in that case you would be picking randomly between the two cups, and you would win approximately 50% of the games. In this scenario, your second guess really would be an independent event with no connection to your first guess, so your comparison of this to one toss in a sequence of coin tosses is reasonable. However...as I said, by picking randomly between the two, you will win about 50% of the time. But if you look back at the games you win, you will find that in most cases your second guess was different from your first. And if you examine the games you lose, you will find that in most cases your second guess was the same as the first. Therefore, you can improve your chances of winning by changing your guess. If you don't believe me, you can confirm it with the formula for calculating the probability of a complex event that you mentioned in your post. If you write down every possible outcome in this game, then for each possible outcome multiply the probabilities of each independent event in the complex event.
For example: P(prize is in cup 1)=1/3 Then, if you add up the probabilities of the outcomes where the player changes guesses and wins the game, the total will be twice the total probability of the outcomes where the player guesses the same cup twice and wins the game. I've done this table and done the math (yes, I really am that obsessed) and I would say that this is irrefutable evidence according to the laws of finite math. Unfortunately I'm not an HTML expert so an attempt to produce a table here would be chaotic I'm sure, but do the table and the math yourself if you want. Thanks for playing, Marion
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Subject: RE: BS: Mathematical Probability Query From: Escamillo Date: 11 Oct 00 - 04:27 AM I simply quoted Dr.Math's demonstration, which starts with this:"We'll start by figuring out the probability that two people have the same birthday." He states that probability is 1/365 and that is false. Why ? Given MY birthday, the probability of a coincidence with yours is indeed 1/365. But given NO date, the probability of we being born the same date is (1/365) * (1/365). It is the same as two persons taking two marble balls from different bags or the same bag in turns. ONCE I took mine, yor probability to take the same is 1/365. But given two persons and no ball in particular, their probability to take the same is 1/365/365. But his results are not very different from mine, because there are so many factors involved. Let's try a calculation by ourselves. This simple Turbo Pascal program will give interesting results, starting from the probability of a NO MATCH and subtracting from 1:
Program test ;
end.
These are some curious results: I think that the apparent unintuitive result (only 22 persons for a 50% probability , 70 persons for nearly a certainty !), appears because it is difficult to visualize the enormous probabilities for many matches. In fact, the probability of just two matches is almost as low as no match at all.
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Subject: RE: BS: Mathematical Probability Query From: Mary in Kentucky Date: 11 Oct 00 - 07:11 AM I see the cup problem now. The word "history" helped me understand it. I think of it as 3 events. I choose, house chooses, then there is a 3rd event in which I choose again. My first choice (if it's empty) survived 1 in 3 probability, but the other cup (if it's empty survived 1 in 2 probability, thus a better choice. |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Bob Schwarer Date: 11 Oct 00 - 07:14 AM If a coin toss comes up heads 17 times in a row it is probably a two-headed coin. Bob S. |
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Subject: RE: BS: Mathematical Probability Query From: Escamillo Date: 11 Oct 00 - 07:32 AM Cups problem: I built this table and can't see anything wrong in the 50% chance:
PRIZE IS IN CUP 1
PRIZE IS IN CUP 2
PRIZE IS IN CUP 3
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PRIZE IS IN CUP 2
PRIZE IS IN CUP 3 I guess (only guess) that the fact that an empty cup is removed does not influence the final selection which becomes a two-cup only game. It is the same as if the house said "Ok, we always take out one empty cup, let's play with only two cups, you make your choice, I ask you if you are sure, and you can keep your choice, or change it" The same would be (always guessing) with 1000 cups, I choose one, they take out 998 empty cups and let me choose between the remaining two. In my line of reasoning, the probability remains 50%. In yours, we would have an enormously high probability of a "bad" choice in the first selection, and then would more strongly recommend a change. Intriguing, but I don't see any mistake in the above tables. Un abrazo - Andrés |
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Subject: RE: BS: Mathematical Probability Query From: Escamillo Date: 11 Oct 00 - 07:43 AM Cups problem : Mary, you say that the house CHOICES, that is, they take a chance ? They don't know where the prize is ? Hey! That would be a very different problem ! I was assuming that the house KNOWS it, and then will take out always the empty cup. Oh, I will have to stand up and think again, I was already sitted down ! Oy
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