Computer Science Canada A math question |
| Author: | Insectoid [ Wed May 28, 2008 5:28 pm ] |
| Post subject: | A math question |
My family had a huge argument about this, so I'm coming to you to see who comes up with the same answer I did, and if nobody does, I'll explain my reasoning in an upcoming post. The story- There are 3 cups set out , upside down, on a table. One has some token inside(hereafter referred to as 'the right one'), and it has not been revealed which one. After one has been selected by Jeremy, one cup is removed and identified as a cup without the token. Now there are 2 cups on the table, one the right one, one a wrong one. Jeremy still doesn't know if his is the right one. Jeremy is given a choice: Change his selection to the other cup, or keep his current selection. The question- Should Jeremy keep the cup he chose when there were 3 on the table, or change it now that there are only two? possible answers- 1. Either; it doesn't make a difference 2. Change it, the other cup has better odds! 3. Don't change it, the one he chose at first is likely the right one! My answer- 2. I wil explain after some other people answer, unless someone else does first. Submit your answer in the poll, explain yourself in a post. PLEASE, explain yourself. +100 bits to the first one to have the same reason as me. |
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| Author: | r691175002 [ Wed May 28, 2008 5:39 pm ] |
| Post subject: | Re: A math question |
The odds are better if you change. I recall writing a program that did a few hundred thousand iterations of this question a while back to "prove" it. You could also show it by identifying all the possibilities since there isn't that many. |
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| Author: | Saad [ Wed May 28, 2008 5:43 pm ] |
| Post subject: | Re: A math question |
Like r691175002 said, the odds are better if you change. This is known as the Monty Hall Problem |
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| Author: | apomb [ Wed May 28, 2008 5:55 pm ] |
| Post subject: | RE:A math question |
I still dont think it matters, he should keep his hand on the one at the beginning... how does the probability change for either cup left on the table now? His has just as likely a probability to have the token than the other one. However, after reading that Wikipedia article, i am completely wrong. lol |
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| Author: | Nick [ Wed May 28, 2008 5:57 pm ] |
| Post subject: | RE:A math question |
every cup has a 33% chance... it doesn't matter sure the odds are the cup he has isn't the right one, but the odds are just the same for each cup |
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| Author: | Mackie [ Wed May 28, 2008 5:58 pm ] |
| Post subject: | RE:A math question |
I saw this on Daily Planet. They told me that you would have better odds if you changed it. So, I'll go with that. Blindly following people I don't know. |
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| Author: | apomb [ Wed May 28, 2008 5:59 pm ] |
| Post subject: | RE:A math question |
Ok, well bevause the game show host KNOWS that there is none behind the door he didnt open, i get it ... your question is a bit ambiguous |
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| Author: | Insectoid [ Wed May 28, 2008 6:51 pm ] |
| Post subject: | RE:A math question |
I believe I used cups, though the original did use doors I believe. Well, I have proof now, what with Daily Planet (great show) AND a wikepedia article. My reasoning: The percent chance of the cups removed * % chance per cup is divided by and added to the odds of number of cups remaining - 1 I.E 3 cups with 33% each. 1 cup removed (1 * 33) / ( 2 - 1 ) = 33. 33 + 33 = 66 Thus, the one cup not chosen has a 66% chance of being the right one, whereas the other cup has 33%. Switch cups man, better odds. A bigger example. 50 cups. choose 1. likely the wrong one, right? remove 48 cup, does your selection magically turn to the right one? no, it's still (likely) the wrong one. Math: 50 cups with 2% chance each, - 48 (48 * 2) / (2-1) = 96 96+2=98% The unselected cup has 98% chance of being the right one. The originally selected cup still has only 2% chance of being the right one. I made this up myself! If this was mentioned in the wiki article or Daily Planet, well, I haven't read/watched those yet. I'm only getting a 71% in math right now. but if I take out all the other students... (31*71) + 71 = 2272%! Woo! Now how to eliminate them... oh yeah, +100 bits to r691175002 EDIT: Don't vote in the poll unless you know (or think you know) what you're talking about. No 'I'll just blindly follow people I don't know' *couph*macki*couph* |
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| Author: | tenniscrazy [ Wed May 28, 2008 8:38 pm ] |
| Post subject: | RE:A math question |
they also did this in the beggining of the movie 21. Thats the one with the smart guy who counts cards in vegas. Edit: right? |
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| Author: | Nick [ Wed May 28, 2008 8:39 pm ] |
| Post subject: | RE:A math question |
your logic is seriously flawed |
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| Author: | tenniscrazy [ Wed May 28, 2008 8:46 pm ] |
| Post subject: | Re: A math question |
man, i keep coming up with different reasonings...i'll have to put some serious thought into this then get back...tho i know it has been proven many times, i still want to prove it to myself. not just follow blindly |
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| Author: | apomb [ Wed May 28, 2008 9:02 pm ] |
| Post subject: | Re: RE:A math question |
Nick @ Wed May 28, 2008 8:39 pm wrote: your logic is seriously flawed
I know it seems flawed, but the actual reasoning, if you read the Wiki is solid... if you make sure you read the question about the host of the show knowing which door to open, it might just click. thats how it happened with me, i thought the same way you did at first. |
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| Author: | rdrake [ Wed May 28, 2008 10:16 pm ] |
| Post subject: | RE:A math question |
Like Saad said above, this is the classic Monty Python Question. It came from the show "Let's Make A Deal" or something like that. |
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| Author: | r691175002 [ Wed May 28, 2008 10:36 pm ] |
| Post subject: | Re: A math question |
Here is a potentially clearer explanation. When you choose your first case, the chance that you chose the right case is 1/3. Here is the key point of the matter. When one case is removed, the probability you chose the right case does not increase. Even after one case is removed, the chance you picked the right case is still 1/3. You made a choice that had a 1/3 chance of being correct. Nothing can change that anymore. Now, since between the two cases, one of them must have the prize, therefore the remaining 2/3 probability goes into the second case. The hinge to this question is whether or not the probability you chose the right case increases as other options are eliminated. |
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| Author: | apomb [ Thu May 29, 2008 12:31 am ] |
| Post subject: | Re: RE:A math question |
rdrake @ Wed May 28, 2008 10:16 pm wrote: Like Saad said above, this is the classic Monty Python Question. It came from the show "Let's Make A Deal" or something like that.
I believe the show was actually called "Faulty Towers" and yes, it did have john cleese, but it wasnt about Monty Python at all!  |
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| Author: | jernst [ Thu May 29, 2008 6:48 am ] |
| Post subject: | Re: A math question |
gah my prof had a whole lecture on this in stats, he spent the entire class talking even though everyone with a laptop just wikipedia'd it and knew the answer in like 2 seconds |
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| Author: | tenniscrazy [ Thu May 29, 2008 7:35 am ] |
| Post subject: | RE:A math question |
ha! i finally got it. you will choose the door it's behind one third of the time. (call this the first door) For the other two doors...say the car is behind the second door. The gameshow host will open the third door if it is behind the second door. And if it's behind the third door than the gameshow host will open the second door. So really, when you choose the door that you didn't choose before, you are choosing the second door 1/3 of the time and the third door 1/3 of the time, so this means that you are actually shosing both doors because one that doesn't have the car is eliminated. just thought i would post that because thats how i made it make sense to myself |
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| Author: | Insectoid [ Thu May 29, 2008 7:50 am ] |
| Post subject: | RE:A math question |
But my math equation works, right? Seems to to me...(not about me getting 2000-ish%, about the cups) It seems to work for me. Getting the right percentage every time! |
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| Author: | tenniscrazy [ Thu May 29, 2008 8:10 am ] |
| Post subject: | Re: A math question |
in your equation Quote: I.E 3 cups with 33% each. 1 cup removed
(1 * 33) / ( 2 - 1 ) = 33. 33 + 33 = 66 i don't get why you divide the first line by 2-1, because that is the same thing as dividing by 1, which does nothing. and you do it in the second one too Quote: 50 cups with 2% chance each, - 48
(48 * 2) / (2-1) = 96 96+2=98% |
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| Author: | Vermette [ Thu May 29, 2008 10:18 am ] |
| Post subject: | RE:A math question |
Don't feel bad if you argued for the incorrect answer (keeping the door you chose); this is a classic problem that has tripped up even experienced statisticians in the past before careful analysis. It was used by one my of profs to demonstrate that in the world of CS/Mathematics, your intuition can sometimes be dangerous. |
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| Author: | Brightguy [ Thu May 29, 2008 10:43 am ] |
| Post subject: | Re: A math question |
Actually the probability is 50-50 for the question as you've posed it. The answer hinges on this line: insectoid @ Wed May 28, 2008 5:28 pm wrote: After one has been selected by Jeremy, one cup is removed and identified as a cup without the token.
That should read one cup is identified as a cup without the token [by someone who knows the token location] and removed. You must also assume that they choose randomly if they have a choice of cups to remove. |
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| Author: | Insectoid [ Thu May 29, 2008 3:03 pm ] |
| Post subject: | RE:A math question |
teniscrazy wrote: in your equation Quote: I.E 3 cups with 33% each. 1 cup removed (1 * 33) / ( 2 - 1 ) = 33. 33 + 33 = 66 i don't get why you divide the first line by 2-1, because that is the same thing as dividing by 1, which does nothing. I did that so that the equation could include a number of scenarios, such as if there were 10 cups and he was alowed to pic 3, then 2 were removed. Then the equation would look like this: ((2*10)/ (8-3))+10= 14 (I think I plugged the numbers in the right spot) |
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| Author: | tenniscrazy [ Fri May 30, 2008 7:59 am ] |
| Post subject: | Re: A math question |
ahk! I was tryna make sense of your equation. it works for the exaample you just did, because each of the other doors would have a 14% chance of having the car behind them. But i'm not sure if you are putting in the same variables for each of the three examples could you post what the actual variables in the equation mean, just to clarify. Cause it would be a cool equation if it worked and made sense |
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| Author: | jernst [ Fri May 30, 2008 9:05 am ] |
| Post subject: | Re: RE:A math question |
Vermette @ Thu May 29, 2008 10:18 am wrote: Don't feel bad if you argued for the incorrect answer (keeping the door you chose); this is a classic problem that has tripped up even experienced statisticians in the past before careful analysis. It was used by one my of profs to demonstrate that in the world of CS/Mathematics, your intuition can sometimes be dangerous.
Yeah in the lecture our prof gave he mentioned how there were math profs who argued that they knew what they were talking about simply because they were "math profs for like 30 years" even though they argued for the wrong side. |
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| Author: | Insectoid [ Fri May 30, 2008 11:24 am ] |
| Post subject: | RE:A math question |
( ( x y ) / ( n - p ) ) + x is the equation, where: x = the base percent chance per cup (100/number of cups) y = number of cups removed n = number of cups remaining p = number of cups chosen initially It could be adjusted to use less variables, but I'm a bit lazy. |
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| Author: | tenniscrazy [ Fri May 30, 2008 1:44 pm ] |
| Post subject: | RE:A math question |
ah! ok i get the equation now, thanks for clarifying |
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| Author: | btiffin [ Fri May 30, 2008 7:06 pm ] |
| Post subject: | Re: A math question |
Just so everyone knows; as stated this is NOT a Monty Hall challenge. The key phrase is "not been revealed". For Monty Hall to pick the cup that skews the odds from 1 in 3 to 2 in 3, he has to know which cup has the prize. As stated, no one knows. It was a fluke that the first cup removed didn't have the token (it had a 1 in 3 chance). Quote: One has some token inside(hereafter referred to as 'the right one'), and it has not been revealed which one.
For a proper Monty Hall problem, there has to be someone that knows the cup with the token, and a desire to not give away where the token is until the very last moment (for that game show tension effect). If there is no human intervention the odds of this problem are purely common sense statistical. After seeing that the first cup removed doesn't have a token does not influence the odds (unless, again, there is a human revealing the cups in an order to heighten the tension). But in honour of the thread; go ahead and assume that someone in Jeremy's house is acting as a game show host. 
More info from the old guy... Monty Hall was the host of a game show called "Let's Make a Deal". The show was famous for having "Door number 1, door number 2, door number 3 ..." One of the doors always had a "zonker", being a goat or somesuch. (Once it was an Osterich but the player that got that zonker actually won a spot in the finals of the show as this zonker was actually pretty expensive). It always ended with Monty walking through the audience and asking people "ok, if you have a raw egg in your pocket I'll give you $500. $200 if it's hardboiled". So the crowd always had huge bags of random stuff with them. He'd ask for hair pins, magazines, clothespins, walnuts, just about anything small and peculiar to carry around. Monty Hall was born in Manitoba by the way - one of the first of the plethora of famous Canadians on mainstream US television. Cheers |
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| Author: | Insectoid [ Sat May 31, 2008 7:48 am ] |
| Post subject: | RE:A math question |
Well, I meant it has not been revealed to Jeremy which one contains the token. Can you really expect me to get the wording right on the first try? I didn't know at first that is was a famous problem, my brother just came home from school one day, asked my mom the question and spawned the whole argument. I had never even heard of monty hall until someone mentioned it and posted the link. So, in order to clarify, this is now about the mony hall problem (though you can still word your answer according to my story.) |
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| Author: | Brightguy [ Sat May 31, 2008 5:42 pm ] |
| Post subject: | Re: RE:A math question |
insectoid @ Sat May 31, 2008 7:48 am wrote: Well, I meant it has not been revealed to Jeremy which one contains the token.
Of course. But you never specified if someone else knows which one contains the token or not. So which question were you having the argument over: one where someone ("Monty") knows which contains the token, or one where no one knows which contains the token? |
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| Author: | Insectoid [ Sun Jun 01, 2008 1:18 pm ] |
| Post subject: | RE:A math question |
Whichever one makes more sence. Okay, monty know where the token is, but jeremy doesn't. Happy? Let's not talk about the story, but the solution! |
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| Author: | Brightguy [ Sun Jun 01, 2008 3:16 pm ] |
| Post subject: | Re: A math question |
Both problems make sense. The point is, you never said anything about Monty in your description of the problem, so your solution is incorrect; the correct answer is "it doesn't make a difference". |
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| Author: | CodeMonkey2000 [ Sun Jun 01, 2008 3:44 pm ] |
| Post subject: | RE:A math question |
Well, when you first start you have a 2 in 3 chances that of picking a cup with no coin. When you realize that one of them doesn't have a coin your chances of having no coin is still the same, since you chose it when there were still three cups (2/3 with no coin). So you can't have a 50/50 chance because of the initial scenario. |
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| Author: | A.J [ Sun Jun 01, 2008 4:27 pm ] |
| Post subject: | Re: A math question |
this (as most of you already know) is a problem whose solution is argued by the best of the mathematicians in the world! both of the point of views make sense, but I say that the other cups has better odds (with similar reasoning to what codemonkey had) |
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| Author: | Insectoid [ Mon Jun 02, 2008 5:01 pm ] |
| Post subject: | RE:A math question |
Geez! Forget my story already and just go with the monty hall one! I don't care if mine was flawed, the question is the same! So my reasoning IS correct from my point of view and I don't care if you think otherwise, just vote in the friggin poll! (prepare for sarcasm)(sarcasm initiated) For a Brightguy, you aren't a very bright guy. (Sarcasm terminated. Normal state restored) |
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| Author: | Brightguy [ Mon Jun 02, 2008 7:55 pm ] |
| Post subject: | Re: A math question |
The question is NOT the same. Question 1: After choosing a cup, one of the other cups is removed (it happens to be one which did not contain the token). Question 2: After choosing a cup, one of the other cups is removed (specifically chosen not to contain a token). What's the probability that the chosen cup contains the token? In Question 1 the probability is 1/2 and in Question 2 the probability is 1/3. The scenario is simple enough that you can draw a probability tree showing every possible outcome: branch on token location and removal choice (you can also branch on original selection choice, but that's irrelevant as following branches will be symmetric). Once you have the probabilities of every possible outcome, take the sum of the ones where the token was chosen and divide by the sum of those consistent with the given information. |
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| Author: | r691175002 [ Mon Jun 02, 2008 8:20 pm ] |
| Post subject: | Re: A math question |
Brightguy @ Mon Jun 02, 2008 7:55 pm wrote: The question is NOT the same.
Question 1: After choosing a cup, one of the other cups is removed (it happens to be one which did not contain the token). Question 2: After choosing a cup, one of the other cups is removed (specifically chosen not to contain a token). What's the probability that the chosen cup contains the token? In Question 1 the probability is 1/2 and in Question 2 the probability is 1/3. The scenario is simple enough that you can draw a probability tree showing every possible outcome: branch on token location and removal choice (you can also branch on original selection choice, but that's irrelevant as following branches will be symmetric). Once you have the probabilities of every possible outcome, take the sum of the ones where the token was chosen and divide by the sum of those consistent with the given information. Intent does not affect probability. Question 1 and Question 2 are the same as soon as it happens to be one which does not contain the token. |
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| Author: | rizzix [ Mon Jun 02, 2008 8:25 pm ] |
| Post subject: | Re: A math question |
r691175002 @ Mon Jun 02, 2008 8:20 pm wrote: Intent does not affect probability. It most certainly does. |
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| Author: | richcash [ Mon Jun 02, 2008 8:40 pm ] |
| Post subject: | Re: A math question |
[quote="r691175002 @ Mon Jun 02, 2008 8:20 pm"] Brightguy @ Mon Jun 02, 2008 7:55 pm wrote: Intent does not affect probability. Question 1 and Question 2 are the same as soon as it happens to be one which does not contain the token. No, they're not the same at all. If you imagine a similar scenario with many, many cups it will become blatantly obvious. |
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| Author: | r691175002 [ Mon Jun 02, 2008 9:17 pm ] |
| Post subject: | Re: A math question |
[quote="richcash @ Mon Jun 02, 2008 8:40 pm"] r691175002 @ Mon Jun 02, 2008 8:20 pm wrote: Brightguy @ Mon Jun 02, 2008 7:55 pm wrote: Intent does not affect probability. Question 1 and Question 2 are the same as soon as it happens to be one which does not contain the token. No, they're not the same at all. If you imagine a similar scenario with many, many cups it will become blatantly obvious. I will say it again, intent does not affect probability (At least in the classical sense). As soon as the incorrect alternate choices have been revealed (Whether by random chance or by purposeful selection) the problem becomes the same. We have three cups. You choose a cup (Probability of 1/3). Now the guy randomly picks out a door and opens it, and it turns out that he opened an incorrect door (Aka does not have the prize). Explain to me how this is in any way different. How does the game host stumbling upon opening the door without the prize change the probability of your selection being correct from 1/3? If the host opens door C, the host has opened door C. Why he did it has no bearing on the question whatsoever. It does not matter if the game host knows ahead of time which choice he is supposed to reveal. If he reveals the correct one in terms of the question (Aka the one that has not been chosen, and that does not contain the prize) the problem is a monty hall problem. Lets look at your questions: Question 1: After choosing a cup, one of the other cups is removed (it happens to be one which did not contain the token). Question 2: After choosing a cup, one of the other cups is removed (specifically chosen not to contain a token). Explain to me how the end result of each of these options is different. In both cases the same cup will have been removed. Probability isn't going to go "Hey! He guessed! That means it doesn't count!". |
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| Author: | tenniscrazy [ Mon Jun 02, 2008 9:28 pm ] |
| Post subject: | Re: A math question |
if you are just picking a random door then you have a chance to pick the one with the prize behind it. when you are picking one that you know doesn't have a prize behind the door than if you switch doors it's like you are choosing both of the doors you hadn't chosen in the first place...get it? so if you randomly choose a door then you could open up the door that has the prize behind it..screwing up the whole question. But if you did happen to randomly choose a door that did not have a prize behind it your chances left would be 50/50. But if it was random than its kinda like you choose 2 doors... |
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| Author: | richcash [ Mon Jun 02, 2008 9:37 pm ] |
| Post subject: | Re: A math question |
It does matter. I'm generally not good at explaining but I'll try with this example. Imagine 100 people getting a number from 1 to 100. They all have a 1% chance each. Now 98 numbers are eliminated and there's 2 left. You're still in it. You and the other guy each have a 1% chance? No, you each have a 50% chance. Now let's imagine it's your birthday and everyone wants to see you in the final 2. A host knows which number won and eliminates 98 people whose number wasn't right. He guarantees to not eliminate your number. You don't have a 50% chance. You still have a 1% chance obviously. That's the difference. I don't know if I explained it well enough.  |
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| Author: | tenniscrazy [ Mon Jun 02, 2008 9:49 pm ] |
| Post subject: | RE:A math question |
don't you have 100% chance  jk |
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| Author: | richcash [ Mon Jun 02, 2008 10:08 pm ] |
| Post subject: | Re: A math question |
tenniscrazy wrote: don't you have 100% chance P jk 
But what I meant was the host knows the winning number from a draw beforehand and then people pick their numbers (or are assigned randomly). The host can just eliminate 98 right off the bat without eliminating you (for suspense purposes  ). But that doesn't make you 50/50, you're still 1% and the guy who survived all those eliminations is 99%.
EDIT-The difference with randomly revealing a cup versus knowingly removing wrong ones is that your cup does have a chance of being eliminated, if that randomly revealed cup had the token. In the other one where the host is knowingly removing a wrong cup, yours had no chance of being eliminated. If you pick a card from a standard playing deck it's 1/52 chance of being Ace of Hearts. If someone then randomly reveals 50 cards and none of them are Ace of Hearts then it's a 50% chance you'll have Ace of Hearts, right? But what if someone was looking at the cards and threw away 50 cards from the deck that were all NOT Ace of Hearts? Then it would still be 1/52 as it was initially, obviously. |
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| Author: | Insectoid [ Tue Jun 03, 2008 7:54 am ] |
| Post subject: | RE:A math question |
All right, it was specifically chosen so as not to be the prize. So shut up and use the monty hall one, and foget it. Just don't mention my story anymore. |
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| Author: | richcash [ Tue Jun 03, 2008 3:22 pm ] |
| Post subject: | Re: A math question |
insectoid wrote: All right, it was specifically chosen so as not to be the prize. So shut up and use the monty hall one, and foget it. Just don't mention my story anymore. I never said your story was wrong once. I was having a friendly debate with r691175002 about the difference between the 2 questions posed by Brightguy. I know you meant Monty Hall, you clarified that pages ago. The discussion has now branched into the difference between these two problems. Why can't we explore the difference between this problem and closely related problems? |
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| Author: | apomb [ Tue Jun 03, 2008 3:24 pm ] |
| Post subject: | RE:A math question |
is this still being debated? who won the bits? |
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| Author: | Insectoid [ Wed Jun 04, 2008 9:04 am ] |
| Post subject: | RE:A math question |
r691175002 did. Still, feel free to debate it. |
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| Author: | apomb [ Wed Jun 04, 2008 9:12 am ] |
| Post subject: | RE:A math question |
Meh, I can understand exactly how and why most people would think the way i originally assumed the question worked, however when i saw the solution and the other ways of stating the question, I just conceded that this is just another one of those debates that will divide message boards and break friendships. Just like the plane on a conveyer debate... its not even a debate, just different ways of stating a question as to make one group of thinkers believe they have fooled the other group. tl;dr : Pointless internet debates are pointless |
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| Author: | Insectoid [ Wed Jun 04, 2008 9:17 am ] |
| Post subject: | RE:A math question |
Hmm, Good answer, maybe it should be added to the poll (not really). I agree with you, the only reason I posted this was to prove to my mom that she was wrong. And now that that has been accomplished, I suppose this thread can die happy. |
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| Author: | Reality Check [ Wed Jun 04, 2008 8:42 pm ] |
| Post subject: | Re: A math question |
Ah the Monty Hall Problem. This is why I love probability. Here's another: You have a neighbour with two kids. Situation 1 You see outside the window and notice that one of the kids is a boy. What are the odds that the other is a girl? Situation 2 You see outside and notice one of the kids is a boy. You strike up a conversation with him and find that he is the eldest child in the family. What are the odds that the other is a girl? Tip: This is conditional probability |
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| Author: | Clayton [ Wed Jun 04, 2008 9:18 pm ] |
| Post subject: | RE:A math question |
1: 50% 2: 75% |
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| Author: | CodeMonkey2000 [ Wed Jun 04, 2008 9:45 pm ] |
| Post subject: | RE:A math question |
Mabey I misinterpreted the question, but isn't the probability for both situations 50-50? How does being the eldest affect gender? Unless I missed something.... |
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| Author: | Brightguy [ Wed Jun 04, 2008 10:19 pm ] |
| Post subject: | Re: RE:A math question |
insectoid @ Tue Jun 03, 2008 7:54 am wrote: So shut up and use the monty hall one, and foget it.
"If your solution doesn't fit the problem, change the problem", to paraphrase Einstein.
insectoid @ Wed Jun 04, 2008 9:17 am wrote: I agree with you, the only reason I posted this was to prove to my mom that she was wrong. And now that that has been accomplished, I suppose this thread can die happy.
You added in the part about Monty afterwards! Your Mom was right. |
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| Author: | Reality Check [ Wed Jun 04, 2008 10:29 pm ] |
| Post subject: | Re: A math question |
They are in fact not the same. Think of all outcomes in a family of two kids. We have: Two boys older boy younder girl younger boy older girl two girls. In other words: BB BG GB GG We are told that one of them is a boy so that rules out the GG prossibility. We now have BB BG GB. So, the probability that the other is a girl for the first situation is in fact 2/3. For the second situation we do the same. GG was ruled out and now that we know the boy is older we can now rule out GB as well. We are now left with BB and BG. Therefore the probability in the second situation is in fact 1/2. Many will disagree I am going to assume... |
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| Author: | btiffin [ Wed Jun 04, 2008 10:41 pm ] |
| Post subject: | RE:A math question |
I apologized to insectoid privately for changing his cool math problem to a "heated" discussion. Was not my intent when I posted; being the old guy I was more trying to fill in the august members of compsci who the ever famous Monty Hall was and the reference to Let's Make a Deal. If you read the wikipedia entry on Monty Hall Problem you will see how hard it is to word this problem to the satisfaction of a mathematician. No disrespect to insectoid was meant or deserved. Great big
Cheers |
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| Author: | apomb [ Wed Jun 04, 2008 10:59 pm ] |
| Post subject: | RE:A math question |
btiffin, it must be the gentlemenly thing to do, i also appologised to insectoid privately for making it seem like this entire thing is pointless, i did not mean for it to be interpreted that way, i meant that the original question was answered, and people continued to argue against the blatantly correct answer. |
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| Author: | Tony [ Wed Jun 04, 2008 11:00 pm ] |
| Post subject: | Re: A math question |
Reality Check @ Wed Jun 04, 2008 10:29 pm wrote: Many will disagree I am going to assume...
Yup. You've missed a couple of possible outcomes. Since you are distinguishing between older/younger sibling, then the order of boy/boy also matters. And so we get a sample space of: B1B2 B2B1 BG GB G1G2 G2G1 Situation 1: one of the kids is a boy. Girl: 2/4 = 50% (B1B2 B2B1 BG GB) |
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| Author: | Reality Check [ Wed Jun 04, 2008 11:18 pm ] |
| Post subject: | Re: A math question |
I've never seen this argued the way you did Tony and it does make sense to me but I had 3 different teachers confirm that it was 2/3. One of which had a PHD in physics... |
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| Author: | A.J [ Wed Jun 04, 2008 11:31 pm ] |
| Post subject: | Re: A math question |
shouldn't we include the possibility that there could be twins (in case 1, where we see one of them being a boy)? I might be just saying something stupid, so ignore me if I'm wrong. |
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| Author: | Tony [ Wed Jun 04, 2008 11:48 pm ] |
| Post subject: | Re: A math question |
Reality Check @ Wed Jun 04, 2008 11:18 pm wrote: One of which had a PHD in physics...
Physics is not Statistics.
Maybe there's another explanation that leads to that answer, but as it stands -- the age of the child either matters for the sample space, or it does not. It seems inconsistent to say that the observed boy could have an older sister, a younger sister, or a brother of either age. |
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| Author: | Brightguy [ Thu Jun 05, 2008 3:21 am ] |
| Post subject: | Re: A math question |
Reality Check: The answer to your problem depends on how exactly you obtained the information "one of the kids is a boy". When you looked out the window, were you able to obtain the genders of both of the kids, or just one? |
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| Author: | Reality Check [ Thu Jun 05, 2008 7:15 am ] |
| Post subject: | Re: A math question |
Yea Tony but this man isn't doesn't just do Physics. He's pure genius that is amazing at Math, English, History, Sciences, and Programming. He is a member of Mensa in fact so while I understand what you are saying, I find it very hard to doubt him. But hey, I suppose he could have been wrong. You know what, I'll ask him again on Monday and present to him what you told me and I'll see what he has to say. |
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| Author: | Insectoid [ Thu Jun 05, 2008 7:58 am ] |
| Post subject: | RE:A math question |
Here's another one: 10 computer nerds argue over a question. they all vote 1 of 2 ways. if 3 vote one way, what are the odds that the others will argue? I suppose I should mention that my mom knew it was the 'wrong' cup that was removed, and that it was intentionally removed as an incorrect cup. This thread's a fighter, that's for sure...(as in, it won't go away...not that it has to or anything...) |
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| Author: | zylum [ Thu Jun 05, 2008 8:45 am ] |
| Post subject: | RE:A math question |
The sample space (in both cases) is: {BB, BG, GB, GG} where the first letter is the first born and second is second born. In conditional probability, you have P(B|A) = P(A n B)/P(A) -> 'n' is the intersection symbol ie. the probability of B such that A is equal to the probability of A intersect B divided by the probability of A. Situation 1: event A = one kid is a boy event B = one kid is a girl P(A n B) = 2/4 = 1/2 P(A) = 3/4 Thus P(B|A) = (1/2)/(3/4) = 2/3 Situation 2: event A = the eldest kid is a boy event B = the younger kid is a girl P(A n B) = 1/4 P(A) = 2/4 = 1/2 Thus P(B|A) = (1/4)/(1/2) = 1/2 This is what you do in the first week of intro stats |
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| Author: | Insectoid [ Thu Jun 05, 2008 9:14 am ] |
| Post subject: | RE:A math question |
Confusing... Well, I guess we could put biology & trends into this to adjust the results (a girl is more likely to be born than a guy, I believe..) But that would be immature (or overly-mature?) and though usually I don't mind acting a bit like a dufus, I feel this thread has had enough twisting and shall refrain. |
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| Author: | Brightguy [ Thu Jun 19, 2008 4:49 pm ] |
| Post subject: | Re: A math question |
Reality Check @ Thu Jun 05, 2008 7:15 am wrote: You know what, I'll ask him again on Monday and present to him what you told me and I'll see what he has to say.
So? Did he concede that Tony is correct?  |
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| Author: | Reality Check [ Thu Jun 19, 2008 5:26 pm ] |
| Post subject: | Re: A math question |
Totally forgot to ask lol. |
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| Author: | I Smell Death [ Thu Jun 19, 2008 6:35 pm ] |
| Post subject: | Re: A math question |
you have to look at it for things that are not specified. i'll look at it a bit at a time After one has been selected by Jeremy, (Good for Jeremy he chose a cup, it doesn't say that he took the cup or looked under it, just that he chose one above the rest) one cup is removed and identified as a cup without the token. (it doesn't say that the cup that was removed was the one that Jeremy chose, infact it doesn't specify which cup it was at all, all that we know is that it is not the right one) Now there are 2 cups on the table, one the right one, one a wrong one. (this would seem obvious that there are two cups on the table with a wrong one having been removed) Jeremy still doesn't know if his is the right one. (here we can determine that Jeremy's cup was infact not the one removed from the table and determined to be the wrong one) Jeremy is given a choice: Change his selection to the other cup, or keep his current selection. (Knowing that a wrong one was removed from the table, leaving the one that he chose and an other one, with one of them being the right one and the other not. Jeremy could choose either of the two remaining cups as there is a 50-50 chance that the cup he chose is the right one) but that's just my view |
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| Author: | Reality Check [ Thu Jun 19, 2008 7:59 pm ] |
| Post subject: | Re: A math question |
I don't know Tony every single site I've searched says the answer is 2/3 and not 1/2 like you suggested. They all use the sample space of: BB BG GB GG. Your solution does make sense but I'm sure there must be a reason they all say 2/3. |
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| Author: | Brightguy [ Thu Jun 19, 2008 9:42 pm ] |
| Post subject: | Re: A math question |
Oh the answer is 1/2 alright. Either you misunderstand the problem or you aren't using the "most plausible" interpretation (this was what my previous question was about). |
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| Author: | Tony [ Thu Jun 19, 2008 10:42 pm ] |
| Post subject: | Re: A math question |
Reality Check @ Thu Jun 19, 2008 7:59 pm wrote: Your solution does make sense but I'm sure there must be a reason they all say 2/3.
What happens with a lot of sites, is that they often repost the same story, if it sounds interesting enough (2/3 sounds counter-intuitive). It'd be interesting if you were to still follow up with your source who could offer his own explanation, not just refer to "all the sites". @zylum -- that is indeed what one does in intro stats. Though your sample space for the first situation is wrong. It's {BB, BG, GG}, where the order of age doesn't matter. Or if you insist on age differentiation, then (by same reasoning as in my previous posts) it should be {B1B2, B2B1, BG, GB, G1G2, G2G1}. Either way, P(B|A) = (1/3)/(2/3) = 1/2 or (2/6)/(4/6) = 1/2. To state my reasoning again, if elder boy in (boy & girl) and younger boy in (boy & girl) count for 2 states, then elder boy in (boy & boy) and younger boy in (boy & boy) should also count for 2 states. |
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| Author: | zylum [ Fri Jun 20, 2008 1:02 am ] |
| Post subject: | RE:A math question |
Tony, I would have to disagree. If your solution space is {BB, BG, GG} then you are saying that the probability that the neighbor has two boys is the same as having a boy and a girl which is a false statement. If you want to use that sample space, then you have to state that the distributions are {0.25, 0.5, 0.25} which is the same as {BB, BG, GB, GG} = {0.25, 0.25, 0.25, 0.25} . |
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| Author: | richcash [ Fri Jun 20, 2008 2:14 am ] |
| Post subject: | Re: A math question |
Reality Check @ Wed Jun 04, 2008 8:42 pm wrote: You have a neighbour with two kids.
Situation 1 You see outside the window and notice that one of the kids is a boy. What are the odds that the other is a girl? Brightguy wrote: .. The answer to your problem depends on how exactly you obtained the information "one of the kids is a boy". When you looked out the window, were you able to obtain the genders of both of the kids, or just one?
Yeah, depending upon interpretation zylum and Tony are both right. If someone looks outside, sees both kids and tells you that "one is a boy" then the probability that the other is a girl is: zylum wrote: Situation 1
event A = one kid is a boy event B = one kid is a girl P(A n B) = 2/4 = 1/2 P(A) = 3/4 Thus P(B|A) = (1/2)/(3/4) = 2/3 However, if you look out and see 1 of the kids, and notice it is a boy (which would be the more common interpretation of the problem statement) then age has nothing to do with this problem. The sample space is {BB, BG, GB, GG} where the first variable is the "kid seen" and the second is the "kid not seen". A = kid seen is a boy B = kid not seen is a girl And then it is the exact same problem as "Situation 2", yielding P(B|A) = 1/2 |
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| Author: | zylum [ Fri Jun 20, 2008 7:24 am ] |
| Post subject: | RE:A math question |
I don't quite understand your reasoning.. Let me redo situation 1 with a sample space where age is not a factor: Sample space = {BB, BG, GG} with probability distribution of {0.25, 0.5, 0.25} Event A = one kid is a boy Event B = one kid is a girl Thus were are looking for the probability that one of the kids is a girl given that you know one of the kids is a boy. Whether you look and see one kid and its a boy or someone else looking outside and seeing both and telling you that one is a boy makes no difference. In both cases all you know is that one kid is a boy. Hence we look for P(B|A) (probability of one kid being a girl such that one kid is a boy). P(A n B) = 0.5 -> P(BG) P(A) = 0.75 ->P(BB) + P(BG) P(B|A) = 2/3 |
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| Author: | richcash [ Fri Jun 20, 2008 11:53 am ] |
| Post subject: | Re: RE:A math question |
You are right for one interpretation of the problem. zylum @ Fri Jun 20, 2008 7:24 am wrote: Whether you look and see one kid and its a boy or someone else looking outside and seeing both and telling you that one is a boy makes no difference. In both cases all you know is that one kid is a boy.
We have to arrange the kids by the order they were seen instead of age. If I look out and see a boy, then the First Kid Seen is a boy, and the Second Kid Seen is an individual event (exactly same problem as Situation 2). If I am just told that one of them is a boy, then I don't know if the First Kid Seen is a boy or girl, I only know that there's at least one boy. If 2 coins are flipped and I see the first one is heads, then the second one has a 1/2 probability of tails. If two coins are flipped and I am told "one of the coins landed heads" then the other one being tails is 2/3 being the same calculation as you showed above. zylum wrote: Sample space = {BB, BG, GG} with probability distribution of {0.25, 0.5, 0.25}
Yes, that is the sample space if we know one of the kids is a boy (order doesn't matter), but order does matter if we look out and the first kid we see is a boy. I'm pretty sure it's just an interpretation difference. I thought the more common interpretation would be that the person looks out and is only able to identify one of the child's genders (which would be First Kid Seen). But I guess you could easily interpret it the other way. |
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| Author: | zylum [ Fri Jun 20, 2008 1:25 pm ] |
| Post subject: | Re: A math question |
 It didn't even cross my mind that it could be interpreted like that. The question doesn't imply anything about order... But yes, if it were interpreted that way then it would be 1/2. |
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