Recommend some good IQ questions

[partygirluk]

My Dad is interviewing some people, one of whom is going to be his replacement. He wants me to come up with roughly 4 IQ questions they can give these guys 10 minutes to work on. He doesn't want them to extremely hard, just to prove that he/she can think on the spot and is sharp.

1 question I suggested is

i) You know a woman has 2 kids. You ask her if she has at least one daughter. She says yes. What is the chance that she has a son?

He thought this was the right level of difficulty for what he is looking for.

Thanks in advance.

[CallMeIshmael]

Are you thinking like 4 completely different areas of thought?

Like, we have 1 Bayes/probability type question, so, no more of those?

[partygirluk]

[ QUOTE ]
Are you thinking like 4 completely different areas of thought?

Like, we have 1 Bayes/probability type question, so, no more of those?

[/ QUOTE ]

That would be best.

[CallMeIshmael]

How about a spatial reasoning test.

Something like this (http://atsrecruitment.navcanada.ca/html/pdfs/spatial.pdf#search='spatial%20reasoning%20test'), but adjusted for how difficult he wants it

[CallMeIshmael]

Also, Im not sure how much you know about the SAT (its a college entrance exam for schools in the States).

They probably have some decent math/verbal reasoning type questions for you.

Do a search for "SAT test questions" or something similar


EDIT: You want the hardest ones. Because, the easier ones are a joke.

[jakethebake]

On average, how much are the sales for an average gas station in New Mexico per year?

[wacki]

[ QUOTE ]

1 question I suggested is

i) You know a woman has 2 kids. You ask her if she has at least one daughter. She says yes. What is the chance that she has a son?

He thought this was the right level of difficulty for what he is looking for.


[/ QUOTE ]

I would say not enough info. Depends on age of mom and kids. More boys are born but they tend to die at a much earlier age so girls eventually outnumber boys later on in life. Race is also a factor as in China the male/female birth rate is 120:100. Of course that is probably due to genercide as the actual biological rate is closer to 108:100.

Still the answer I think your dad is looking for is 50% due to the probability law of independece.

Is this the answer he was looking for? It seems really easy for trying to find a replacement.

[Jman28]

I think he's looking for 66.6%.

Possible combinations:

b/b
g/b
b/g
g/g

All equal chance.

Then, you know it's not b/b, so...

b/g
g/b
g/g

All equal. P(b/g or g/b) = 2/3


Edit:

I don't think this is a great question, unless mathmatical reasoning/logic is related to the job (and same can be said for word problems if they are unrelated).

Dumb people who have been exposed to bayesian probability can get this right, while some smart people who haven't been will get it wrong. I don't think it shows much about natural intelligence, unless you know what the answerer has learned in their life.

[wacki]

[ QUOTE ]
I think he's looking for 66.6%.

[/ QUOTE ]

Your answer is wrong. The conditions were that one child is a girl. This does not alter (mathematically) the chance that the second child is a boy/girl. Your mistake is in your assumptions.

[CallMeIshmael]

[ QUOTE ]
[ QUOTE ]
I think he's looking for 66.6%.

[/ QUOTE ]

Your answer is wrong. The conditions were that one child is a girl. This does not alter (mathematically) the chance that the second child is a boy/girl. Your mistake is in your assumptions.

[/ QUOTE ]

Im nearly certain that he is correct.

Two links:

1 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=1949103&page=&view=&s b=5&o=)
2 (http://mathforum.org/dr.math/faq/faq.boy.girl.html)

[wacki]

Whatever. If you flip a coin and get heads is the next flip more likely to be a tail? No, it isn't.

The conditions were "one child was a girl". Given that, the odds of the second child being a boy are still 50%. This is a mistake stats people make all the time and it costs a lot of companies a lot of money.

[krimson]

[ QUOTE ]
Whatever. If you flip a coin and get heads is the next flip more likely to be a tail? No, it isn't.

The conditions were "one child was a girl". Given that, the odds of the second child being a boy are still 50%. This is a mistake stats people make all the time and it costs a lot of companies a lot of money.

[/ QUOTE ]

Both children were born prior to the question being asked whether she has a girl. So your coin-flip analogy is incorrect.

You flip two coins, coin A and coin B, one is heads, what is the probability that the other is tails?

The coin flips have 4 outcomes

AB
HH
HT
TH
TT

If one is heads than we see that 2 of 3 instances involve the other coin being a tails. Hence 66.7%

[wacki]

Here are some real life statistics.

http://www.in-gender.com/Gender/Gender_Odds.aspx

Although we often hear the "statistic" that you are 30% or even 70% more likely to keep having the same gender, this is just an old wives tale. It is NOT a fact. The truth is, your odds stay pretty close to 50% for each child and only vary slightly. If you have had 2 or 3 boys, you are only about 2% to 6% more likely to have another boy. If you have had girls, you are slightly more likely to have a boy next.

It's an old wives tale resulting from bad math. The slight variation is due to biological reasons. Still the "REAL LIFE" statistics show the odds are always 50%.

Freedman has several chapters about this concept in:

Linky (http://www.amazon.com/exec/obidos/tg/detail/-/0393970833/qid=1120074457/sr=8-1/ref=pd_bbs_ur_1/103-7697750-8828625?v=glance&s=books&n=507846)

[Jman28]

[ QUOTE ]

You flip two coins, coin A and coin B, one is heads, what is the probability that the other is tails?

The coin flips have 4 outcomes

AB
HH
HT
TH
TT

If one is heads than we see that 2 of 3 instances involve the other coin being a tails. Hence 66.7%

[/ QUOTE ]

Well explained, but, isn't that what I said...
[ QUOTE ]
I think he's looking for 66.6%.

Possible combinations:

b/b
g/b
b/g
g/g

All equal chance.

Then, you know it's not b/b, so...

b/g
g/b
g/g

All equal. P(b/g or g/b) = 2/3



[/ QUOTE ]

Wacki, you would be correct if it was asked if the first child is a girl, rather than if either were.

[CallMeIshmael]

You're confusing the concepts here.


No one is saying: IF you're first child is a boy, the second one is more or less likely than 50:50 to be a boy.

We are saying that there are more ways for her to have a boy given she has at least one girl, than there are ways to have no boys.

[CallMeIshmael]

[ QUOTE ]
Wacki, you would be correct if it was asked if the first child is a girl, rather than if either were.

[/ QUOTE ]

[CallMeIshmael]

The following is a C/C++ program that gets the result of 66% (note: I suck at programming, so its poorly written)

#include <iostream.h>
#include <stdlib.h>


int numGirls(int *kids);

void main()

{
int kids[2], numB = 0, total = 0;

srand(62);



for (int i = 0; i < 10000; i++)
{
kids[0] = rand()%2;
kids[1] = rand()%2;

if (numGirls(kids))
{
total += 1;

if (numGirls(kids) == 1)
numB += 1;
}


}

cout << "Number with 1 boy : " << numB << " out of a total of: " << total;

}


int numGirls(int *kids)

{
if (kids[0] && kids[1])
return 2;
else
if (kids[0] || kids[1])
return 1;

return 0;
}

[Patrick del Poker Grande]

Yes. Let's please play this game again. Now what if you asked her if she has a girl named Sarah?

[CallMeIshmael]

[ QUOTE ]
Yes. Let's please play this game again. Now what if you asked her if she has a girl named Sarah?

[/ QUOTE ]

that was an awesome thread

[BruceZ]

Assuming boys and girls are equally probable, in families of 2 children having at least one girl, 2/3 of those families have 1 boy, and 1/3 of those families have 2 girls.

Your coin flip problem would be a correct analogy if the problem stated that the youngest child were a girl. Or if the problem were stated in any way that made the two children distinguishable from each other, such as if you saw only one of the children playing in the yard, and it was a girl. Then the probability that the other child is a girl would be 50%. That is a very different problem.

[wacki]

[censored] me I'm retarded. Lack of sleep, dealing with loans, dealing with mechanics that are fixing my alternator for the 3rd time this month, and training Hidden Markov Models. Well if I screwed anything up today I'm glad it was the least important one.

My info was right, just reading the situation wrong. [censored].

[Spaded]

[ QUOTE ]
I think he's looking for 66.6%.

Possible combinations:

b/b
g/b
b/g
g/g

All equal chance.

Then, you know it's not b/b, so...

b/g
g/b
g/g

All equal. P(b/g or g/b) = 2/3


[/ QUOTE ]

What a load of horse crap. b/g and g/b are the same f'ing thing if she says she already had one girl. Why would b/g and g/b be different? One boy and one girl.

The chance that the other child is a boy if you already have a girl is 50/50 , assuming she isn't an identical twin. Actually, since an identical twin is a possibility, i'd say the chances are something like 50.0001/49.9999 (pulled it out of my ass) of having a girl.

So this is what you have:
b/b (impossible)
b/g and g/b (boy and girl)
g/g (2 girls)

So it is between b/g and g/g, 50/50.

If you already have a girl, your balls don't suddenly kill all sperm bearing an X chromosome.

[CallMeIshmael]

[ QUOTE ]
What a load of horse crap. b/g and g/b are the same f'ing thing if she says she already had one girl. Why would b/g and g/b be different? One boy and one girl.

[/ QUOTE ]

You are not right.

[Jman28]

[ QUOTE ]
What a load of horse crap. b/g and g/b are the same f'ing thing if she says she already had one girl.

[/ QUOTE ]

It's funny how angry you are at me for being right.

Reread the question and the rest of the thread. It's been explained already.

[Spaded]

Sorry bro, just steamed from a bad beat /images/graemlins/grin.gif

[West]

link (http://www.wiskit.com/marilyn/boys.html)

[edtost]

have him give one of the impossible reasoning type questions, like "how many gas stations are there in LA" or "how many traffic lights are there in nyc" and observe the way they try to figure out an approximation.

edit: try to avoid questions where getting the correct answer relies on figuring out some trick, which basically means that candidates who've seen it before will get it right and the others won't. you want to evaluate their thought processes, not their quantity of random knowledge.

[MtnDave]

Back to the orginal question.
Have him check out here (http://www.techinterview.org/index.html)
to see if there is anything there he can use.

[Spaded]

[ QUOTE ]
link (http://www.wiskit.com/marilyn/boys.html)

[/ QUOTE ]

Good link! It all depends on how the question is worded. The original poster's wording of the question suggests that one of the childrens' genders is pre-determined, so it should be 50%.

[CallMeIshmael]

The way PGUK worded it leaves 0 doubt. The correct answer to his problem is 66%.

EDIT: "You ask her if she has at least one daughter. She says yes."

Clearly states that we are not asking her about a specific child. Simply, that at least one of her kids is a girl.

[MortalNuts]

This one is easy, and sort of classic:

You're a monk in an isolated monastery, with 99 other monks. You have all taken vows of silence -- you can't say a word, or communicate by any other means (written notes, etc.) with any of the other monks. You spend all your days in a room by yourself, in total isolation, with one exception: you eat dinner with all the other monks. Every night at 6 pm, all of you sit down at a big round table (where you can see all 99 other monks) and quietly have dinner, which takes a couple hours. Afterwards, you all file back to your individual private chambers to spend the night in prayer and repose.

One night, you get a message from God in a dream. You're sure it's God, not just some renegade monk playing a joke on you or bad gruel or whatever. Anyway, God tells you that at least one of you (the 100 monks in this monastery) must kill himself. He (big G) has handily indicated which of you needs to do the deed -- he's placed a large red dot on the forehead of everyone who needs to commit suicide. He tells you that if you figure out you have a red dot on your head, you must commit suicide alone in your chambers that very night, at say 11 pm. Furthermore, he tells you that all of the other monks have been informed of this very same fact. He also reiterates that at least one of you is marked in this fashion.

How do you figure out whether or not you have a red dot on your head? Assume that all the other monks are logical, and that anyone who figures out they're marked will indeed commit suicide.

There are no reflective surfaces, no chance of communicating with outsiders, no collecting some rainwater and looking into that, etc, etc. Basically, no tricks. This is a logic problem.

This is really easy for anyone with any kind of math or computer science background, so I could see it (or some variation) being useful if that's the sort of position you're talking about. I.e., it would be a mild red flag to me if someone with that kind of background couldn't figure out the basic idea here pretty quickly.

later,

mn

[SpearsBritney]

I'm assuming this message is received after dinner? But then why the dinner info? Wouldn't you just look for the dot at the table? Im confused, and by confused, I mean stupid. Also, why's god such an A-hole?

Edit: Ok, the solution to this problem (atleast the way you worded it) couldn't be anything other than to kill yourself if there is no dot on anyone else at dinner(assuming all 99 are there). How is this an IQ question, or am I missing something?

[MortalNuts]

[ QUOTE ]
I'm assuming this message is received after dinner? But then why the dinner info? Wouldn't you just look for the dot at the table? Im confused, and by confused, I mean stupid. Also, why's god such an A-hole?

[/ QUOTE ]

the lord moves in mysterious ways, apparently.

but yeah, you get the message after dinner. so the next time you'll see anyone is the next day (at 6 pm). the only reason I attached times to everything was to re-emphasize the point that you figure things out at dinner, then afterwards go kill yourself. i.e., you can't just linger over dessert hoping someone will suddenly leap up and off themselves or something.

note a couple other things (maybe these are obvious to you, but just in case): the minimum number of dots is one, but there can be many more than that. also, you might not be able to figure out that you're marked in just one monk-dinner.

later,

mn

[SpearsBritney]

[ QUOTE ]
the minimum number of dots is one, but there can be many more than that

[/ QUOTE ]

My bad. I thought there was only one dot.

[ QUOTE ]
he's placed a large red dot on the forehead of everyone who needs to commit suicide.

[/ QUOTE ]

Missed that. I guess I failed. /images/graemlins/blush.gif

Edit: ...and that...

[ QUOTE ]
Anyway, God tells you that at least one of you (the 100 monks in this monastery) must kill himself.

[/ QUOTE ]

...and that...

[ QUOTE ]
He also reiterates that at least one of you is marked in this fashion

[/ QUOTE ]

This is a good lesson; if you don't know the answer to a question, STFU!

[PairTheBoard]

You try a door and it doesn't open. What do you do?

answer: Try the door again.

PairTheBoard

[RacersEdge]

I would go with the actual interview questions consulting companies (mainly) use. Like "Why is a manhole cover round?" or "How many jelly beans does it take to fill a 757 airplane"?

These get at more logical reasoning and there is no actual right answer, but it's all about figuring out how people think. There are actual books out there with these questions in them.

[brassnuts]

FWIW, I think probability questions aren't good choices for this type of test. A lot of the time, they are so counter-intuitive while seemingly intuitive that a very intelligent person might just not think it through well enough. Case in point, Wacki.

I like the idea about spatial reasoning, and they can get very difficult, as seen here. (http://www.highiqsociety.org/noflash/nonmembers/iqtests.htm) I suggest you check out the ultimate IQ test on that page and copy the forms of some of the spatial reasoning and other questions as well.

[Spaded]

[ QUOTE ]

How do you figure out whether or not you have a red dot on your head?

[/ QUOTE ]

People will look at your forehead instead of your eyes, and people will have sad expressions on their faces that will indicate to you that you have a dot. Or you could just pinch a piece of your forehead skin off and look at it /images/graemlins/grin.gif

But I know it's not THAT easy, usually in IQ tests there is only ONE answer that they will accept, and it is always the one that they think is the hardest to get.

[Jman28]

This is a good question two. I've heard it before.

The problem with questions like these are that if the interviewee has heard of them, it doesn't mean he's smart. I could tell the answer to my dumb friend and then he would pass the interview test.

The only thing you can tell from this question is if someone is not a super-genius. If they don't know the answer, they aren't one of the top .1% smartest people on earth. I believe even very smart people (but not top .1%)who have never been exposed to questions like this may still get them wrong.

Because of this, these questions can't tell you much about the person.

Anyone agree? disagree?

[popniklas]

I can't solve this problem and that really bothers me....

[wacki]

[ QUOTE ]

FWIW, I think probability questions aren't good choices for this type of test. A lot of the time, they are so counter-intuitive while seemingly intuitive that a very intelligent person might just not think it through well enough. Case in point, Wacki.


[/ QUOTE ]

Thx for sticking up for me brassnuts.

In my defense I have been analyzing DNA nonstop for some time now. I've been creating complex position specific scoring matrices and many forms of genomic alignments. Much of what I do is sequential probability and I was stuck in that mode when reading partygirls post. Make sure your dad understands the reasoning behind all of the answers. Everything I've said in this thread is true, I just didn't read (or look) at the question right. Depending on your dads needs, I should either be fired, or even hired.

P.S. I hope this thread gets buried soon. /images/graemlins/blush.gif

[MortalNuts]

[ QUOTE ]
I can't solve this problem and that really bothers me....

[/ QUOTE ]

A hint (in white): <font color="white"> think about what happens if only one person has a dot on his forehead. How would you determine whether you are that person? Okay, now think about what happens if there are two people with dots. Now three, and so on. </font>

later,

mn

[MortalNuts]

[ QUOTE ]

The problem with questions like these are that if the interviewee has heard of them, it doesn't mean he's smart. I could tell the answer to my dumb friend and then he would pass the interview test.

The only thing you can tell from this question is if someone is not a super-genius. If they don't know the answer, they aren't one of the top .1% smartest people on earth. I believe even very smart people (but not top .1%)who have never been exposed to questions like this may still get them wrong.

Because of this, these questions can't tell you much about the person.

Anyone agree? disagree?

[/ QUOTE ]

Well, yeah. To be perfectly honest I think most questions of this type generally aren't all that useful, even if you can guarantee the candidate has never heard them before, except for picking out the extremes in the distribution of candidates. That is, it's pretty easy to come up with a question that only a very bright person will be able to answer quickly, or to come up with a question that all but very dumb people should be able to answer. It's much, much tougher to come up with questions that can meaningfully differentiate between people in the middle of the scale. I think the question I quoted does better than most at this, but it's still not that great.

In practice, the situation is even worse, since (as you mentioned) some people may have heard the question before and others may not have. For this reason alone, I would never use a "stock" question like this as a way of interviewing someone. You at least have to change it so that the basic procedure is the same (e.g., in this problem, find the base case and generalize from it), but the details are different.

If you're looking to test someone's ability in a particular subject, it's possible to do a lot better, and actually write questions that (imho) differentiate between candidates much more finely.

later,

mn

[jason_t]

Answer in white:
<font color="white">
CallMeIshmael (6:22:35 PM): http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=2754742&amp;page=0&amp;view=c ollapsed&amp;sb=5&amp;o=&amp;vc=1
CallMeIshmael (6:22:37 PM): have u seen that?
jason_t (6:22:47 PM): it's long
jason_t (6:22:49 PM): /images/graemlins/frown.gif
CallMeIshmael (6:22:57 PM): the solutiuon is long?
jason_t (6:23:00 PM): the post
CallMeIshmael (6:23:03 PM): lol
jason_t (6:23:05 PM): i'm very lazy today
jason_t (6:23:08 PM): &lt;--- woke up at 5 pm
CallMeIshmael (6:23:09 PM): its a logic problem
CallMeIshmael (6:23:13 PM): wow
CallMeIshmael (6:23:21 PM): i went to bed at like 4
CallMeIshmael (6:23:23 PM): 3
jason_t (6:23:42 PM): i went to be around 5:30 am, i got a phone call at 7 am and was on the phone until 8:30 am
jason_t (6:23:48 PM): then slept until 5 pm
CallMeIshmael (6:24:01 PM): wow
CallMeIshmael (6:24:03 PM): that is absurd
CallMeIshmael (6:24:38 PM): well
CallMeIshmael (6:24:41 PM): read it sometime
CallMeIshmael (6:24:45 PM): cause, i dont know teh answer
CallMeIshmael (6:24:49 PM): but, i have a guess
jason_t (6:24:55 PM): ok, i will later
CallMeIshmael (6:24:59 PM): /images/graemlins/smile.gif
jason_t (6:27:07 PM): &lt;--- read it, solved it
CallMeIshmael (6:27:15 PM): waht do u do?
jason_t (6:29:41 PM): well, let's just be silly about it
jason_t (6:29:46 PM): if there is only one dot
jason_t (6:30:02 PM): the person with the dot will realize the next time he sees the group he is the only one with a dot
CallMeIshmael (6:30:13 PM): and kill himself
jason_t (6:30:13 PM): becuase he won't see a dot on anyone else
jason_t (6:30:15 PM): right
jason_t (6:30:19 PM): say there are two dots
jason_t (6:31:13 PM): consider one of the guys who has a dot on his head
jason_t (6:31:53 PM): he'll see a dot on another guy
jason_t (6:31:59 PM): and think he's in the clear on the first night
jason_t (6:32:16 PM): all the people with no dots will see the two dots and think they are in the clear
jason_t (6:32:22 PM): the next night they'll see each other
jason_t (6:32:26 PM): and no one will be dead
jason_t (6:32:59 PM): now the guys with the dots on their heads will realize it must be because there are two dots in the group
jason_t (6:33:04 PM): and they each have it on their head
jason_t (6:33:08 PM): they'll kill themselves
jason_t (6:33:28 PM): similarly, for three dots, after two nights the three people will realize and kill themselves
jason_t (6:33:29 PM): etc.
jason_t (6:33:30 PM): done
CallMeIshmael (6:34:04 PM): sounds good /images/graemlins/smile.gif
jason_t (6:34:09 PM): &lt;--- supergenius
CallMeIshmael (6:34:15 PM): &lt;--- knew that was coming
jason_t (6:34:24 PM): ^should have announced pre-event
jason_t (6:34:32 PM): post-event it's not interesting</font>

Further comment in white:
<font color="white">For what's it worth, this is how to approach any problem of this type: try small n and see if the reasoning generalizes.</font>

[RacersEdge]

lol, as usual with these things, I miss the premise..

as a hint to any others who made the same mistake as me, read the white:

<font color="white"> The decisions about who has the dots doesn't have to be made after just one dinner... /images/graemlins/tongue.gif </font>