Wannabe an engineer? Test yourself here

greg66_tri_v2.0

http://www.telegraph.co.uk/education/10 ... rview.html

Knock yerselves out...

daviesee

60%

When do I start?

Wheelspinner

A measly 4/10. Too lazy to do the maths in my head on a couple, so am not too displeased.

However, the Professor should be ashamed. In Q4, his office contains a "two draw" filing cabinet.

Really? Does that mean you can fit a couple of sketches in it?

Instant credibility fail I think.

msmancunia

Oh dear - I got 0%.

I bet they can't type at 80wpm like what I can though.

tgotb

90%, stupidly overthought the die one and got it wrong.

Enjoyed the horizon one; didn't realise that calculation was relatively straightforward until I tried it...

greg66_tri_v2.0

TGOTB wrote:

90%, stupidly overthought the die one and got it wrong.

Enjoyed the horizon one; didn't realise that calculation was relatively straightforward until I tried it...

I must admit to still not getting the dice question (which in fact means I think the answer given is wrong, obviously).

stu-bim

50%, guess I need to stick to accountancy

kamiokande

70% got the dice one, the balloon one and the air one (didn't even attempt it) wrong.

I'm not convinced about the dice once...and the balloon one as well, anyone care to explain that to me?

daviesee

The dice was the easy one.
The balloon baffled everyone here. Google has the answer though.

greg66_tri_v2.0

The balloon one: It's not just about the balloon. Think about what the air does as well.

Right: let's hear the explanation for the dice answer.

daviesee

Greg66 Tri v2.0 wrote:

The balloon one: It's not just about the balloon. Think about what the air does as well.

Right: let's hear the explanation for the dice answer.

The dice has 6 sides. The odds on it landing with a particular face up is 1 in 6.

Maybe I read the question wrong and got lucky.

suzyb

20% by guessing 2 correct

And I realised I'd made a mistake on the metal one after I'd clicked the button Which was the only one I even had a clue how to start working out the answer.

msmancunia and I seem to be re-enforcing female stereotypes here :oops:

tgotb

Greg66 Tri v2.0 wrote:

The balloon one: It's not just about the balloon. Think about what the air does as well.

Right: let's hear the explanation for the dice answer.

Imagine throwing it 60 times; on average you'll get ten 6's. How many times do you need to throw it to average one 6?

greg66_tri_v2.0

Hmm. Ok. Here we go.

The question is "An unbiased cubic die has numbers 1 to 6 inscribed on each side. On average, how many rolls will you need in order to get a 6?"

As I read that, the question is asking me how long I will have to wait, on average, to throw a six. It's not "what are the chances of throwing a 6?"

So, I throw the dice. I may get a dice on the first throw, or on the sixth throw, or somewhere in between. There is a precisely 1/6 chance that I will get a six in any of the first six throws. So on average, I am going to get a 6 at the precise midpoint of my six throws - after 3.5 throws.

Which is, apparently, not the right answer. Unless it is (it obviously is. Obviously).

greg66_tri_v2.0

TGOTB wrote:

Greg66 Tri v2.0 wrote:

The balloon one: It's not just about the balloon. Think about what the air does as well.

Right: let's hear the explanation for the dice answer.

Imagine throwing it 60 times; on average you'll get ten 6's. How many times do you need to throw it to average one 6?

See above - as I read the question, it is how many times, on average, do you have to throw the dice before you get a 6.

So take your 60 throws and divide them into ten sets of six. In each set of six yo are guaranteed to get a six at some point. But once you have a six, the set ends - you don't need the remaining 1, 2, 3, 4, 5 throws, and so you don't need to throw the dice 60 times.

I suspect that the question has been slightly garbled in the translation to print.

tgotb

Greg66 Tri v2.0 wrote:

TGOTB wrote:

Greg66 Tri v2.0 wrote:

The balloon one: It's not just about the balloon. Think about what the air does as well.

Right: let's hear the explanation for the dice answer.

Imagine throwing it 60 times; on average you'll get ten 6's. How many times do you need to throw it to average one 6?

See above - as I read the question, it is how many times, on average, do you have to throw the dice before you get a 6.

So take your 60 throws and divide them into ten sets of six. In each set of six yo are guaranteed to get a six at some point. But once you have a six, the set ends - you don't need the remaining 1, 2, 3, 4, 5 throws, and so you don't need to throw the dice 60 times.

I suspect that the question has been slightly garbled in the translation to print.

I agree - poorly-worded question. Whoopee! I should have got 100%

Wouldn't be an issue in a real Cambridge interview; if you answered 3.5 they'd ask you to justify it, and any decent explanation would get you the credit...

daviesee

Greg66 Tri v2.0 wrote:

Hmm. Ok. Here we go.

The question is "An unbiased cubic die has numbers 1 to 6 inscribed on each side. On average, how many rolls will you need in order to get a 6?"

As I read that, the question is asking me how long I will have to wait, on average, to throw a six. It's not "what are the chances of throwing a 6?"

So, I throw the dice. I may get a dice on the first throw, or on the sixth throw, or somewhere in between. There is a precisely 1/6 chance that I will get a six in any of the first six throws. So on average, I am going to get a 6 at the precise midpoint of my six throws - after 3.5 throws.

Which is, apparently, not the right answer. Unless it is (it obviously is. Obviously).

Trying to hard.
The odd's are 1:6 so on average out of millions of throw's, one in six will come up as a 6. It really is that easy.

Or do we have to have a thread on average, mean and median?

tgotb

daviesee wrote:

Greg66 Tri v2.0 wrote:

Hmm. Ok. Here we go.

The question is "An unbiased cubic die has numbers 1 to 6 inscribed on each side. On average, how many rolls will you need in order to get a 6?"

As I read that, the question is asking me how long I will have to wait, on average, to throw a six. It's not "what are the chances of throwing a 6?"

So, I throw the dice. I may get a dice on the first throw, or on the sixth throw, or somewhere in between. There is a precisely 1/6 chance that I will get a six in any of the first six throws. So on average, I am going to get a 6 at the precise midpoint of my six throws - after 3.5 throws.

Which is, apparently, not the right answer. Unless it is (it obviously is. Obviously).

Trying to hard.
The odd's are 1:6 so on average out of millions of throw's, one in six will come up as a 6. It really is that easy.

Or do we have to have a thread on average, mean and median?

I disagree - Greg's got a point. He's answered the question they asked, whereas you've answered the question they meant to ask...

Greg's answer (and accompanying justification) would have made a better impression in a Cambridge interview than the "correct" answer.

daviesee

Who said you got 6 rolls?

Runtothehills

Greg66 Tri v2.0 wrote:

Hmm. Ok. Here we go.

The question is "An unbiased cubic die has numbers 1 to 6 inscribed on each side. On average, how many rolls will you need in order to get a 6?"

As I read that, the question is asking me how long I will have to wait, on average, to throw a six. It's not "what are the chances of throwing a 6?"

So, I throw the dice. I may get a dice on the first throw, or on the sixth throw, or somewhere in between. There is a precisely 1/6 chance that I will get a six in any of the first six throws. So on average, I am going to get a 6 at the precise midpoint of my six throws - after 3.5 throws.

Which is, apparently, not the right answer. Unless it is (it obviously is. Obviously).

You seem to be assuming that you will get a 6 in the first 6 throws but why should this be the case? There's a decent chance (33% or so) that you'll have to throw the dice more than 6 times. The mean number is pulled up by this.

By the way the mode (most likely result) of the number of throws is 1 and the median (middle result, if you put 101 results into order this would be the 51st) is 4

Runtothehills

TGOTB wrote:

daviesee wrote:

Greg66 Tri v2.0 wrote:

Hmm. Ok. Here we go.

The question is "An unbiased cubic die has numbers 1 to 6 inscribed on each side. On average, how many rolls will you need in order to get a 6?"

As I read that, the question is asking me how long I will have to wait, on average, to throw a six. It's not "what are the chances of throwing a 6?"

So, I throw the dice. I may get a dice on the first throw, or on the sixth throw, or somewhere in between. There is a precisely 1/6 chance that I will get a six in any of the first six throws. So on average, I am going to get a 6 at the precise midpoint of my six throws - after 3.5 throws.

Which is, apparently, not the right answer. Unless it is (it obviously is. Obviously).

Trying to hard.
The odd's are 1:6 so on average out of millions of throw's, one in six will come up as a 6. It really is that easy.

Or do we have to have a thread on average, mean and median?

I disagree - Greg's got a point. He's answered the question they asked, whereas you've answered the question they meant to ask...

Greg's answer (and accompanying justification) would have made a better impression in a Cambridge interview than the "correct" answer.

See above- He didn't answer the question they meant to ask as 3.5 isn't the mode, median or mean. Daviesee has got a really good way of visualising it:

Imagine you roll the dice 6 million times.

Then split the results after every 6 so that you get a load of sets that always end with a 6 and have no other sixes in them, the number of dice in each set is the number you had to throw before getting a 6. Some of the sets will be 1 long, some 2, some might be 50 or more. Now what is the average length of each set, ei the number of dice rolled before a 6 was achieved. It's the total number of dice (6 million) divided by the number of sets (1 million) or 6 dice.

stu-bim

Greg66 Tri v2.0 wrote:

So take your 60 throws and divide them into ten sets of six. In each set of six yo are guaranteed to get a six at some point. But once you have a six, the set ends - you don't need the remaining 1, 2, 3, 4, 5 throws, and so you don't need to throw the dice 60 times.

The average is 6. There is no certainty that there will be a six in every 6 throws. The average number of throws is 6 as this is the likelihood of a six coming up. No one set of 6 throws is dependent on any others so 60 throws is not 10 sets of throws.

Every time you pick up the dice up the same result is as likely as any other so no 3.5 is not correct and there is no logic to support it. Now if you said each time you get a number it is excluded then maybe 3.5 would work (then again that is a different type of probability calculations, I think).

All that said I got 50% but I do like logical puzzles

tgotb

Runtothehills wrote:

Imagine you roll the dice 6 million times.

Then split the results after every 6 so that you get a load of sets that always end with a 6 and have no other sixes in them, the number of dice in each set is the number you had to throw before getting a 6. Some of the sets will be 1 long, some 2, some might be 50 or more. Now what is the average length of each set, ei the number of dice rolled before a 6 was achieved. It's the total number of dice (6 million) divided by the number of sets (1 million) or 6 dice.

That's very elegant, I stand corrected :-)

veronese68

daviesee wrote:

Greg66 Tri v2.0 wrote:

Hmm. Ok. Here we go.

The question is "An unbiased cubic die has numbers 1 to 6 inscribed on each side. On average, how many rolls will you need in order to get a 6?"

As I read that, the question is asking me how long I will have to wait, on average, to throw a six. It's not "what are the chances of throwing a 6?"

So, I throw the dice. I may get a dice on the first throw, or on the sixth throw, or somewhere in between. There is a precisely 1/6 chance that I will get a six in any of the first six throws. So on average, I am going to get a 6 at the precise midpoint of my six throws - after 3.5 throws.

Which is, apparently, not the right answer. Unless it is (it obviously is. Obviously).

Trying to hard.
The odd's are 1:6 so on average out of millions of throw's, one in six will come up as a 6. It really is that easy.

Or do we have to have a thread on average, mean and median?

I agree with Greg on this one. If that's the answer they worded it badly. I know the odds of throwing a 6, that's not what they asked though. I'm not just clutching at straws because of all my incorrect guesses honest.

greg66_tri_v2.0

Runtothehills wrote:

Then split the results after every 6 so that you get a load of sets that always end with a 6 and have no other sixes in them, the number of dice in each set is the number you had to throw before getting a 6. Some of the sets will be 1 long, some 2, some might be 50 or more. Now what is the average length of each set, ei the number of dice rolled before a 6 was achieved. It's the total number of dice (6 million) divided by the number of sets (1 million) or 6 dice.

Ok - that makes sense. I get it now.

stu-bim

Veronese68 wrote:

I agree with Greg on this one. If that's the answer they worded it badly. I know the odds of throwing a 6, that's not what they asked though. I'm not just clutching at straws because of all my incorrect guesses honest.

To disprove the theory that on average you will get a six in 3.5 throws: what is the average for a five? or four?

They can't all be 3.5 if there are 6 numbers

UndercoverElephant

stu-bim wrote:

Veronese68 wrote:

I agree with Greg on this one. If that's the answer they worded it badly. I know the odds of throwing a 6, that's not what they asked though. I'm not just clutching at straws because of all my incorrect guesses honest.

To disprove the theory that on average you will get a six in 3.5 throws: what is the average for a five? or four?

They can't all be 3.5 if there are six numbers

I thought they worded it fine, though I was a little unnerved that it was so easy.

If the odds are 1 in 6, you would average 6 rolls to get any particular result. This is why posh people shouldn't go to casinos.

veronese68

stu-bim wrote:

Veronese68 wrote:

I agree with Greg on this one. If that's the answer they worded it badly. I know the odds of throwing a 6, that's not what they asked though. I'm not just clutching at straws because of all my incorrect guesses honest.

To disprove the theory that on average you will get a six in 3.5 throws: what is the average for a five? or four?

They can't all be 3.5 if there are 6 numbers

Yes, got it now. This moved on a lot quite quickly and I now see your point and stand corrected. Pesky customers phoning whilst I'm trying to post rubbish on the internet.
I'm going to make myself a pointy hat with a large D on it and sit in the corner now.

daviesee

UndercoverElephant wrote:

This is why posh people shouldn't go to casinos.

tgotb

The above discussion illustrates perfectly the reason they ask questions like this.

sketchley

"An unbiased cubic die has numbers 1 to 6 inscribed on each side. On average, how many rolls will you need in order to get a 6? "

If "on each side" it has "numbers 1 to 6 inscribed" that means there is a 6 on each side and also a 1, 2 ,3 ,4 & 5. Therefore you would get a six on every throw. So the answer is one.