Occam. - 11 October 2010 12:50 PM

You are both right, however, you missed my comment about the logical assumption.  I assumed a hole with zero diameter thus making the ring height and the sphere diameter the same.  Then I used the arithmetic equsion domokato listed.

Occam

Ya, the width of the drill was not stated so I just assumed it was negligible.

This isn’t a riddle, but it is a freaky fun math check.  What’s your result?  It worked like a charm for me.

It didn’t work for me, JITP. I am not really sure why it should…

Here’s One I Posted At My Home Forum:

This is interesting and absolutely based on a video I am watching, I just thought it might be worth doing it here to see what results we get

Oh yeah, since it’s stats I’m putting it in science.

Two people are tossing coins and recording the whether it goes head or tails, each is looking for a particular sequence:
[list][*]HTH (HEAD - TAILS - HEAD)
[*]HTT (HEAD - TAILS - TAILS)[/list]

Consider the following possibilities:

A: The average number of tosses until HTH is larger than the average number of tosses until HTT.

B: The average number of tosses until HTH is the same as the average number of tosses until HTT.

C: The average number of tosses until HTH is smaller than the average number of tosses until HTT.

Which of the above, in your opinion,  is correct and why?

I will give the answer and show the video in a while.

Keke

B. Since both sequences start with HT, we can look at just the last letter, which is either H or T, and the chance of getting H or T are equal. So assuming a sequence ending in HT has been tossed, either H or T will come up next with equal probability, so both chances are equal.

Here’s a fun one. It’s been around a while, so please don’t give it away if you already know the answer.

Three people check into a hotel. They pay $30 to the manager and go to their room. The manager suddenly remembers that the room rate is $25 and gives $5 to the bellboy to return to the people. On the way to the room the bellboy reasons that $5 would be difficult to share among three people so he pockets $2 and gives $1 to each person. Now each person paid $10 and got back $1. So they paid $9 each, totalling $27. The bellboy has $2, totalling $29. Where is the missing $1?

I love that one. It’s so tricky

The answer is 42.

It’s a logical error.  The author pulled the ‘ol switch-a-roo.  Good one!    ( Switching an addition for a subtraction!  Clever.  If we begin with the original total of $30, and consider that each of three roomies recieved $1, then we subtract each dollar from the $30 to arrive at $27, their final price.  But the price they should have had was partly in the bell boy’s pocket, so subtract his $2 to bring the price down to the manager’s price of $25.  The puzzle makes one think that the bell boy’s $2 should be added to the $27, which is the reverse operation that is needed. )

At first I thought that it was an off-by-one error, but no.  It was clever.  I’ve heard sales pitches as sneaky as that.  I wish I had sent this before DM figured it out, I’m just too slow!

Kyuuketsuki UK - 15 October 2010 11:10 AM

A: The average number of tosses until HTH is larger than the average number of tosses until HTT.

Keke

I don’t know this one.  ( I’ll guess A because the HTT seems more random, because it has more repetition, and so I think that the less likely HTH will need more tosses before we see it. )

Domokato & JITP:

I thought as you (Domokato) do too ... I’ll wait a bit longer (Sunday?) to reveal the answer

Keke

GdB - 23 September 2010 11:25 PM

Ok.

Next riddle:

GdB

Um…the first riddle was solved? Was the answer “hexadecimal?” 

Count every “F” in the following text:

FINISHED FILES ARE THE RE
SULT OF YEARS OF SCIENTI
FIC STUDY COMBINED WITH
THE EXPERIENCE OF YEARS…

How many did you get? If you got any other number than (6), count it again. And again… 

Kyuuketsuki UK - 16 October 2010 12:15 AM

Domokato & JITP:

I thought as you (Domokato) do too ... I’ll wait a bit longer (Sunday?) to reveal the answer

Keke

Dang, I always sucked at probabilities :(. Too counter-intuitive

OK, on the coin-tossing thing:

Effectively C is correct because when throwing to get the sequence HTT and getting it wrong on the last throw (HTH) you already have the first toss in the sequence “in your pocket” whereas failing on HTH you still have to start the desired sequence Smile

Video here: TED: Peter Donnelly shows how stats fool juries

I thought it was brilliant

Keke