Brain Teaser No 1

There are 3 ways you could have drawn a red side:
the red side of the red/black card
one red side of the red/red card
the other side of the red/red card

These are equally likely. So it is twice as likely that you have the red/red card as the red/black card. Therefore bet on the reverse side being red.

On a cycling trip my friends and I ended up in a conversation about whether weight differences in bikes designed for downhill racing really made any difference since by definition you were going downhill and the riders phyiscal effort was pretty much irrelevant. During this my friend Penny suggested that if we all lined up at the top of a hill, set off together and coasted downhill Dion -- being 17 stone of solid Maori -- would accelerate away.

However, half remembering my schoolboy physics and something about how Galileo dropping cannonballs off the Leaning Tower of Pisa proved that heavy objects don't acccelerate faster we would in fact all accelerate at the same rate and stay roughly level. And if there were any slight differences they would be related to air resistance and so Dion -- with his improbably large cross sectional area -- would presumably, if anything, be slower. So we would all start level until we hit a top speed and then Dion would accelerate away becuase of something to do with the fact that you can't kill mice by throwing them down coal mines.

So we had a bet and then conducted a rigourosly controlled scientific experiment at the next hill. Much to my dismay Dion shot off into the distance. And not just crept away slowly either but shot off in a manner which suggested that some very large differnce was responsible. i.e. most likely his enormous mass. We repeated this a number of times with the same results and even swapped bikes in case Dion's uber expensive Zirconium pedals really did make a difference (They didn't).

So the question is why do fat blokes on bikes seemingly accelerate faster when coasting downhill?

Matthew

PS Penny's body is perfectly formed. Mine and Dion's definitely not.

quote:

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Either it is the other red side of the red/red or the black side of the red/black (the black/black don't feature, nor does it matter how you come to draw the card that has the red side, the important thing is that you have before you a red side) - right?

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RR
RR
RB
BR
BB
BB

The left column is the side you see first. Each of these is equally likely. We only gamble on

RR
RR
RB

and still each of these is equally likely. But the back being red occurs twice as often as black.

Have you ever noticed that 'but' really means 'and'?

Paul

quote:

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where does your third(fourth) card come from?

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Because there are two ways to draw the red/red card and see red.

RichardN explains it much better.

Paul

quote:

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So the question is why do fat blokes on bikes seemingly accelerate faster when coasting downhill

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Resistance - Bodies of different weights accelerate at the same rate given the same force (mass time gravity) where there is no resistance.

Fat blokes, tandems, and recumbent bikes do better than the average cyclist down a hill for this reason, the same reason that a speeding cyclist gets into a tuck to remove wind resistance,

The fat bloke has about the same resistance as a regular rider, but has more force (mass times gravity times the sine of the incline) than the regular rider.

Same with a tandem - same resistance, over double the weight and therefore over double the force.

Recumbents have the same amount of weight (force), but have less air resistance.

Cheers

Keith.

But I thought the whole point was that if resistance is the same then mass is irrelevant and bodies of different mass will accelerate at the same rate? Which is why Galileo's different size cannonballs fall at the same rate.

Matthew

quote:

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But I thought the whole point was that if resistance is the same then mass is irrelevant and bodies of different mass will accelerate at the same rate?

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The resistance has to be proportional to the mass.

Newton suggested that f=ma, force is equivalent to mass times acceleration. The force down the slope is mgsinA where A is the angle of the slope from the horizontal. If the resistance were 'R' then,

mgsinA - R = ma

So if R is constant (or at least independent of mass, as it largely is when considering drag on pushbikes) then the fat bloke has more m and consequently more a.

Paul

Equating potential energy (which is what you have at the top of the hill) to kinetic energy (which is what you generate on your journey down) we have:

m*g*h = 1/2*m*v*v

where:
m = mass
g = acceleration due to gravity
h = starting height
v = velocity

So, the big guy has more mass and thus more potential energy. He has to 'disperse' his PE on the journey down and the only non-constant in the equivalance equations is velocity.

Best Regards,
Mark Dunn

OK - consider that you only have the two cards - R/R and R/B.

You pull them randomly from a hat and ONLY have to guess the other side when you see Red on the side you pull.

So...

It IS 50/50 as to which card you pull from the hat BUT you will replace the R/B card half of the time that you draw it (because you see the Black face). Thus, when you are guessing, you will be holding the R/R card much more often and you should guess Red.

Is this clearer? These things can be tricky to explain.

Any anyone please let me know if I've got this wrong.

- GregB

Insert Witty Signature Line Here

Is this clearer? These things can be tricky to explain.

Nice explanation!

Doubt if it'll convince Declan, but still a nice explanation!

Cheers

Don

I have constructed a spreadsheet in Excel to simulate the 3 card R/R, R/B, B/B game of chance.

The spreadsheet randomly selects card R/R, R/B, B/B and randomly selects which face is shown.

The spreadsheet then rejects or "Aborts" all those selections that show a Black face upwards.

It then "calls" the downward face "Red" and checks to see if this is "Right"

Finally the spreadsheet counts the number of times "Red" called on the downward face is "Right" and compares this with the count of "non-abortive" trials.

The spreadsheet runs the simulation 2,000 times. This generates about 1,000 "non-abortive" trials each run. The results of 10 such runs represents about 10,000 trials.

Based on the first such 10,000 trials, the average probability of getting a RED downward face given that you already have a RED upward face is 0.669. Which is kindof close to 2/3.

Bloody sickening, isn't it!

Cheers

Don

quote:

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Originally posted by Matthew Robinson:
Keith,

But I thought the whole point was that if resistance is the same then mass is irrelevant and bodies of different mass will accelerate at the same rate? Which is why Galileo's different size cannonballs fall at the same rate.

Matthew

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Galileo's balls fell at the same rate because the resistance was negligible for the force applied to them. Far greater was the difference in release time between them as he assumed that they were let go at the same time.

The same is not true for cyclists going down a hill. Air resistance is a huge factor, so there are two solutions to go faster down that hill:

• Add weight
• Become more aerodynamic - remove resistance

The best model for this is the equivalent "test" that an astronaut performed in the vacuum of the moon's surface - he dropped a feather from one hand and a ball of some sort from the other. The feather fell just the same as the ball - no air resistance.

Cheers

Keith.

"...something to do with the fact that you can't kill mice by throwing them down coal mines"
I think this is potentially very cruel and should not be encouraged in this part of the forum. A mouse may survive the fall but will find the experience very traumatic and will surely be unable to find food once there. Condemned to slow starvation and fruitless attempts to climb back out to reunite with familly and friends. Oh the shame of it!

I took a class in grad school on Monte Carlo simulations. We would write Fortran programs so simulate all kinds of things and it would loop 10,000 or whatever times and keep track - same as you're doing with the spreadsheets.

I don't have a Fortran compiler here and I'm not up to speed on programing in Excel so no simulations for me these days Would you send me one of your spreadsheets so I can "see the guts" as it were and shorten the learning curve? E-mail is gobeatty@yahoo.com.

- Greg

Insert Witty Signature Line Here

Cheers

Don

Apologies, I forgot to 'compress' the last spreadsheet that I sent you. The first three were reduced to about 12kB each. The last one was the full 1.2MB.

They took about 10 minutes to send, so will take about 10 minute to receive! sorry

Cheers

Don

Twas the night before Christmas and in the Naim factory not a creature was stirring except for... In an attempt to discover the secrets of Naim's new NAP500 a team of Sony engineers (the ones mentioned in that other post who collect the musical electrons off the floor that have fallen out of unsoldered speaker cable strands) decided to gain illegal entry to the Naim factory. Oh the shame of it!

They arrive at the factory in the dead of night and decide to try to gain entry via the roof. Naturally they aren't carrying a ladder because this would arouse suspicion among the locals in Salisbury. So they search for one and eventually find an old wooden, slightly rotten set of rungs. The ladder has length L. The factory roof is hard to access, even with a ladder, and on their first attempt the ladder crackles and threatens to break. They stop climbing and wonder what to do. Scouting for a way in they find a wall against which a large, cubic crate has been left of side 1m. They reason that if they position the ladder so that it touches the ground, the wall and the edge of the crate that this will provide sufficient support to prevent the ladder from breaking. Good.

Being exacting engineers and slaves to planning they first decide to calculate the exact distance that they must place the foot of the ladder away from the side of the crate so that it will touch both the wall and the crate edge. Can they work it out before sunrise?

In terms of the ladder length L what is the horizontal distance the foot of the ladder must be placed away from the side of the crate? Assume the ground is horizontal, the wall vertical and the crate a perfect cube.

The winner will have the most elegant formula.

Merry Christmas and enjoy,
BAM

The winner will have the most elegant formula.

No ELEGANT solutions just yet.

Clearly the MINIMUM length of ladder is 2*(sq rt of 2) and is placed 1m from the cube. For all other ladders (which must be longer) there are two possible solutions for the distance from the cube, each with a complimentary height up the wall.

A little graph, (produced with the aid of Excel), is easy to plot for given values of distance. Then from the graph (or using Excel to do the iterations) the distance can be interpolated for any length of ladder.

Turning equations around to make x the dependant variable (rather than y) always seemed dead easy at school!!!! Guess I shall have to go back to school over Xmas, or find a maths teacher.

Perhaps, we should simply declare when we have a solution, e-mailed to bam to confirm its validity and degree of elegance. This will increase the chances of somebody being driven 'up-the-wall' by Bam's little teaser.

Merry Xmas Everbody (Just in case I don't get an elegant solution beforehand)

Cheers

Don

You have followed exactly the same train of thought as I did - ie an infinite number of scenarios of ladder lengths (in excess of the minimum length) and heights up the wall.

However, I think the problem has a single solution, ie a fixed ladder length "L" reaching a fixed height up the wall (ie to reach the roof) say "H".

Could Bam confirm this please.

I've got lots of equations, but nothing approaching a solution nor elegance!

STEVE D

Nothing elegant but bloody tantalising!!

L=Length of ladder
d=distance from the cube

L^2=(1+d)^2 + (1+(1/d))^2 (Pythag)

L^2=d^2 + 2d + 2 + 2/d + 1/(d^2)

Which has a degree of 'symetry'

Also Let y = (d + 1/d) and hence y^2 = d^2 + 2 + (1/(d^2))

L^2=y^2 + 2y

0=y^2 + 2y - L^2

But this doesn't have simple roots!!!!!

So, next I might try:- Let -y = (d + 1/d) ???

On the other hand I might get back into the kitchen before she misses me to help Mrs D with Xmas Eve Diner before our guests arrive. I might even slip a shortened version of Bam's Xmas teaser into each cracker........

Merry Xmas

Don

Don wrote: "I might even slip a shortened version of Bam's Xmas teaser into each cracker........"
Cruelty indeed! Keep downing the Christmas spirit it's bound to generate some extra creative dopamine. That's my excuse and I'm sticking to it!

Good cheer to all,
BAM

Depends how much you've had to drink as to which one that is! The one with the ladder almost horizontal looks fine to me tonight!

BTW, 2* (sq rt of 2) = 8^0.5 which is the minimum point on my graph and the rest of the graph does give two solutions for any value of L greater than this. L tends rapidly towards infinity as d>>0 and also tends towards d+1 in the other direction.

Elegant! Now how elegant is elegant I wonder?

Cheers

Don

For long(ish) ladders the following gives a half-decent approximation:-

d=L-1-(1/L) (if you're drunk)

or (1/d)=L-1-(1/L) (if you don't suffer from vertigo)

as L>infinity, so d>L-1 OR (1/d)>L-1 either of which which is correct!

ok, I SAID it was crude!!!(I used it as the starting point of my iteration type solutions

Cheers

Don