Brain Teasers ? or 50 Years On........... ?

Well.................I did say it was scruffy, but.................

I do hope it is legible and that it helps.

I will try to find time soon to print and post the associated spreadshaeet

Cheers

Don

PS the version in flickr is probably easier to read

I should have said that the probability of the outcome of each branch is got by multiplying the fractions along each branch.

Selecting an Ace is denoted by an "A"

Selecting a card other than an Ace is denoted by a blank

excellent illustration of the issues from first principles Don. of course for this simple problem, one can jump straight to the combinatoric formulae to count # of elements in sample space and therefore subsets, i.e., using nCr = n! /r! (n-r)! with appropriate choice of r and n

but sometimes, for example in the case of the so called "Mony Hall Problem", it really does help to enumerate the sample spaces from 1st principles as you have done here.

enjoy

ken

Thanks ken. You are right regarding the use of combinations, which is what I used first. Its much easier to find the combination for "Zero-Aces" first and deduct this from "One/100% probability" to find the balance of "at least one ace". My Prob Tree shows why this is the case.

The real reason I produced the Prob Tree was to help winky and sjbabbey to identify the various combinations in which "at least one ace" could be generated, so that they could see how and why their orignial calculations were slightly out of line with myself, yourself and Mulbury. Its very difficult to explain this without the Prob Tree.

Also, I think the Prob Tree helps to show that the "Probability of Exactly One Ace" is no more difficult to find than any of the other combinations of Aces.

I will have a little browse through my A Level books tonight to see if I can find any other interesting probability questions.....................

ok a bit more challenging, I think !.............but still A-level from 50 years ago

In a game of darts, the probability that a particular player hits a treble-twenty with one dart is 0.4. How many throws are necessary so that the probability of hitting the treble-twenty at least once exceeds 0.9 ?

Five.

By the fifth throw the odds of not having thrown a treble twenty would be 0.6^5 i.e. 0.07776 so chances of having thrown at least one treble twenty would be 0.92224

Spot on sj. Five it is.

For the benefit of others who might be interested, my evaluation is as follows...........although sj's is much more straight-forward..............

The prob of NOT hitting a treble twenty is 0.6 on one throw  and on two throws is (0.6)^2 etc

The prob of hitting a treble twenty at least once in "n" throws is thus 1 - (0.6)^n

So to exceed 0.9 in "n" throws...........1 -(0.6)^n > 0.9

or (0.6)^n < 0.1

n log 0.6 < log 0.1

n > (log 0.1)/(log 0.6)

n > 4.5

Hence five throws are necessary

Hope this brings back fond memories of by-gone days at school...............

SJ - Simple when you see the method ! Well done.

Don - Your more detailed solution did get me thinking about a related side-issue of inequalities containing negative numbers. I agree with your answer, but have a little trouble going from:

n log 0.6 < log 0.1 to n > (log 0.1)/(log0.6) (note the emphasis on going from "<" to ">"

I am aware in the general case that if you multiply both sides of an equation by "-1" then the sign of the equality needs to change:

So for X > Y becomes -X < -Y

But I am struggling to reconcile an example such as:

5 * 2 > 4 (TRUE)

5 > 4/2 (TRUE)

Now multiply both sides by -1:

5 * -2 < -4 (TRUE)

5 < (-4) / (-2) (NOT TRUE)

Am I missing something obvious ?

Regards,

Peter

Peter,

In the second equation, you have not multiplied both sides by -1.

Surely it should be -5<-4/2, which is still True?

Steve D

Ok - I now see it !

Don has divided both sides of the equation by log (0.6). And since log (0.6) is a negative number, then the sign of the inequality goes from "<" to ">".

In my example the last inequality is achieved by dividing both sides by "-2" and so the last line should read:

5 > (-4)/(-2) (TRUE)

Sorry for yet another schoolboy error !

can avoid dividing by log as follows:

0.1 > 0.6^n

10^-1 > (6 * 10^-1)^n

now take log (base 10 for convenience -- but without loss of generality)

-1 > nlog6 -n

-1 > -n(1-log6)

i.e.

-n < -1/(1-log6)

or 

n > 1 /(1 - log6)

enjoy

ken

Good point Peter, and many thanks to steveD and ken c for explaining what was going on.

I have now added in a bit of info that I really should have included in my original text. Apologies.

Resistance is Futile...........

The diagram shows a resistance pyramid, of the sort Naim might (not) be considering for their new discrete range of Mini Statements...........

Q1. If there are "n" levels constructed from discrete resistors each of value R, what is the total resistance across the pyramid ?

Q2. What is the resistance of an infinite pyramid of this sort ?

Can we assume that a knowledge of electrical circuits is unnecessary and that this is a simple geometric progression?

If so then Σn = R(2^n -1)

So as n →∞ then ∑n →∞

no, the resistors are in parallel, so at any level n, the equivalent single resistance is 

R* (1/2)^(n-1)

so that at level 1, the resistance is R

at level 2, the resistance is R/2

at level 3, the resistance is R/4 etc

You are connect though that the total resistance is the sum of the equivalent single resistors, i.e the sum of a GP.

2R(1- 0.5^n) so that sum to 1 level is 2R*0.5 = R, etc etc

as n →∞

the total reststance tend to 2R

enjoy

ken

I see Ken. In this case the ratio of the series is 1/2 rather than 2. So the series will tend to a sum of 2R i.e. R( 1 + 1/2 + 1/4 + 1/8......)

Thanks for that.

Well done ken, and sj

You just saved Naim a small fortune on their Mini Statements............

My turn. I believe this is a fairly simple one?

The probability that Don an solve a certain brain teaser is 3/4

The probably that Ken can solve it is somewhat less, 3/5

What is the probability that the brain teaser will be solved if both Don and Ken attempt it s a joint effort?

enjoy

ken

The "modern" AS/A2 book in which I saw this next one, simply offered the advice........."Sometimes you just have to do anything you can think of and Hope !"

Simplify 4x + 4x/(x + 1) - 4(x + 1)

Let's Hope that somebody can still manipulate fractions

Ooops !

Sorry ken, I got interrupted whilst typing and didn't realise you had posted.

I'm sure that two sums on the go at once won't cause the Padded Cell to "crash"

I reckon it is

-4/(x + 1)

chances of me solving it are still only 3/4.

That's enough homework for me.

Well, it looks like your negative fractions are up to scratch sj

I recon ken and I only stood a 90% chance of solving it !

With those odds ken, if we tackled it independently I recon the probability the problem would be solved is 0.9, but together ......................................it would be dead cert.

My logic..................

Prob Don NOT solving Brain Teaser = 1/4

Prob ken NOT solving Brain Teaser = 2/5

Prob neither Don nor ken solving Brain Teaser = 0.25 x 0.40 = 0.1

Prob the Brain Teaser being solved = 1 - 0.1 = 0.9

But i'm never totally confident that I get these things right !

However, with colaboration, two (or three) heads are MUCH better than one, and as we have seen in the past, between us all, we have managed to solve ALL the brain teasers, hence the "Dead Cert".

BTW I think ken is being rather polite with his probability estimate for "Don" at 3/4 and has equally under-estimated his assessment for "ken" at 3/5

that is indeed correct Don.

what is somewhat 'depressing' is that if neither of us can solve the problem with certainty, (prob <1) then although combining efforts gves us a better proability, its by no means a dead cert ...

enjoy

ken