Brain Teaser No 1

Not sure I like this question.


You have my sympathy, neither did I when I first encountered it !!

Let me help yourself and any others who might be considering this one.

First, we have already established a couple of principles. When we had 10 wrong guesses, we knew the last card had to be red. When we had 11 wrong guesses, we knew the last card had to be black.

Suppose we had 12 wrong guesses or 14......

Suppose we had 13 wrong guesses or 15.....

OK, this leads to the already quoted position that even numbers mean RED and odd numbers mean BLACK.

Put another way, if the number of wrong guesses is 2N, then there would be N of each (wrong guesses) and the last card is red. Also if the number of wrong guesses is 2N+1 then there would be N+1 wrong black predictions and N wrong red predictions and the last card is black.

Back to "the suppose..."

10, 12, 14.....wrong implies red
11, 13, 15.....wrong implies black

Now when Paul R made his comment wouldn't 'at least 11 wrong' be more interesting? he was really teasing....Stick to 10 to start with.....

It helps to keep the scores even....

Cheers

Don

This time got the spread sheet out etc and assumed that the general case is 0.5 chance of getting any individual card accurate.

I have carried out my calculation and come up with a probaility of 0.500047 of the last card being red and therefore red has the higher probability. Please at this stage just say right or wrong.. I can of course discuss my methodology...

Regards

Belsizepark

ok don i have tried one more time

This time got the spread sheet out........ Please at this stage just say right or wrong..

"Right"

Cheers

Don

PS...but probably for the wrong reasons........

As noted above.....

Back to "the suppose..."

10, 12, 14.....wrong implies red
11, 13, 15.....wrong implies black


In how many ways can these happen ?

BIG HINT

Don't try to work out the actual number of ways, unless you have a computer.......

but it would help enormously if you wrote down the number of ways of getting 10 predictions wrong.....and 12......and 14........and also 11.......and 13.......and 15.......

Now I hope I haven't given too much away.....

Cheers

Don

Don .. that is basically what I did.. I even worked out how to use the factorial function on Excel.. never used that one before..I extended it however to calculating the probabilities of getting each of those numbers. I calculated even numbers seperately from the odds as if an even number incorrect the 51 cards already down are 26 black and 25 red and if there is an odd number of incorrect answers then the 51 cards are 26 red and 25 black.

There may be a much simpler way of calculating the answer than I have as you have suggested it is solveable without a computer (and I guess you also mean calculator). However I fail to see it.

I dont necessarily believe that it is intuitively obvious. I should be grateful if you would now show your answer in full....

I would be interested in seeing all your workings .. because quite frankly I have had enough of this

Thank you for your time. I have spent quite a bit of effort myself on this blasted problem and if there is a logical answer to the problem in the same way as to the first 2, I shall very angry with myself for not spotting it.

Despite the fact that I believe that I have solved the problem, I still dont like it...

Regards

Belsizepark

[This message was edited by belsizepark on THURSDAY 12 December 2002 at 00:45.]

The envelope switching problem.

Part 1

Tom holds 1 envelope in each of his hands, and tells Harry that he can have either one of them at his choice. Harry picks one and then Tom tells Harry that one envelope holds twice as much money inside it as the other. Before Harry can open the envelope and see what is inside, Tom offers Harry the chance to switch envelopes. The question is should Harry switch and does it matter?

Won't make the slightest difference whether Harry sticks or changes his mind. If Tom repeats the trick a hundred times (or a thousand or a hundred thousand etc)it won't make any difference whether Harry always sticks, always changes his mind or randomly sticks/changes his mind.

Harry's prospects are 1.5 times the lesser contents or 0.75 times the greater contents, in each case multiplied by the number of games. Both prospects are equal.

Or perhaps the maths has changed in the 40 odd years since I went to school......

Cheers

Don

I should be grateful if you would now show your answer in full....

because quite frankly I have had enough of this....

Thank you for your time. I have spent quite a bit of effort myself on this blasted problem....

I still dont like it...

Oh !!!! You prefered Bam's ladder problem that kept us ammused for about 6 months....? (see page 13 IIRC)

you have suggested it is solveable without a computer (and I guess you also mean calculator). Well, at least you have correctly deduced this bit....

FYI, at the moment I haven't worked out the probability of it being red; just that red is more probable than black, ie there are more ways it could be red than there are ways of it being black......


Now I just know you don't really want me to publish my solution just yet. It is soooooooooo much more satisfying to get there under your own steam or with the help of a couple of friends on the Forum......

Cheers

Don

Belsizepark did alot of hard work and said....

I have carried out my calculation and come up with a probaility of 0.500047 of the last card being red

Don did bugger all and said...

FYI, at the moment I haven't worked out the probability of it being red;....

Since then I felt guilty.....so used the old excel sheet to churn out the actual probabilty, which isn't necessary to answer the question.... and I get.....

0.500000678

But i haven't checked it...

yet!!

Cheers

Don

You are probably correct on the probability . it now seems an awful long time since my University days and one of my first year stats modules. Intuitively to me it was very close to 0.5 from the word go (assuming that the guy is not psychic etc. I may have even done it corectly myself, the underlying numbers get so large, it wouldnt have taken much in decimal place rounding by slight different methods to get an answer that it as different as yours to mine if we did the same thing in different ways. But in all honesty I have clue about what Excel does with rounding on calculations. (i ddid mine in numerous seperate stages although I linked the cells as opposed to rounding anything myself)..

I think I will leave it.. sorry but not taking it any further..I liked the first 2 parts to the problem, but I think I will give any further work on this one a miss. I assume you have a good solution, my email is on my profile,(delete the nospam bit) I would be delighted to receive a copy of your spreadsheet calculation as well as know your logical answer.

Sorry if I have appeared rude in some of these posts on this question.. I may have been frustrated by the question, but that is no excuse for rudeness.

I notice however that you haven't commented on my envelope switching question that I posted the other day...

Regards

Belsizepark

I notice however that you haven't commented on my envelope switching question that I posted the other day...


Oooops!!

Whats the probability of finding a comment to part 1, four posts above this one?

Cheers

Don

Anyone got any ideas about AL4N's Teaser?

Cheers

Don

I have counted up (one, two, three.....) and find there are six hundred and seventy two million ninty four thousand five hundred more ways in which the last card could be a red than a black (IMHO)...........

Cheers

Don

PS But thats not how I got the answer to the question

I spent the summer in Canada this year and bought a beautifully made douglas fir presentation box with 4 'mint' Canadian coins displayed in it.

The value of each coin is proportional to its radius and the total value of the four, in pounds is 28.

The presentation pack is cleverly designed as a triangle with sides 13, 14 and 15 cm. The largest coin just fits in the box touching all three sides.

There are three pieces of very thin card, each just touches the large coin and is parallel to a side of the box. This creates three smaller triangles, one in each corner.

The three remaining coins fit into the corner triangles such that each just touches the three sides of its triangle.

I was showing some friends the coins the other night, (along with my 6 hours of holiday video and 252 slides in an interesting AV night.......) when in shear desperation it seemed, one of my friends asked how much the smallest coin was worth in ££££.

Well, he couldn't work it out, so had to suffer the rest of the holiday snaps.

Can anybody help him out?

Cheers

Don

PS I had prepared a picture of the box and coins to attach but find the attachment facility has been disabled........

Hopefully the attachment facility is back on, well according to PS over in the hi-fi corner it is.

So here is the picture of my presentation box

Cheers

Don

Especially for Belsizepark, I felt I should start to publish some solutions.

The trilogy part iii. Can be solved by using a spreadsheet, buteven before spreadsheets it was pssible to demonstrate that red was more likely than black.

The first part of my solution follows as an attachment, I hope its legible....

Cheers

Don

OK, even I couldn't read the attachment.....

Lets hope this bigger version, in two parts, works.

Part 1

Cheers

Don

Cheers

Don

You omitted a coin...

Paul

Paul,

You omitted a coin...


For some time now, i've kind of felt our libraries might contain a few duplicate titles. Someday we might find out...

I thought i'd lost the coin, but fortunately found it, so i can now publish a proper picture...

Cheers

Don

I suspect the smallest coin is worth £1.71 or to be precise £(504/295).

Or course I could well be wrong, rather an involved calculation!

Matthew

I suspect the smallest coin is worth £1.71 or to be precise £(504/295).

Or course I could well be wrong, rather an involved calculation!

'fraid your wrong on this at the moment Matthew.

Try working with ratios (for all four coins) for as long as possible.......

Cheers

Don

The last line shows that R is bigger than B so there must be more ways in which Troy's last card could be red.

Note, this last line doesn't allow you to calculate the probability of the last card being red, just that its more likely to be red.

Hope you all found this one absolutely rivetting and fascinating...........

well, perhaps not....

Cheers

Don

Whoops, that'll be radius not area.

A round £2 will be the correct answer then!

Matthew

Matthew,

A round £2 will be the correct answer then!


Assuming you mean 'a nice round number' rather than 'something close to' I'm afraid you're still wrong. Unfortunately, I couldn't even say that 'something close to £2' would be right. But then that depends on the definition of 'close to'.

I think you must be on the right lines, however, and i'm sure a quick check of the arithmetic, rather than the principles, should deliver the right answer.

I hope so anyway.

Cheers

Don

Now i've looked at Belsizepark's envelope switching problem again and am having doubt's about the best tactics.

On the face of it, both part 1 and part2 are simple 50/50 choices.

OTOH, being told that the first choice holds £100, means that the other hand holds either £50 or £200, which seems to suggest that the prospects on switching improve to £125 whereas the prospects on not switching remain at £100.

Any thoughts ?

Cheers

Don