Puzzles with numbers and things

[davis_greatest]

... meanwhile there has been a second solver of round 372... well done...

[davis_greatest]

...and now a 3rd solver of round 372, and 2nd solver of round 371 (all to be named shortly, I'm sure) - congratulations!

[davis_greatest]

Rounds 371 and 373 - please would anyone (no one is excluded! ) please put solutions to either of these up on the thread.

Round 372 - I am aware that at least one person is working on this, and I think is nearly there, so this will remain open a little longer!

Another round may come shortly... I shall have a think....

[abextra]

Mini Round 373 -

Charlie, Gordon and Oliver know that my box contains 3 red balls and 2 pink balls, and each ape takes one ball, which he looks at (unseen by the others). Gordon then secretly shows Oliver his (Gordon's) ball, at which point Oliver announces that he knows which ball Charlie has too.

What colour ball does each ape have?

I would say, if Gordon and Oliver both had pink balls, then little Oliver could be sure that Charlie's ball is red (as there were only two pink balls)... oh, explaining is not for me.

[davis_greatest]

Yes, indeed, abextra!

And so, this goes to show, that even though there were several possibilities of balls for the apes, and even though we did not know what balls Oliver saw, the fact that we know that what he did see was sufficient for him to work out the other balls, is in itself sufficient for us then to work out what balls Oliver saw and what the other balls were!

Confusing?

That is the same principle that was needed for round 372! It seems that everyone got a bit confused by that at first, but all have now either got there or are getting close. Congratulations so far to snookersfun, abextra and Monique on round 372! (Still open though.)

[davis_greatest]

OK, I'm not sure whether this one will seem hard or not! Let's give it a try!

Round 374 – Break Breakdown the Second

Gordon has just been practising making total clearances again, and made one every frame! (His total clearances each consist of 15 reds, each with a colour, and then the 6 colours.)

He then came home and told Oliver the following about his practice session, which consisted of at least 2 such frames:

(a) For each of the six colours, the total points he had scored from that colour (during the entire session) equalled the total points he had scored from another colour; and also happened to equal the total points he had scored from yet another colour!

(For example, the total points scored from blues might have been the same as the total points scored from greens; and also the same as the total points scored from blacks – or whatever. Note again – this was during the entire session, not necessarily during any one particular frame.)

(b) Due to the high hourly cost at the club, he had played the fewest possible number of frames that could make (a) possible.

(c) Each time he had made a total clearance, his break was at least as great as any that had gone before – and his final break was 50 greater than his first.

(d) The number of yellows he had potted in each frame was a square number.

(e) During one frame, at least, he had potted the same number of yellows as browns.

It wasn’t long, of course, before smart Oliver was able to tell Gordon the breaks that Gordon had made in every frame – Oliver even knew how many balls of each colour had been potted by Gordon during each clearance. :snooker:

So.... what were they? Answers by Private Message please.

[snookersfun]

Rounds 371 and 373 - please would anyone (no one is excluded! ) please put solutions to either of these up on the thread.

still on a high of having solved a probability puzzle in one go

so here is my explanation:

First, the probabilities of
Charlie pots=x, misses=1-x,
Gordon pots=1-x, misses x

The probability of A=Charlie winning being at the table was calculated with that abacus as 6:1 i.e. A=6/7.
That probability is actually an infinite geometric series composed of all options of Charlie winning, i.e. by potting his ball (x), plus missing his ball, Gordon missing and Charlie then potting (x^2(1-x)), plus Charlie miss, Gordon miss, Charlie miss, Gordon miss, and Charlie then potting (x^3(1-x)^2),.... etc., like this:

A= 6/7= x + x^2(1-x) +x^3(1-x)^2+…..

the probability for Gordon winning while Charlie is at the table (1-A = 1/7)
(starting with Charlie miss, Gordon pots +... )

(1-A)=1/7= (1-x)^2 +x(1-x)^3+ x^2(1-x)^4+….. = B*(1-x) (*)

Now the probability of B=Gordon being on table winning is (a bit like Charlie's first, just now of course switching the probabilities)

B= (1-x) +x(1-x)^2+ x^2(1-x)^3+…..…..

and (1-B) Charlie winning

(1-B) = x^2 + x^3(1-x) +x^4(1-x)^2+….. = A*x =6/7x (*)

(*)looking at the two options (A and B) I can see that one can factor out x or (1-x)

and I decided to use (1-B)=A*x=6/7x
i.e. the probability of Charlie still winning when Gordon is at the table is that of Charlie potting times the probability of him winning while he (Charlie) is on the table

In order to obtain x, I used the fact that the probability that Charlie is winning while on table = probability that he pots + probability that he misses * probability Gordon misses * probability Charlie wins

so :
6/7= x + (1-x)x*6/7
or rearranging:
x^2 -13/6x +1 =0

x= 2/3 (for possible x=0.5-1)

using x in the formula above, therefore

(1-B) = 6/7*2/3= 4/7 is the probability of Charlie winning when Gordon is at the table. Meaning Charlie is still more likely to win.

I seriously hope I didn't mess anything up or forgot anything...

[davis_greatest]

Round 371 update - we can close this. Charlie is indeed still more likely to win, even when Gordon is at the table - Charlie's chances now being 4/7 to Gordon's 3/7. Congratulations to snookersfun and Monique!

snookersfun's explanation is correct - well done! (However, there is no need to sum any geometric progressions and one can jump directly to this:

6/7= x + (1-x)x*6/7
or rearranging:
x^2 -13/6x +1 =0

x= 2/3 (for possible x=0.5-1)

The last line should say for possible x in the range 0
and follows from rearranging the previous line to the form (x-2/3)(x-3/2) = 0

Having identified that Charlie has a 2/3 chance of potting the black each attempt, and Gordon hence a 1/3 chance, the rest follows! Well done.)


Round 372 update - we can close this round too. Congratulations to snookersfun, abextra, Monique and moglet! Answers can go on the thread please!

Round 374 update two correct answers in so far! This is now the only round still open (I think - at least, the only round I have still open).

[snookersfun]

The last line should say for possible x in the range 0 < x < 1 ...

why doesn't this show in your post? Or in mine unless I add spaces in. Strange!
Anyway, I would agree... for the general case, but we had been told that Charlie's chances at potting are better than Oliver's and together with the x, 1-x respective probabilities, we can conclude that x>0.5, can't we?

[davis_greatest]

why doesn't this show in your post? Or in mine unless I add spaces in. Strange!
Anyway, I would agree... for the general case, but we had been told that Charlie's chances at potting are better than Oliver's and together with the x, 1-x respective probabilities, we can conclude that x>0.5, can't we?

I don't know why it doesn't show - as I certainly typed 0 < x < 1 (without the spaces), and it seems to make it appear as 0 < 1 (which again doesn't appear properly if you type it without the spaces).

Re the last sentence, yes we can!

[davis_greatest]

R374 update

Three correct answers have been received to round 374 so far. I think there is one person working on it, and so the names will be published and solutions invited after that. To confirm one point, following a query - the red balls in this question do not count as a "colour".

[davis_greatest]

Round 375 - Clockwatching

Charlie and Gordon have just finished a frame of snooker, with the beautiful Talia Mabb refereeing once more. All the balls (except the white) are now in one of the snooker table's six pockets. Charlie says "Gordon - look at this, as I walk around the table clockwise, as many times as I want, each pocket contains either balls of a greater value* than the previous pocket, or a greater number of balls (or both)."

"I doubt that very much," says Gordon. "Must be impossible."

"No, I don't think it is," points out Miss Mabb.

*the value of a ball is its usual value in snooker - 1 point for each of the 15 reds, 2 for the yellow,..., up to 7 for the black


Who is right? Either prove that it can be done, or prove that it is impossible. Answers by Private Message please.

[davis_greatest]

Round 374 update - I think it is time to close this one. Congratulations to solvers abextra, snookersfun, Monique and moglet. Please put your answers up on the thread!

[Monique]

4 frames.

72 (colours - 15 reds + 15 yellows - 16 yellows in total)
85 (colours - 15 reds + 8 yellows +1 green + 6 browns - 9 yellows in total)
87 (colours - 15 reds + 3 yellows + 9 greens + 3 browns - 4 yellows and 4 browns in total)
122 (colours - 15 reds + 2 blacks + 3 pinks + 8 blues + 2 browns - 1 yellow in total)

6 blacks, 7 pinks and 14 greens all sum up to 42
12 blues, 15 browns and 30 yellows sum up to 60

[moglet]

4 frames, breaks 72,85,87,122 (∆=50)
Yellow numbers 16,9,4,1
4 yellows with 4 browns in third frame
Totals from each colour:
Blacks 7x6, 42
Pinks 6x7, 42
Greens, 3x14, 42
Blues, 5x12, 60
Browns, 4x15, 60
Yellows, 2x30, 60

Balls per break
1st all yellow clearance
2nd 8 yellows, 1 green, 6 browns, final colours
3rd 3 yellows, 9 greens, 3 browns, final colours
4th 2 browns, 8 blues, 3 pinks, 2 blacks, final colours