Puzzles with numbers and things

[snookersfun]

or 6) better: 2-2^-39

[snookersfun]

1) What is the expected number of turns required to mark 1 ball? 1 turn
2) What is the expected number of turns required to mark both balls? infinity???

3) What is the expected number of balls marked after 1 turn? 1
4) What is the expected number of balls marked after 2 turns? 1.5
5) What is the expected number of balls marked after 3 turns? 1.75
6) What is the expected number of balls marked after 100 turns? 1.9999999999... would have to check that better

[chasmmi]

ok im a bit confused but...

1) 1
2) 3
3) 1
4) 1.5
5) 2
6) 2

???

[davis_greatest]

I've just switched the order of the questions, so please refresh, but otherwise there is no major difference!

[davis_greatest]

Round 69 - just 2 balls

Here is a simplified version of snookersfun's round 66, which may shed more light on it:

Charlie has a bag with 2 balls.

Gordon repeatedly takes one ball out of the bag, at random, marks the ball with some essence of banana, and then replaces the ball into the bag. Each time he does that, it is called a "turn".

Questions

1) What is the expected number of turns required to mark 1 ball?
2) What is the expected number of turns required to mark both balls?

3) What is the expected number of balls marked after 1 turn?
4) What is the expected number of balls marked after 2 turns?
5) What is the expected number of balls marked after 3 turns?
6) What is the expected number of balls marked after 100 turns?

[davis_greatest]

Here is the updated scoreboard after round 66.
I decided to give 1 point each to d_g and Chasmmi (who braved calculating under adverse circumstances). Elvaago - here is another honorable mention for understanding the question and participating (at one point I shall add these up to something useful scoreboard wise)


snookersfun……………………….…..31
abextra...............................16
davis_greatest.....................13½
Vidas..................................12½
robert602.............................6
elvaago...............................6
chasmmi..............................5
The Statman……………………..……4
Semih_Sayginer.....................2½

Round 67 is closed, so that is the scoreboard after round 67, but before round 68, which I have already posted, and before round 69, which I shall post in a few minutes.

[chasmmi]

hooray i'm moving up and i've already got a cracking break in big dumb ape break or whatever it is i think

[snookersfun]

points for d_g and chasmmi

Here is the updated scoreboard after round 66.
I decided to give 1 point each to d_g and Chasmmi (who braved calculating under adverse circumstances). Elvaago - here is another honorable mention for understanding the question and participating (at one point I shall add these up to something useful scoreboard wise)


snookersfun……………………….…..31
abextra...............................16
davis_greatest.....................13½
Vidas..................................12½
robert602.............................6
elvaago...............................6
chasmmi..............................5
The Statman……………………..……4
Semih_Sayginer.....................2½

[davis_greatest]

Round 68 - Ape Break Madness

Would you like another Ape Break round?

I'm sure you are familiar with the rules by now. You need to find the highest break you can (without using a free ball).

This time, it's a bit different. It's like normal snooker, still with 15 reds, but now there are 4 extra colours. You have to pot red, colour, red, colour etc, just like in normal snooker, but then pot the 10 colours in order (instead of the usual six).

The colours are:

yellow: 2 points (if you can't read that, it says yellow: 2 points)
green: 3 points
brown: 4 points
blue: 5 points
pink: 6 points
black: 7 points
orange: 8 points
silver: 9 points
olive: 10 points
purple: 11 points


There are now 10 pockets (one extra pocket added on each edge of the table). From top left, going clockwise, the pockets are:

purple, brown, orange, green, pink, silver, blue, olive, yellow, black

So the table looks a bit like this - I've put the value of the colour of each pocket to help you see, in case you are colour blind.

11------4------8
!......................!
!......................!
!......................!
7.....................3
!......................!
!......................!
!......................!
2.....................6
!......................!
!......................!
!......................!
10------5------9


The rules are

a) Once a colour has been potted, the same colour cannot be potted following the next red, nor following the red after that. (Once the 15th red and colour have been potted, this rule no longer applies - the final colours may and must be potted in the usual order of yellow, green, brown, blue, pink, black, orange, silver, olive, purple, regardless of the colours potted with the final reds.)

Example 1: Red Brown Red Yellow Red Blue Red Brown IS allowed
BUT
Example 2: Red Brown Red Yellow Red Brown IS NOT

Example 3: For the 14th and 15th reds,
Red Black Red Yellow Yellow Green Brown Blue Pink Black Orange Silver Olive Purple IS allowed


b) Whenever a colour has been potted into a corner pocket, the following colour cannot be potted into any pocket that lies along the same edge - that means it may not be on the same side (left or right) of the table, nor at the same end (top or bottom).

Example A: after potting a colour into the orange pocket, it would not be permissible to pot the next colour into the orange, green, pink or silver pockets (same side), nor into the purple or brown pockets (same end).

c) Whenever a colour has been potted into a pocket that is not a corner pocket, the following colour must be potted into a corner pocket, but must not lie on the same edge.

Example B: after potting a colour into the yellow pocket, the following colour can only be potted into the orange or silver pockets.

Example C: after potting a colour into the blue pocket, the following colour can only be potted into the orange or purple pockets.

d) Rules b) and c) apply even when down to the final 10 colours after all the reds have gone.

e) None of these rules apply to reds. It makes no difference into which pockets reds are potted.

f) And this is the important bit: potting a colour into a pocket of the same colour as the ball (e.g. pink into pink pocket) scores double points (in this example 2 x 6 = 12).


As ever, your question is: what is the highest break (ignoring free balls) you can make?

You don't need to say the highest theoretically possible - you just need to send me the highest break that YOU can find by Private Message by the Initial Deadline of 18:00 GMT on Sunday 3 December.


If you have any questions, please ask them on the thread.


You should explain how you get your break. For example, you might say:

Red
Green into Yellow pocket (or whatever)
Red
Pink into Orange pocket (or whatever)
....
...

and after all 15 reds and colours...

Yellow into Yellow pocket (or whatever)
Green into Silver pocket (or whatever)
Brown into Black pocket (or whatever)
....

[davis_greatest]

snookersfun's round 66

quick update:
I received a valuable contribution at least worth a 1/2 point from Chasmmi (I love him, he understands me) for my mangled round 66. I will extend the deadline to tomorrow evening 19:00 GMT for anybody waiting for further clarification. Here is the question again, incorporating d_g's inputs :


Suppose you had a bag with 600 red snooker balls and would at each turn randomly take 15 of them, mark them and, here is the important part, return them to the bag. Picking the balls randomly (or blind) how many balls would be expected to be marked after 40 turns and how many turns would one need in average to mark 90% of the balls.
Excel or similar solutions worth 1/2 point, others 1 point. (by PM)
Deadline Tuesday-19:00 pm (GMT)

I think the deadline has now passed and snookersfun asked me to post.

First part

Expected number of balls marked after 40 turns is

600 x [1-(1-15/600)^40] = 382.06

Prooof:
If there are n balls in the bag
and each time you mark m balls
then the probability that any given ball is not marked each time is 1-m/n
and the probability that it is not marked after k turns is (1-m/n)^k
and the probability that it IS marked within k turns is 1-(1-m/n)^k

So the expected number of balls marked after k turns is n [1-(1-m/n)^k]

Put n=600, m=15, k=40 and you get the above result.

Second part

The second part of the question is harder - in fact, I couldn't immediately see a nice formula. The problem is that the expected number of turns required to mark 90% of the balls is NOT in general the same as the number of turns after which the expected proportion of balls marked is 90%.

So you cannot simply write 1-(1-m/n)^k = 90% and solve for k.

One way you can solve it numerically, without too much effort, is to find iteratively the probabilities p(j,k) of j balls being marked after k turns, for j=0,1,2,...,600 and k=0,1,2,3,...
This isn't hard.

Then the expected minimum number of turns required to mark 90% of the balls is:

sum from k=1 to infinity of (Probability that at least k turns are required to mark 90% of the balls)

= sum from k=0 to infinity of (Probability that fewer than 90% of the balls are marked after k turns)

which can be calculated using the p(j,k)


....Anyway, all that is far too hard for this thread, so let's get back to some serious Ape Break Madness....

[elvaago]

Of course. Otherwise the answer is too easy.

[snookersfun]

Elvaago, but did you take into account, that the marked balls are always returned into the bag, before 15 balls are chosen in the next turn?

[elvaago]

I'm sticking to my original PM, snookersfun, however wrong it might be! (Chance was the only exam I ever failed in high school math, so wrong is very likely!)

[snookersfun]

quick update:
I received a valuable contribution at least worth a 1/2 point from Chasmmi (I love him, he understands me) for my mangled round 66. I will extend the deadline to tomorrow evening 19:00 GMT for anybody waiting for further clarification. Here is the question again, incorporating d_g's inputs :


Suppose you had a bag with 600 red snooker balls and would at each turn randomly take 15 of them, mark them and, here is the important part, return them to the bag. Picking the balls randomly (or blind) how many balls would be expected to be marked after 40 turns and how many turns would one need in average to mark 90% of the balls.
Excel or similar solutions worth 1/2 point, others 1 point. (by PM)
Deadline Tuesday-19:00 pm (GMT)

[davis_greatest]

Points for round 67 also awarded to The Statman and to chasmmi:

HERE IS THE SCOREBOARD AFTER ROUND 65 BUT BEFORE ROUND 66 AND WITH POINTS INCLUDED SO FAR FOR ROUND 67 FOR ELVAAGO, SNOOKERSFUN, THE STATMAN AND CHASMMI

snookersfun……………………….…..31
abextra……………………………..…...16
Vidas……………………………………….12½
davis_greatest…………………..……12½
robert602…………………………………6
elvaago...............................6
The Statman……………………..……4
chasmmi ………………………………….4
Semih_Sayginer.....................2½