Puzzles with numbers and things

[davis_greatest]

Actually, chasmmi, your first answer is correct! Well done!

Since the ape who started has neither advantage nor disadvantage (each ape being equally likely to win the frame), his chance of potting the black to win must be the same as his chance of fouling to lose.

[davis_greatest]

Round 199 - Hold that black!

Just before Gordon plays his shot in round 198, I realise that I had my probabilities wrong - the two apes were not equally likely to win.

In fact, each ape really has a 20% chance of potting the black on any shot and a 10% chance of fouling (and therefore a 70% chance of his oppponent coming back to the table).

So, now what is Gordon's chance of winning the frame?

Answers / guesses on the thread please!

[Monique]

R200 Improving human snooker ... follow up

Having gathered the necessary amount of bananeuros Howardus has indeed organised his improvement classes in Crete, a place that suits both humans - it has sea, sun and wine - and apes - it produces good bananas and a wealth of other fruit. Howardus invited all the 32 top human players to attend, as well as Jess Vanraas and Tallia Mabb the referees. Ebert Petdon however declined: he's got enough practice with his good mate Harald Dove ... so he said anyway.
After a month of intensive coaching, the coaches (all apes) were quite disapointed: no noticable progress had been made by these humans, at least by apes' standards. The humans however were quite pleased. Rollie O'Sunnyman and Jess Vanrass had been practising together and have succeeded in compleeting a maximum in less that five minutes on several occasions in practice. Davy Steeves was seduced by Charlie's colorfull methods to keep scores, and now wants to make this the standard way in snooker plus! Damon Grott did not get any slaps for a full month and thinks much of the apes, these lovely creatures (he did however slip on countless banana skins ... strange!). Jasper Rott who attended as the BBC pundit, has now found a new passion: Oliver's mathematical puzzles are a lot harder to solve over breakfeast than those stupid crosswords of The Times. And Tallia has got the most sexy of tans (and a remarquable collection of bikinis!)
Well, everything comes to an end. Now the apes have organised a snooker tournament to close the event; only the human players were to participate. Oliver, Gordon and Charlie have designed a very original format ... The players had been arranged in teams, each team playing as a league (each player playing each other player in the league once, in a single frame match). Each player participated in several leagues. The way the apes did it, it happens that at the end of the tournament each player had in fact played each other player just once. Also the audience was very pleased: it was possible to watch all the matches as each league played on a different day. It would have been difficult to do otherwise anyway: taken two leagues - whatever the choice - there was always one player that was supposed to play in both!

How long did the tournament last?
How many matches were played every day?
Barry The Baboon came to watch on the first and last day: how many different players did he see in action?

Highlight: Rollie made a maximum in 4'48" on his last match (He was playing Robert Nilson the Aussie). Jess was refereeing. However the WSA (the human World Snooker Asocciation) refused to consider it as official, because both Rollie and Jess were wearing runners ...

Answers initially by PM please ...

[Monique]

R199

I' d say 53% ...
Here is why: each ape has the same chances of winning when at the table because both have the same probability of potting and fouling. However the one who is at the table has an advantage ...

Gordon being at the table

P(win) = 0.2 (potting) + 0.7*P(Charlie does not win - Charlie at the table)
P(win) =0.2 + 0.7(1-P(win))
P(win) = 0.9/1.7 = 0.53 (approximatively)

[davis_greatest]

I' d say 53% ...
Here is why: each ape has the same chances of winning when at the table because both have the same probability of potting and fouling. However the one who is at the table has an advantage ...

Gordon being at the table

P(win) = 0.2 (potting) + 0.7*P(Charlie does not win - Charlie at the table)
P(win) =0.2 + 0.7(1-P(win))
P(win) = 0.9/1.7 = 0.53 (approximatively)

Congratulations, Monique! 9/17 is the correct answer!

Or, in terminology that Willie Thorne would understand, Gordon's chances are 9 to 8 on, and Charlie's are 9 to 8 against.

[Monique]

R200

Correct answer already entered by D_G! Congratulations!

[Monique]

R200 Improving human snooker ... follow up

And abextra also. Well done

Next answers on the thread please.

[davis_greatest]

While we await someone to post an answer to round 200...

Round 201 – Snooker Ping Pong

I am watching the quarter finals of the World Snooker Championships in the Crucible Theatre, with a nice central seat giving me an excellent view of both tables. I am enjoying the session, for Steve Davis on table 1, and Stephen Hendry on table 2, are both in the process of making total clearances (15 reds, each with a colour, followed by the 6 colours). Funnily enough, they are playing at the same pace, but Davis is about 2 seconds ahead of Hendry (so Davis pots his first red, then Hendry his, then Davis his first colour, then Hendry his, etc.)

I am keeping an eye on the breaks, looking back from one table to the other, and I notice a couple of funny things:

* From the moment that Davis has potted his first colour, Davis’s and Hendry’s breaks are at no point equal.

* Also, every time that Davis has just potted a ball, the difference between Davis’s and Hendry’s breaks is always a square number (e.g. 1, 4, 9, 16, 25,…) – and this continues right up to (and including) the point when Davis has potted the final yellow, immediately after Hendry has potted a black following his 15th red.

Give an example of the colours that Davis and Hendry could have potted with each of their reds in compiling their breaks.

Answers initially by Private Message please...

[abextra]

Very good, d_g, nice to hear about Steve and Stephen again!

I wonder if stupid questions are allowed... One of them is in hidden text below (in case you don't want to see any ).

Is there any possibility that Davis's break was higher than Hendry's one?

[davis_greatest]

...Is there any possibility that Davis's break was higher than Hendry's one?

Hendry's break was higher than Davis's when I was watching!

But any possibility that Davis's could be higher?... Er, I think probably not, but haven't thought about it in enough detail!

[davis_greatest]

Round 201 - there are many possible answers to this - congratulations to Monique, abextra and chasmmi who have all found some of them!

Next answer on the thread please...

Then we'll get on with round 202.

[snookersfun]

Round 201 – Snooker Ping Pong

I am watching the quarter finals of the World Snooker Championships in the Crucible Theatre, with a nice central seat giving me an excellent view of both tables. I am enjoying the session, for Steve Davis on table 1, and Stephen Hendry on table 2, are both in the process of making total clearances (15 reds, each with a colour, followed by the 6 colours). Funnily enough, they are playing at the same pace, but Davis is about 2 seconds ahead of Hendry (so Davis pots his first red, then Hendry his, then Davis his first colour, then Hendry his, etc.)

I am keeping an eye on the breaks, looking back from one table to the other, and I notice a couple of funny things:

* From the moment that Davis has potted his first colour, Davis’s and Hendry’s breaks are at no point equal.

* Also, every time that Davis has just potted a ball, the difference between Davis’s and Hendry’s breaks is always a square number (e.g. 1, 4, 9, 16, 25,…) – and this continues right up to (and including) the point when Davis has potted the final yellow, immediately after Hendry has potted a black following his 15th red.

Give an example of the colours that Davis and Hendry could have potted with each of their reds in compiling their breaks.

Answers initially by Private Message please...

Ok, I went for lots of pinks in the end, just because of the 'at no point equal' statement (is that in time, after # of balls or what?):

Davis starts out with red brown and takes all the other reds with pinks, while Hendry takes all his reds with pinks except the last one with a black.
So the scores after each of Davis' balls are:
1-0 -1st red
5-1
6-7 -2nd red
oh, that is already cumbersome... never mind, I'll go on:
12-8
13-14 -3rd
19-15
20-21 -4th
26-22
27-28-5th...
...
until
103-99 after the pink with the 15th red
105-106 after Davis' yellow and Hendry's black.

Hopefully

[davis_greatest]

Ok, I went for lots of pinks in the end, just because of the 'at no point equal' statement (is that in time, after # of balls or what?):

It means at no point in time may the breaks be equal in value. The answer above is almost there, but unfortunately can't quite be accepted because when Davis pots the final green, he would then be on the same score as Hendry (108 points each, Hendry having potted yellow but not yet the green, being 2 seconds behind Davis), which is not allowed. The values of the breaks must never be the same at any time during the frames (after the first colour has been potted).

So, just a small tweak needed, I think...

[abextra]

Hehehe... I made the same mistake, didn't check the scores while potting the final colours...

[davis_greatest]

I think snookersfun is giving up! Time to close round 201. abextra, please would you post your picture answer here...