Puzzles with numbers and things

[davis_greatest]

...2 wins, 4 losses: 5 arrangements: wwllll, wlwlll, wllwll, lwlwll, lwwlll ...

I think I make this one 4 arrangements, but I wouldn't recommend trying to solve the problem this way - you will find it very difficult with this approach.

PS Your first arrangement wwllll would have resulted in Michael conceding after the first two frames.

[davis_greatest]

1/5 is close as well

Yes, it is! And it is the right answer! ... but I'm only going to award the point if you can explain why.

(And, of course, if you can't - the point will then be opened up to anyone who can. But I'm sure you will explain delightfully )

[snookersfun]

I think I make this one 4 arrangements, but I wouldn't recommend trying to solve the problem this way - you will find it very difficult with this approach.
I know
PS Your first arrangement wwlll would have resulted in Michael conceding after the first two frames.

Ahhhhhhh (spotlight on), forgot one tiny little vital part . I had assumed that only Ron is conceding...

[elvaago]

I will give a clue for #5.

The clue is in the title of this thread.

Puzzles with numbers and things.

Maybe it's not merely a natural progression of numbers, but of things that are not numbers, but can be translated into numbers. Y'all have until tomorrow. (PS, I have seen this puzzle elsewhere before on an IQ quiz. I didn't get it back then, either.)

[snookersfun]

Yes, it is! And it is the right answer! ... but I'm only going to award the point if you can explain why.

(And, of course, if you can't - the point will then be opened up to anyone who can. But I'm sure you will explain delightfully )

ermmm, except that it fits the 'progression' above, no explanation yet.
Watching the snooker, so good night! If anybody has a good explanation, shoot

and Elvaago, thanks so much for the hint. Now I started twisting my brain again

[davis_greatest]

Ahhhhhhh (spotlight on), forgot one tiny little vital part . I had assumed that only Ron is conceding...

Hehe - that would have been quite a silly question if only Ronald were able to concede - because then the chance that Ronald is the one to concede would be 100%! The question says "if either player ever falls two frames behind his opponent, he will concede..."

[snookersfun]

Hehe - that would have been quite a silly question if only Ronald were able to concede - because then the chance that Ronald is the one to concede would be 100%! The question says "if either player ever falls two frames behind his opponent, he will concede..."

I guess it only slipped to the back of my mind.

But more important, here are some calculations: they are in need of more words though, I admit.

1/3 x 1/3 + 2/3 x 2/3 x Prob

= 1/9 + 4/9 x Prob


So 5/9 x Prob = 1/9

So Prob = 1/5

[davis_greatest]

I guess it only slipped to the back of my mind.

But more important, here are some calculations: they are in need of more words though, I admit.

1/3 x 1/3 + 2/3 x 2/3 x Prob

= 1/9 + 4/9 x Prob


So 5/9 x Prob = 1/9

So Prob = 1/5

Yes! You need to put "Prob = " at the start, of course!

Just explain please the "2/3 x 2/3 x Prob" part for the point.

[snookersfun]

Yes! You need to put "Prob = " at the start, of course!

Just explain please the "2/3 x 2/3 x Prob" part for the point.

I'll try...
P(giving up) = P(loosing the first two frames) + P(loosing later on)

= 1/3*1/3+ P(winning first or winning second game and then getting into the 2 frame loss situation)

=1/9 + 2/3*2/3*P(giving up)

[davis_greatest]

I'll try...
P(giving up) = P(loosing the first two frames) + P(loosing later on)

= 1/3*1/3+ P(winning first or winning second game and then getting into the 2 frame loss situation)

=1/9 + 2/3*2/3*P(giving up)

I'll give you the point, snookersfun. I don't understand what you mean by "winning first or winning second game and then getting into the 2 frame loss situation" but the calculation looks OK - as long as your 2/3 * 2/3 is really 2 x 1/3 x 2/3 which comes to the same thing.

I'll fill in some of the words soon...

[elvaago]

While I've gotten several correct /answers/ to question number 1 of puzzle 83, I'm disappointed that a forum full of snooker fans hasn't gotten more creative with the explanation, apart from a single individual!

[davis_greatest]

Here are the missing words for snookersfun's answer to round 84!

OK. Let’s say that:
Ronald wins each frame with probability R = 2/3
Michael wins each frame with probability M = 1/3

Then consider the possible situations after the first 2 frames. There are 3 possibilities:

A) Michael wins both frames, so Ronald concedes. Probability M x M = 1/3 x 1/3 = 1/9

B) Ronald wins both frames, so Michael concedes. Probability R x R = 2/3 x 2/3 = 4/9

C) They win one each, so the scores are level and we are back where we started.
Probability 1 – M x M - R x R = 4/9 (you can also write this as 2 M x R)

Probability(Ronald concedes) = A + C x Probability(Ronald concedes)
= 1/9 + 4/9 x Probability(Ronald concedes)

So, 5/9 x Probability(Ronald concedes) = 1/9

So, Probability(Ronald concedes) = 1/5


In fact,
Probability(Ronald concedes) = M^2 / (M^2+R^2)
Probability(Michael concedes) = R^2 / (M^2+R^2)

[davis_greatest]

SO HERE IS THE SCOREBOARD AFTER ROUND 84 BUT BEFORE POINTS FOR ROUND 83

snookersfun……………………….…..41½
abextra...............................23
davis_greatest.....................18½
Vidas..................................12½
elvaago...............................8
chasmmi..............................8
Sarmu.................................7
robert602.............................6
The Statman……………………...……5
Semih_Sayginer.....................2½
austrian_girl and her dad.........2½
April Madness........................1

[davis_greatest]

Round 85 - Sharing out the balls

You are organising a game at Oliver's party. 27 of the guests have gone home, having had enough cake, so there are 10 apes left to play.

You need to take some snooker colours - as many as you want, but no reds, and there must be at least one yellow, one green, one brown, one blue, one pink and one black (so there can be more than one of each).

Then you must share the balls out between just two apes (Oliver and Gordon), so each has balls of the same total value.

For example, you might have one yellow, two greens, one brown, one blue, one pink and one black, and split them as follows:

Oliver: one yellow, one pink, one black - total value 2+6+7 = 15
Gordon: two greens, one brown, one blue - total value 3+3+4+5 = 15.


Then, you take some colours again (not necessarily the same ones as before), and must share them among 3 apes, so each has balls of the same total value.

Then again, but sharing among 4 apes, then 5, 6, 7, 8, 9 and finally among 10 apes. Each time, you take a new set of balls. So, in total, you must distribute 9 sets of balls.


You need to try to do this using as few balls, in total, as possible. Make your bids here, of the fewest total number of balls you think you can use (e.g. just bid "100" or whatever).

Once there has been enough time for the bids to come in, you will then be asked which balls you chose for each round and how they were distributed. But you should not give that information yet.

[davis_greatest]

I'll put chasmmi's questions here and my answers, so everyone gets the same information.

"A few questions:

Can the total value cahnge from set to set or does it have to be say 15 every single time?"

Yes, the total value can change every time. Each time you need to take a new set of balls (the only requirement is that each time only colours are allowed, and there must be at least one of each colour).


"Is there anything to stop me just giving say all six colours to every ape every time?"

Nothing to stop you doing that. It probably wouldn't result in the fewest balls though.