Puzzles with numbers and things

[chasmmi]

is it anything like 4/81?

[davis_greatest]

is it anything like 4/81?

Um.... not really....sorry

[elvaago]

For Ronald to concede the match, Michael needs to win two frames in a row. The chance of that is 1/9th.
This can happen at any moment, after 2 frames, but also after a billion frames.
If it happens after 2 frames, the chance is 1/9th.
If it happens after 3 frames, Ronald would have to win one frame, chance is 2/3rd, and Michael needs to win 3. chance is 1/27th.
So there is a formula. The chance for X frames is 2/3rd^(x-2) * 1/3rd^x
And since you have to add up all the chances you need to take the limit of x = 1 to infinity.

My guess is: 2/63

[davis_greatest]

For Ronald to concede the match, Michael needs to win two frames in a row. The chance of that is 1/9th.
This can happen at any moment, after 2 frames, but also after a billion frames.
If it happens after 2 frames, the chance is 1/9th.
If it happens after 3 frames, Ronald would have to win one frame, chance is 2/3rd, and Michael needs to win 3. chance is 1/27th.
So there is a formula. The chance for X frames is 2/3rd^(x-2) * 1/3rd^x
And since you have to add up all the chances you need to take the limit of x = 1 to infinity.

My guess is: 2/63

Interesting idea... I'm afraid not right though. Sorry Ronald is more likely to concede than that. This can be done without needing to look at any infinite sums

[davis_greatest]

For Ronald to concede the match, Michael needs to win two frames in a row.

Hey... be careful. Ronald will concede if he is 2 frames behind.... but will not necessarily concede if Michael has just won 2 frames in a row.

Example: If Ronald wins the first frame, and then Michael wins two frames in a row, then Michael will be 2-1 ahead. Ronald would not concede at that point. (He would concede if Michael also won the next frame to lead 3-1.)

Your statement though is strictly correct. If Ronald concedes, then Michael must have won the 2 previous frames (and, as per the example above, possibly as many as 3).

[davis_greatest]

Round... I always forget. 83?

Here's a neat one. Complete the following five series of numbers. Answers and explanations by PM only please! Some of them are easy, some of them are hard. If you get all 5, you get a point, if you get 4, you get half a point, 3 or
less, sorry! If no one gets them all before Monday 7 PM CET, I get a point. :-)
1) 1 - 8 - 9 - 16 - 17 - ???
2) 1 - 4 - 9 - 16 - 25 - ???
3) 1 - 2 - 3 - 5 - 8 - ???
4) 1 - 2 - 3 - 7 - 16 - ???
5) 3 - 3 - 1 - 2 - 4 - ???

Has anyone got number 5, elvaago? If not, I think we need a clue.

[snookersfun]

Hey... be careful. Ronald will concede if he is 2 frames behind.... but will not necessarily concede if Michael has just won 2 frames in a row.

Example: If Ronald wins the first frame, and then Michael wins two frames in a row, then Michael will be 2-1 ahead. Ronald would not concede at that point. (He would concede if Michael also won the next frame to lead 3-1.)

Your statement though is strictly correct. If Ronald concedes, then Michael must have won the 2 previous frames (and, as per the example above, possibly as many as 3).

I didn't really wanted to get into that again (showing my ignorance about probabilities). But anyway, how about 4/27?
1/9 the probability for Ron loosing 2 matches in a row and 1/27 for 3 matches in a row.

[davis_greatest]

I didn't really wanted to get into that again (showing my ignorance about probabilities). But anyway, how about 4/27?
1/9 the probability for Ron loosing 2 matches in a row and 1/27 for 3 matches in a row.

Sadly, no. The solution is slightly more complicated than that... but the answer is much simpler than 4/27!

[chasmmi]

is it just 1/3?

[davis_greatest]

is it just 1/3?

No - but a nice try.

[snookersfun]

I am afraid, I am getting stuck at infinite sums again

let me just think aloud here:
first possibility to loose two games is 1/3x1/3=1/9
but there are now much more arrangements possible, each with two loosing games more than winning games,
e.g. 1 win 3 losses: 2 arrangements: wlll, lwll and I suppose the probability for that should be then 2*2/3*(1/3)^3 = 4/81
2 wins, 4 losses: 5 arrangements: wwllll, wlwlll, wllwll, lwlwll, lwwlll = 5*(2/3)^2*(1/3)4 = 20/729
3 wins, 4 losses......

I have a feeling all these will add up to 1/10 (max 1/9th???)

so allover 1/9th +1/10th ???, nah, rather 1/9 +1/9 =2/9

[davis_greatest]

Well... let me see... I suppose 2/9 isn't a million miles away...

but still no correct answers.

[chasmmi]

my last attempt for the day: 1/4?

[davis_greatest]

my last attempt for the day: 1/4?

Ooooooh.................























no.

[snookersfun]

1/5 is close as well