Puzzles with numbers and things

[The Statman]

ROUND 54:

You are invited to a party.

You are the first to arrive, so at that time there is just you and the host. Gradually, and individually, other invitees turn up.

At what point does the chance of any two people at the party sharing a birthday reach 50%?

By which I mean, when the n-1th person arrives the chacne is less than 50%, and when the nth person arrives, the chance has reached/exceeded 50%.

[elvaago]

You know, my original answer was 0.2 + 4/11, which is almost correct. I won't feel so bad about this one. :-)

[davis_greatest]

Question 4, on the first page of this thread, I had asked on the BBC website and you had answered it correctly. On Sunday night (message 632, http://www.thesnookerforum.com/showt...2-page_64.html) I decided that it was right that you should have a point for it to be included here!

[The Statman]

A slight query which you might wish to investigate.

I appear now to have three points but I am sure I have only answered, let alone correctly answered, two questions and I doubt either was worth more than a point!

Otherwise, I am happy to be on 3!

[davis_greatest]

Congratulations, The Statman...

27/55 is the correct answer! There are many ways to solve this, yours being just as good as any of them!


SO HERE IS THE SCOREBOARD AFTER ROUND 53

snookersfun……………………….…..21½
abextra……………………………..…...11
Vidas……………………………………….10½
davis_greatest…………………..……6
robert602…………………………………6
elvaago...............................4
The Statman……………………..……3
Semih_Sayginer.....................1½

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)

[The Statman]

Let's try again

Charlie comes along, and has ten positions to choose from, two of which would give two apes in a row: That is 2 in 10. Then, no matter where Gordon sits, the criteria will be met.

There is also an 8 in 10 chance that Charlie doesn't. Then Gordon comes along. He will have 11 positions to choose from, 4 of which will result in two apes in a row. That's 4 in 11 0.363636etc.

So I think the answer is:
(0.2×1)+(0.8×0.363636etc.)

Which is 0.490909090909090etc.

Or, if you prefer it in fractions, twenty-seven fifty-fifths!

[davis_greatest]

....Then Gordon comes along. He will have 22 positions to choose from...

22?

[davis_greatest]

the chance has to be bigger than that, I would guess close to 1/2 (might be utter nonsense as well)

Assuming you are answering the original question and not elvaago's post about 3 apes sitting in a line then, yes, it is very close to 1/2. That should help!

[davis_greatest]

And before you ask, the chances of 3 apes sitting in a row are 0.2 * 3/11 = 0.0545454.

Yes, it is - but I wasn't going to ask that

[davis_greatest]

It's early morning. Time for elvaago to make an idiot out of himself again with his feeble math skills!

There's already one ape sitting in the circle. There are 10 people in the circle, so there's 10 available spots for the second ape to sit. Sitting on the left or on the right of Oliver means there's two apes in a row, so the chance of the first ape joining sitting next to Oliver is 2 in 10 or 1 in 5 or 0.2.

If the first ape that joins isn't sitting next to Oliver, then there's two apes sitting in the circle. This makes the number of available spots for two apes next to eachother four, with eleven seating spots available. That's a chance of 4 out of 11 or 0.363636.

So I guess the chance is 1 out of 5 * 4 out of 11 = 0.2 * 0.363636 = 0.0727272?

Afraid not, elvaago. Also, you say there are 10 people in the circle There are 9 people and 3 apes...

[The Statman]

Recall that in round 52, Oliver, my pet orang utan was sitting in a circle during his birthday party with 9 of his friends from this forum. For the next game, Charlie the chimpanzee and Gordon the gorilla decide to join in, so they each squeeze in somewhere into the circle, their positions chosen at random. (So there are now 12 in the circle.)

What is the chance that at least two apes are sitting next to each other? (not counting forum members as apes)

Charlie comes along, and has ten positions to choose from, two of which would give two apes in a row: That is 2 in 10. Then, no matter where Gordon sits, the criteria will be met.

There is also an 8 in 10 chance that Charlie doesn't. Then Gordon comes along. He will have 22 positions to choose from, 4 of which will result in two apes in a row. That's 4 in 22 or 2 in 11 or 0.181818etc.

So I think the answer is:
(0.2×1)+(0.8×0.181818etc.)

Which is 0.345454545454545454545etc.

Or, if you prefer it in fractions, nineteen fifty-fifths!

[elvaago]

I think I see what you are saying. But Oliver can easily have 0 balls at some point. For example, snookersfun may start by passing balls anticlockwise (3 to robert602, 2 to April Madness, 1 to Oliver). Then Oliver will have 12 balls. He can then pass 6 clockwise (3 to April Madness, 2 to robert602, 1 to snookersfun), and then pass 6 balls again, and he'll have none left!

True, but I thought everyone went in turn and in sequence, which was my mistake. I don't like this chaos that you propose!

[snookersfun]

the chance has to be bigger than that, I would guess close to 1/2 (might be utter nonsense as well)

[elvaago]

And before you ask, the chances of 3 apes sitting in a row are 0.2 * 3/11 = 0.0545454.

[elvaago]

It's early morning. Time for elvaago to make an idiot out of himself again with his feeble math skills!

There's already one ape sitting in the circle. There are 10 people in the circle, so there's 10 available spots for the second ape to sit. Sitting on the left or on the right of Oliver means there's two apes in a row, so the chance of the first ape joining sitting next to Oliver is 2 in 10 or 1 in 5 or 0.2.

If the first ape that joins isn't sitting next to Oliver, then there's two apes sitting in the circle. This makes the number of available spots for two apes next to eachother four, with eleven seating spots available. That's a chance of 4 out of 11 or 0.363636.

So I guess the chance is 1 out of 5 * 4 out of 11 = 0.2 * 0.363636 = 0.0727272?