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## #1 Re: Puzzles and Games » Post more bigger numbers » 2013-03-05 13:39:23

I present to you...

I prefer an N.

## #3 Re: Puzzles and Games » New Puzzles 4 » 2013-03-05 13:27:40

Ah, thanks for reminding me, my simple less-rigorous proof isn't adequate if i'm only proving the number of solutions, rather than the value. Removing it now. Hope the more complex proof doesn't make anyone's eyes glaze over.

## #4 Re: Puzzles and Games » New Puzzles 4 » 2013-03-04 05:23:07

anonimnystefy wrote:

Hi Thurhame

Can you specify where it was claimed that 1+1/(1+1/(1+...)) is equal to 2?

Oh, my mistake, it just said there were two solutions to the formula. However, that's still wrong; my proof shows that any solution must be equal to the golden ratio, i.e. there are at most 1 solutions.

Edited my post above. Thanks!

## #5 Re: Puzzles and Games » Probability of a Boy » 2013-03-04 01:33:08

MathsIsFun wrote:

"I tell you I have two children and that (at least) one of them is a boy, ... your right answer to my question is not 1/2 but 1/3."  Agree!

"I tell you I have two children, and (at least) one of them is a boy born on a Tuesday. What probability should you assign to the event that I have two boys?" he gives the answer 13/27  ... Really?

Assume that boy/girl are equally likely and each day of the week is equally likely for each child.

Case 1: Both children were born on the same day of the week (1/7).
This day could be Tuesday (1/7) or NotTuesday (6/7).
Genders could be BB (1/4), BG (2/4), GG (1/4).

Case 2: The children were born on different days of the week (6/7).
Case 2a: Exactly one of the children was born on a Tuesday ((6 choose 1)/(7 choose 2) = 2/7).
Gender of the Tuesday child could be B (1/2) or G (1/2).
Gender of the other child could be B (1/2) or G (1/2).
Case 2b: Neither child was born on a Tuesday ((6 choose 2)/(7 choose 2) = 5/7).
Genders could be BB (1/4), BG (2/4), GG (1/4).

This is the sample set. Applying the restriction "one of the children is a boy born on a Tuesday" leaves us with
3/196 from Case 1, including 1/196 that both are boys.
24/196 from Case 2, including 12/196 that both are boys.
The probability of both being boys given the restriction is thus (1/196 + 12/196)/(3/196 + 24/196) = 13/27.

To put it another way, saying the boy mentioned was born on a Tuesday has a 12/13 chance of identifying the boy mentioned (1/2 chance of the other being a boy), and a 1/13 chance of leaving the boy unidentified (1/3 chance of both being boys).

## #6 Re: Puzzles and Games » New Puzzles 4 » 2013-03-04 00:04:21

There is a problem with the "Three Of The Best" puzzle.
The puzzle claims there are two solutions to "1+(1/(1+(1/...)))". However, it actually converges to the golden ratio

, meaning there is only one solution.

MORE COMPLEX PROOF:
The formula is the limit as n goes to infinity of the sequence

defined by
.
Assume some
exists (otherwise the sequence would be undefined).
Define a sequence
.
We proceed by induction.
Base step:

Inductive Hypothesis:

Thus we have proved by induction the relation
for all
.
Since
is a sequence of the form
, this ratio converges to the golden ratio.