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#337738 How big was the dinner table?

Posted by bonanova on 08 March 2014 - 10:33 AM

At a family party, a grandfather, a grandmother, two fathers, two mothers, four children, three grandchildren, one brother, two sisters, two sons, two daughters, one father-in-law, one mother-in-law, and one daughter-in-law, sit at a table. What is the fewest number of chairs required to seat all of them? Each had his/her own chair.

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#337737 Dueling cards

Posted by bonanova on 08 March 2014 - 06:26 AM

Moe and Joe had some time on their hands, so they played a simple game called "Dueling Cards."


In this game, each player has a deck of cards, shuffled and without jokers.

A hand consists of a single card taken from the top of each player's deck.

Suits and face values are both ranked. No two cards in a deck have the same rank.

A player wins the hand by holding a card that outranks the card of his opponent.

The bank pays 1 chip to the winner of a hand.


It's a simple game. No poker hands. No betting. Just high card wins.

After 52 hands the decks are re-shuffled, and play continues as long as desired.


Moe and Joe played through their decks 10 times, then stopped to see who won.


As they counted their chips Moe (a statistician) said, I wonder what the most likely winning score is.

Joe (a simpler person) replied, I wonder what the most likely combined score is.


I'm sure Bushindo can answer Moe's question.


I can't. So instead this puzzle asks Joe's question.

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#337727 Orderly numbers

Posted by bonanova on 07 March 2014 - 03:00 PM

Spoiler for Hint

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#337617 Triangles inside circles

Posted by bonanova on 28 February 2014 - 12:18 AM

A few puzzles posted in this forum have related to random triangles inside a circle.


By evaluating nasty integrals, or by my preferred method, simulation, it can be shown, perhaps surprisingly, that triangles constructed from sets of three uniformly chosen points within a circle cover only about 7.388% of the circle's area on average. After looking at Phil's recent problem on the subject, I simulated 1 million triangles to determine the median area. It turns out to be about 5.335% of the circle's area. Read: a random triangle has a 50% chance of being smaller.


If the distribution of random-triangle areas has a mean of about 7.4% and a median of about 5.3%, what value might you expect for the mode?

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#337573 Orderly numbers

Posted by bonanova on 27 February 2014 - 08:48 AM

Consider the numbers from one to one million: 1, 2, 3, ..., 999998, 999999, 1000000.

What is remarkable about the numbers 40, 8, and 2202?

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#337517 The kindest (shortest) cut of all

Posted by bonanova on 25 February 2014 - 10:39 PM

Here is a piece of plywood in the shape of an isosceles triangle.

The side lengths are 1, 1, sqrt(2) units.


Quick and dirty representation:



|  \

|     \

|        \

|           \       



The angle at B is a right angle.


We'd like to cut this into two pieces of equal area.

There are many ways to do this with a single cut.

Which cut has the shortest distance?

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#337512 Making 271

Posted by bonanova on 25 February 2014 - 09:13 PM

Is this with regards or without regards to orderings of the summands?


I would guess "ways" means combinations.

The permutations of 271 1's are not that interesting.

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#337443 Binary tic-tac-toe

Posted by bonanova on 20 February 2014 - 11:07 AM

In the all-digital future, X and O are banished from the game of tic-tac-toe.

They are replaced by 1 and 0, the the result of such a game might look like this:


1 | 0 | 1

- + - + -

0 | 1 | 1

- + - + -

0 | 1 | 0


Under the usual rules that require getting 3-in-a-row, it would be a draw.

But this is the digital age, and there are different rules for winning.


If we sum the eight rows of three numbers we get 2, 2, 1 (horizontally) 1, 2, 2 (vertically) and 2, 2 (diagonally).

Six of the sums are even, and two are odd.

The final parity of the board is thus even, and the game is said to have an even outcome.

If there were more odd sums than even, the game would have an odd outcome.

If there were four even (and therefore four odd) sums, the game would have a neutral outcome.


The game is played as follows:


The winner of a fair-coin toss (call him player A) chooses whether to play first or second.

The other player (call her player B) decides whether she wants an odd, even, or neutral game outcome.


On each turn, a player places his choice of either a 1 or a 0 on any unoccupied place on the grid.

As in normal tic-tac-toe, players alternate turns; but here on each turn a player may play either a 0 or a 1.

When the places are filled, the board is examined to determine whether it is odd, even or neutral.


If the final board parity matches player B's choice, player B wins; otherwise player A wins.


The questions to answer are:

  1. Is there an advantage to winning the coin toss?
  2. Is there a winning strategy for either player?

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#337442 The clock in the mirror

Posted by bonanova on 20 February 2014 - 10:07 AM

A while ago, BMAD posted a problem about a clock with indistinguishable hour and minute hands and asked at what times of day, between the hours of noon and midnight, it was impossible to unambiguously determine the time. The hands moved continuously.


This puzzle asks a related question. At what times of day, between the hours of noon and midnight, is it impossible to distinguish the hands of such a clock from those of its mirror image? Clearly noon is one of these times, but not in general thereafter -- since the clock's hands will move clockwise while the hands of its mirror image will move counterclockwise.

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#337283 Betting on red

Posted by bonanova on 30 January 2014 - 02:46 PM

Here's  a variation on a previous puzzle that let you make multiple bets while choosing the color of the next card.


I shuffle an ordinary deck of playing cards and then turn over the top card sequentially so that you can see it. At any time you may ask me to stop and place a $1 bet that the next card to be exposed will be red. If you never ask me to stop, you will automatically bet on the last card. To summarize:

  1. You can bet on only one card.
  2. You don't get to choose the color.
  3. You must bet the card will be red.

What is your best strategy?

How much better than even can you do?

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#337125 Second 2014 puzzle. Dissection of a square

Posted by bonanova on 01 January 2014 - 11:13 PM

Place a point P at coordinates (6, 7) in a square with diagonal vertices (0, 0) and (12, 12).
From P draw lines to the vertices and perpendiculars to the sides.
This defines eight triangles that meet at P.

Ignoring permutation of identical pieces, how many other ways can these triangles form a square?

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#337062 Word play. Because they're fun to play with

Posted by bonanova on 21 December 2013 - 11:55 PM

If progress is moving forward, then what is congress?
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#337061 Pun alert - II. Somebody stop me!

Posted by bonanova on 21 December 2013 - 11:49 PM

Moe: Did you hear about the actress who stabbed her boyfriend?
Joe: No, I didn't.
Moe: Yeah, it was, uh, Reese ...
Joe: Witherspoon?
Moe: No, with a knife.

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#337010 Infinite Flips

Posted by bonanova on 16 December 2013 - 01:02 AM

Agree. Better simply to say simply that in any large ensemble of N outcomes, where 60% will be H, the outcomes can be partitioned into groups that are totally fair (H=T) and totally biased (all H.). The fair partition will be N x 2 p(T) in size.
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#333964 3 x 4 Connect-The-Dots Game

Posted by bonanova on 08 July 2013 - 05:57 PM

I don't think that works. We are saying that the green curve below, connecting two green dots,by virtue of sharing a single point (touching but not crossing)isolates the lower red dot from the upper red dot. (The green characters do not actually touch, they can't, but they represent touching lines.) adO|fsO a\|  fs/ sa\e| / asd\|/ asd/|\ as/d|f\ s/ad|fa\ /cxv|cxz\ \cxvO|cz/ a\s/|f\/ s/\a|f/ /cx\_/  Touching and crossing cannot be distinguished, once the line has been drawn.It's exactly the same locus of points. But let's say it's all about how the lineswere drawn, not how they end up. a line cannot be drawn in a manner bywhich an existing line is crossed by the pencil. The "touching but not crossing" line still does not isolate the red dots. The red dots can still be connected by a red line that also sharesthe common touching point without crossing either the left part or the right part of the green line. Thus: adO|fsO a\ds|fs/  sa\e| / asd\|/ asd/|\ as/d|f\ s/ad|fa\ /cxv|cxz\ \cxvO|cz/ a\s/|f\/ s/\a|f/ /cx\_/ So I still contend you can isolate a portion of the plane, for the purposes of this game,but only by using a line segment of infinite extent or by crossing an existing line.

This debate about touching vs crossing was the heated debate for which I spoke of earlier :)The difference in this problem's contest is subtle. Think of a drawn line as a physical object with defined spaced (which is key to understanding how my homeland uses the words). If one draws a line and touches a line they simply stood in the other line's space like standing on a doorstep (the drawn line) to knock on a door and left if no one answers, while If one draws a line that crosses the line they knocked on the door and went inside.So I think key to solving this problem is understanding that it treats a line segment as an object with physical properties, area, length, etc. and not as a cutting tool.I believe, one of my friends best described this idea as an old Tron game where two people cut off the board drawing segments in motorbikes.

The red line just stands on the doorstep as well.
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