Maths problems (interesting ones)

[FizMan]

I'll start thinking aboot the coin one, but right now with the fruit, I'm pretty sure you take just two pieces of fruit.

Each guard you give him half (1/2 of two fruit is 1 piece) then he gives you one piece back (back up to two). You never lose any fruit.

[Kris Verdeyen]

Look at it like this: there are two children, one who opens the door, and one who doesn't. Before she opens the door, the possibilities are:

Door Opener / Non Door opener

Boy / Girl
Boy / Boy

Girl / Girl
Girl / Boy

Once she opens the door, the first two possibilities are eliminated, leaving only the options that the shy child is either a Boy or a girl. In other words, the sex of the one child has nothing to do with the sex of the other.

Prove me wrong.


EDIT:

Even better, look at it like this:

FAMILIES WITH TWO CHILDREN:
25% have two boys - A girl will answer the door 0% of the time
50% have a boy and a girl - A girl will answer 50% of the time
25% have two girls - a girl will answer 100% of the time

If we take the product of the two percentages in each row, we get some more numbers to cloud the issue:

PROBABLILITY THAT A GIRL WILL ANSWER...
...and the family has two boys - 0%
...and there's a boy and a girl - 25%
...and the family has two girls - 25%

So, a girl will answer (0% + 25% + 25%) = 50% of the time.
Exactly half of those responses will come from families with two girls, and exactly half will come from families with a boy and a girl.

Or I could be completely off base - let me know.

[Astronouth7303]

so the odds are somewhere between 1/3 and 1/2?
Somebody start knocking!

[Joe Ross]

diobsidian,

I've heard the first, so I won't answer.

As for the second, you forgot to disclaim that the person can't carry more then reasonably possible, or something like that. For example, FizMan's answer is the most efficient way to get 2, but if you want 3, all you have to do is start with 130. Or for 4, start with 258.

Another thing of interest is that if you were to start with 1, you'd cut it in half for the first guard, and then he'd have to somehow give you 1, and you'd have 1.5. After the next guard, you'd have 1.75. If there was an infinite number of guards, you would have 2, but it would probably be spoiled from having been cut open so long

[FizMan]

I think you're off Kris... I'm going to trust my highschool Finite math on this one...


We'll do it your way:


Door Opener / Non Door opener

Boy / Girl
Boy / Boy

Girl / Girl
Girl / Boy


Okay, a girl opens the door... so you say that eliminates the first two possibilities...

Why? Remember, two combinations of boy/girl exist, and you've eliminated one of them for the wrong reason.

Whoever opens the door does not determine what children exist in the family (with the exception of the boy/boy possibility)

If you want to keep your method of reasoning, we have to add more possibilities. That is, the order of children. I'll mention the first born child, the other is safely assumed to be the second born:


Door Opener / Non Door opener

Boy (1st born) / Girl
Boy (1st born)/ Boy

Boy / Girl (1st born)
Boy / Boy (1st born)

Girl (1st born)/ Girl
Girl (1st born)/ Boy

Girl / Girl (1st born)
Girl / Boy (1st born)


Eliminate the Boys answering the door, you're left with the last four. We can eliminate one of the extraneous Girl/Girl possibilities as well.

Girl (1st born)/ Boy

Girl / Girl (1st born)
Girl / Boy (1st born)

Voila, 1/3.

[mtrawls]

Okay. I, too, have heard those before, so I'll propose my own (well, as you might guess, it is by no means original with me). Suppose that there is a game where we flip a coin. If the first time the coin is flipped it lands on heads, I pay you $2 (2^1), if the first time it lands on heads is the second toss, I pay you $4 (2^2), if the first time it lands on heads is the third toss, I pay you $8 (2^3), ad infinitum. Using simple probability, tell me how much you would pay me to play this game and still be able to make a decent profit (but enough to tempt me to agree to the game).

[FizMan]

You mean, for example, if I said, "Hey mtrawls, I'll pay you $20 to play <insert above mentioned game here> with me"?

[Joe Ross]

Originally Posted by Kris Verdeyen

Prove me wrong.

Well, I tried. First I tried analytically, and had the same explanation as FizMan. However, when I tried to figure out why yours was wrong, I couldn't. So, I wrote a program to do it a bunch of times. In short (unless my algorithm is wrong), you're right, it is 1/2.

I've included the code below in case someone wants to review my work.

/** @mainpage license
 * Copyright (c) 2004 Joseph Ross
 * 
 * Permission is hereby granted, free of charge, to any person obtaining a 
 * copy of this software and associated documentation files (the "Software"), 
 * to deal in the Software without restriction, including without limitation 
 * the rights to use, copy, modify, merge, publish, distribute, sublicense, 
 * and/or sell copies of the Software, and to permit persons to whom the 
 * Software is furnished to do so, subject to the following conditions:
 *
 * The above copyright notice and this permission notice shall be included 
 * in all copies or substantial portions of the Software.
 * 
 * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS 
 * OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, 
 * FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL 
 * THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER 
 * LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING 
 * FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER 
 * DEALINGS IN THE SOFTWARE.
 */

/** @file smiths.c
 *
 * @author	Joseph Ross, christmasboy_81@rossesmail.com
 * @date	7/7/2004
 * 
 * This program attempts to solve the smith's children problem as defined as
 * problem #8 in this thread on chiefdelphi.com: 
 * http://www.chiefdelphi.com/forums/sh...3&page=1&pp=15
 *
 * The problem is as follows;
 * Mr and Mrs Smith have two children. You ring at their door, and one of their 
 * two children, a girl opens the door. What is the probability that the other 
 * child is also a girl?
 * 
 * The problem is also stated at 
 * http://www.julienstern.org/riddle_sol.php3?id=7 and gives the answer as 1/3
 *
 * The algorithm used is as follows;
 *
 * @li assign two kids as boys or girls randomly and idependently of each other
 * @li pick one of the two to answer the door.
 * @li if the one picked is a girl, check if the other one is a girl and record
 * the number of times of each.
 * @li divide the number of girls from the first pick by the number of girls 
 * from the second pick.
 *
 * The answer, as given by this program is 1/2.
 */ 

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

/** The 1 is a Girl (0 is a boy) */
#define GIRL 1

/** Calculates the answer to the smiths children problem 
 * 
 * @param argc the number of arguments to the program.
 * @param *argv[] array of pointers to the arguments of the program.
 * 
 * @retval 0 no error.
 * @retval 1 no argument given.
 * @retval 2 invalid number of iterations.
 * 
 */
int main(int argc, char *argv[])
{
	int i;			// counter.
	int kids[2];		// array of two kids.
	int its = 0;		// number of iterations to run.
	int second_girls = 0;	// number of times the second kid was a girl.
	int first_girls = 0;	// number of times the first kid was a girl.
	int first_kid = 0;	// which kid answered the door.
	
	if (argc != 2)
	{
		printf("usage: smiths <number of iterations>\n");
		return(1);
	}
	its = atoi(argv[1]);
	if (its <= 0)
	{
		printf("usage: smiths <number of iterations>\n");
		printf("number of iterations must be greater then 0\n");
		return(2);
	}
	
	srandom(time(NULL));

	for(i=0;i<its;i++)
	{
		//pick the gender of the kids
		kids[0] = random() % 2;
		kids[1] = random() % 2;

		//pick which kid opens the door
		first_kid = random() % 2;

		//count first girl and check second kid if first kid is girl
		if (kids[first_kid] == GIRL)
		{
			first_girls++;
			//check second kid
			if (first_kid == 0 && kids[1] == GIRL)
			{
				second_girls++;
			}
			else if (first_kid == 1 && kids[0] == GIRL)
			{
				second_girls++;
			}
		}
	}

	//print results
	printf("iterations:\t%i\n", its);
	printf("first girls:\t%i\n", first_girls);
	printf("second girls:\t%i\n", second_girls);
	printf("%% second girls:\t%f\n", (double)second_girls/first_girls);
	
	return(0);
}

[Denman]

ok ok enough debate
the answer is 1/3




Right, 4 possiblities, as most of you got

boy/girl
girl/boy
boy/boy
girl/girl
from genetics
but if you find its a girl, you only eliminate one of them (boy/boy) not 2.
thus the probability - 1/3 (2 where the other is a boy and 1 where other is a girl) .

i quote an answer someone else wrote
You know that they have two children, that is four possibilities (Boy-Boy, Girl-Boy, Boy-Girl and Girl-Girl). Because you know that one of them is a girl, the first possibility is ruled out. So the probability that the second one is a girl is 1/3.


Another one :
Three cowboys are fighting together in a meadows. The first one, John, is a good shooter, he kills his target 2/3 of the time, and miss one third of the time. The second, Slim, is an average shooter, whose chances to kill and miss or even. The third, Bob, is a poor shooter, he only kills his target 1/3 of the time. They look at one another in silence... They still look... Sweat starts pearling one their faces... OK. Anyway. Their rule is the following. They shoot in rounds. One shot per round. They always shoot at the best shooter but themselves. They do as many rounds as neccesary so that either they all die or only one remains alive. Who has the most chances to survive? (Thanks to Matthias for this one)

oh and joe, dont answer this one lol seeings as you probably know where to find the answer lol


A grandfather clock chimes the appropriate number of times to indicate the hour, as well as chiming once at each quarter hour. If you were in another room and heard the clock chime just once, what would be the longest period of time you would have to wait in order to be certain of the correct time?


Let us go into heroic fantasy land. The story takes place in a antic land where dwarves are mining to extract gold. The king is visiting the mine when he ears voices around the corner: "I have a neat trick: we have to melt gold bars the size of this mold, that is exactly 100 grams. I remove just a tiny fraction to all my bars and they all weight 99 grams. Their scales are not precise enough, they don't see anything..." The king tries to run after then, but they ear him and manage to flee among the corridors. Annoyed, the king summons the gods, who grant him a gift from the future: a digital scale. The digital scales works as follows: you put whatever you want on the scale, (nothing happens then), you press a button and the exact weight of what is on the scale appears. Then, the scale vanishes. Note that before you have pressed the button, no weight appears and once you have pressed the button the weight won't change even if you add or remove things. Armed with this mighty scale, the king summons all his 100 workers. How will he find the cheater?

[FizMan]

Yes, it IS 1/3. REMEMBER what I told you? The door does NOT, I repeat, had NO bearing on the possible combinations of children. IF you CHOOSE to go that route of using the door for logic, you ARE EXCLUDING possibilities.


And it's too early in the morning for me to figure out your code right now, but I'm pretty sure it's flawed too since again, you're using the door for reference and therefore probably excluding possibilities.


Grandfather Clock Question:
1 hour and a half. Five possibilities:

Either it hit a quarter of an hour (on some hour OTHER than 12 o'clock) OR it hit a quarter of an hour (15, 30, or 45 minutes) on 12 o'clock OR it hit 1 o'clock

In the first possibility, all you have to wait is 45 minutes, in which case you'll hear it chime more than once.

In the second possibility (15 minutes) you'd have to wait 1:45 for the clock to chime more than once.

In the third possibility (30 minutes) you'd have to wait 1:30 for the clock to chime more than once.

In the fourth possibility (45 minutes) you'd have to wait 1:15 for the clock to chime more than once.

In the fifth possibility (1 o'clock) you'd have to wait 1 hour for the clock to chime more than once.


As you wait, if the clock still chimes once at an HOUR AND A HALF, you've elminated all the possibilities but one (number 2) and you know for a fact that it's CURRENTLY 1:45 and when you heard the first chime, it was 12:15.



Cowboy question:
I'm going to go with a 54.78% chance that Bob (the poorest shooter) will win. I hope I'm right; good ol' notepad:

.83 ->
  .66 ->
	.5478 win for 1/3 guy
  <- .33
	.50 ->
	  .33 ->
		.0451935 all die
	  <- .66
		.090387 1/2 wins
	<- .50
	  .33 ->
		.0451935 win for 1/3 guy

I kind of stopped after that when I realized that the possibilites won't top the first one (in which Slim and Bob manage to shoot down John. And John shoots down Slim)

[Denman]

yes right for grandfather , however i havn't worked out the cowboy one yet as i am meant to be working

Grandfather clock answer

You would have to wait 90 minutes between 12:15 and 1:45. Once you had heard seven single chimes, you would know that the next chime would be two chimes for 2 o'clock.

[Alan Anderson]

Originally Posted by diobsidian

You have 12 identical-looking coins. One of the coins is a fake, yet you do not know which one it is nor if it weighs more or less than the other 11. Using 3 turns with a regular balance scale, determine which coin is the fake and if it weighs more or less than the other 11.

The details are simple but hard to explain without graphics. Label the twelve coins A through I. Weigh ABC against DEF, then ABC against GHI. The results will tell you which group of three is different and whether it is heavy or light. Then you weigh any two of the different group's coins against each other to find out which is the odd one; if they match, the third one is the different one.

if ABC < DEF and ABC < GHI, one of ABC is light
if ABC < DEF and ABC = GHI, one of DEF is heavy
if ABC = DEF and ABC < GHI, one of GHI is heavy
if ABC = DEF and ABC > GHI, one of GHI is light
if ABC > DEF and ABC = GHI, one of DEF is light
if ABC > DEF and ABC > GHI, one of ABC is heavy

The other three combinations are not possible if exactly one coin is different.

Once it's narrowed down to three coins, just compare two of them. If they don't balance, you already know whether the fake is heavy or light; just choose that one. If they do balance, the other must be the fake.

Originally Posted by diobsidian

There is a tree with exemplary fruit in the middle of a courtyard that a farmer has decided must be guarded from people trying to take the fruit. There are 7 circular fences around the tree with a guard at each. Wanting to get at the fruit so badly, you go up to the first guard and tell him that if he lets you through, when you return you shall give him half of all the fruit you have taken, but then the guard must return one piece to you. You continue to bribe every guard you reach until you have reached the tree, whereupon you take the fruit and upon leaving fulfill your deal with all of the gaurds. How many pieces of fruit did you take?

If I take two, then I can give one (half of my two) to each guard and according to the agreement he'll give it back to me, leaving me with the two I took.

[Steve Howland]

Originally Posted by Alan Anderson

The details are simple but hard to explain without graphics. Label the twelve coins A through I. Weigh ABC against DEF, then ABC against GHI. The results will tell you which group of three is different and whether it is heavy or light. Then you weigh any two of the different group's coins against each other to find out which is the odd one; if they match, the third one is the different one.

if ABC < DEF and ABC < GHI, one of ABC is light
if ABC < DEF and ABC = GHI, one of DEF is heavy
if ABC = DEF and ABC < GHI, one of GHI is heavy
if ABC = DEF and ABC > GHI, one of GHI is light
if ABC > DEF and ABC = GHI, one of DEF is light
if ABC > DEF and ABC > GHI, one of ABC is heavy

The other three combinations are not possible if exactly one coin is different.

Once it's narrowed down to three coins, just compare two of them. If they don't balance, you already know whether the fake is heavy or light; just choose that one. If they do balance, the other must be the fake.

I hate to break it to you after all your work, but there are 12 coins, and lettering them A through I is only 9.

If I take two, then I can give one (half of my two) to each guard and according to the agreement he'll give it back to me, leaving me with the two I took.

Yep you got that one!

Keep working on the 12 coin problem - its a real toughie!

[Alan Anderson]

Originally Posted by Kris Verdeyen

PROBABLILITY THAT A GIRL WILL ANSWER...
...and the family has two boys - 0%
...and there's a boy and a girl - 25%
...and the family has two girls - 25%

So, a girl will answer (0% + 25% + 25%) = 50% of the time.
Exactly half of those responses will come from families with two girls, and exactly half will come from families with a boy and a girl.

Or I could be completely off base - let me know.

You're missing the fact that there are two different ways in which the door may be answered by a two-girl family. Your analysis is looking at the probability of a girl answering, which doesn't matter for this problem, because a girl answering is given.

[Kris Verdeyen]

Originally Posted by Denman

ok ok enough debate
the answer is 1/3

Right, 4 possiblities, as most of you got

boy/girl
girl/boy
boy/boy
girl/girl
from genetics
but if you find its a girl, you only eliminate one of them (boy/boy) not 2.
thus the probability - 1/3 (2 where the other is a boy and 1 where other is a girl) .

i quote an answer someone else wrote
You know that they have two children, that is four possibilities (Boy-Boy, Girl-Boy, Boy-Girl and Girl-Girl). Because you know that one of them is a girl, the first possibility is ruled out. So the probability that the second one is a girl is 1/3.

Originally Posted by FizMan

If you want to keep your method of reasoning, we have to add more possibilities. That is, the order of children. I'll mention the first born child, the other is safely assumed to be the second born:


Door Opener / Non Door opener

Boy (1st born) / Girl
Boy (1st born)/ Boy

Boy / Girl (1st born)
Boy / Boy (1st born)

Girl (1st born)/ Girl
Girl (1st born)/ Boy

Girl / Girl (1st born)
Girl / Boy (1st born)


Eliminate the Boys answering the door, you're left with the last four. We can eliminate one of the extraneous Girl/Girl possibilities as well.

Girl (1st born)/ Boy

Girl / Girl (1st born)
Girl / Boy (1st born)

Voila, 1/3.

I understand that the answers you've looked up or been given in finite math class are what you've demonstrated so far. And I understand your answer, and see why it works. What I don't see is why mine was wrong. Your proof above was a bunch of arbitrary hand waving. Why do we need to "eliminate one of the extraneous Girl/Girl possibilites as well" ? If you don't eliminate that - the probability is (surprise surprise) 1/2.

Think for yourselves and come up with the answer! Don't tell me that your high school math teacher told you this was the answer, and therefore it's so. There is more than one way to look at any problem - the challenge here is to determine, at a fundamental level, what is different between the way your math teacher looks at the problem and the way I've outlined it above. I don't see anything fundamentally different, but just because I don't doesn't mean it isn't there.

My proof and Joe's simulation both show the answer to be 1/2. I've also included an Excel simulation that does the same - why are we wrong?