The worlds first riddle!

It is possible that the level will drop if the rock is tossed into the boat.

If the additional weight of the rock causes the boat to sink, the boat will displace its volume.

If the boat is relatively large with a very thin but dense hull and is barely afloat, the amount of water it displaces will drop significantly. The drop may be more than the volume of the rock.

[size=8]BRIDGE
There are C(36,13) hands that can be made from 2-10.
There are C(52,13) hands that can be made from a full deck.

Therefore, the answer is C(36,13) / C(52,13) = 0.0036[/size]

Oh good, an answer before 9am.



There are many people who have the same birthday as you. With only 366 birthdays to go around, and over 6,000,000,000 people in the world there has to be quite a bit of sharing. But what if you are in a room with a few people? How likely is it that there is at least one shared birthday?

The old question was: "How many people do you think there would have to be in a room to have a better than even chance that at least two people share a birthday?"

The new question is:

How many people do you need for the probability to be greater than 90%

Ok, this is confusing so I'm going to work it out...

D=M/V
R_D = 2.5*W_D

R_V = 125 inches^3 = (125 inches^3)*(1 ft^3 / 1728 inches^3) = 0.072338 ft^3 (rock volume)
W_D = 62.4 lbs/ft^3 (water density)

R_D = 2.5(62.4) = 156 lbs/ft^3 (rock density)
R_M = R_D*R_V = (156 lbs/ft^3 )*(0.072338 ft^3) = 11.2847 lbs

Ok this much is in agreement.



R_V*2.5 = 312.5 in^3, but why did you do this calculation? What equation tells you the volume of displacement for the rock in the boat?

A floating object displaces its weight (not volume) in water.

Therefore, R_W / W_D = displaced volume.

[size=8]BIRTHDAYS
The probability is 0.903 with 41 people.[/size]

All I know my fine gentlemen is that my birthday is in 7 months and I am expecting gifts!!

Is it April 11?

That's a long way off. I might forget by then. Put a note under my pillow the day before.

Ok so more like 6 months and some change.....March 19th. Will that be soon enough. I am not so good at the reminders!

Mark:
BIRTHDAYS
The probability is 0.903 with 41 people.

It is clear to me sweet Shari gave you the answer.


One way to solve this is to turn the problem around and think about how likely it is for there to be NO matches in a group of a given size. If there is only one person a room there can be no shared birthdays since there is no one to share with. The probability of not having a match in this case is 1. Events that are certain are said to have a probability of 1. At the other extreme, with 367 people in the room, it is certain that there will be at least one shared birthday since there aren't enough birthdays to go around.
Now imagine that a second person walks into the room. The probability of that person not having the same birthday as the first occupant of the room is 365 / 366 or 0.997. There are 366 possible birthdays and only one of them is a match.

Now if the first two people in the room have different birthdays and a third person walks in, there are two days used up so the probability of the third person not sharing a birthday with either roommate is 364 / 366 and the probability of no sharing amongst the three of them is 1 * 365 / 366 * 364 / 366 = 0.992, which is still over 99%. So with 2 or 3 people in the room there is less than a 1% chance of a shared birthday.

You can continue to calculate the chances of not having a shared birthday for any number of people:
1 * 365 / 366 * 364 / 366 * 363 / 366 * 362 / 366 ...
Things change quickly as the number of people increases. With 10 people in the room there is a better than 10% chance of a match. When there are 23 people in a room the chance of a shared birthday is slightly greater than 50% and it rises above 90% with 41 people.


Shari (whoever she is) wrote, "my birthday is in 7 months and I am expecting gifts!!"

Mark replied, "Put a note under my pillow the day before."


Why don't you two get a room already!!!


Shari riddle - It is well known amongst LA socialites that Shari has a short fuse. To further complicate matters; the rate at which a fuse burns is irregular. If an m-minute fuse is cut into two pieces, one of the pieces will burn for n minutes and the other for m-n minutes, but there is no way of determining n.

Suppose you have two 60-minute fuses, is it possible to clock 45 minutes



A pot contains 75 white beans and 150 black ones. Next to the pot is a large pile of black beans.
A somewhat demented cook removes the beans from the pot, one at a time, according to the following strange rule:

He removes two beans from the pot at random. If at least one of the beans is black, he places it on the bean-pile and drops the other bean, no matter what color, back in the pot. If both beans are white, on the other hand, he discards both of them and removes one black bean from the pile and drops it in the pot.

At each turn of this procedure, the pot has one less bean in it. Eventually, just one bean is left in the pot. What color is it

[size=7]yes...light both ends of fuse A and one end of fuse B. after 30 minutes fuse A is gone and fuse B has 30 min left. light other end of fuse B and it wll too be gone in 15 more minutes, so they burn for atotal of 45 min

a white bean will be left. white beans can only be removed 2 at a time and theres an odd numbr of them [/size]

"Shari (whoever she is) wrote, "my birthday is in 7 months and I am expecting gifts!!" "


I see some things never change.....cocky as ever!!

[size=8]FUSES

T = 0
60, 60
Light both ends of one and one end of the other.

T= 30
0, 30
Light the other end of the remaining fuse.

T=45
0, 0

BEANS
The parity of white beans never changes. Therefore, there will always be an odd number of white beans. Zero is even; so you can't eliminate all of the white beans. One is odd; so you can end up with a single white bean.[/size]

Why is she so familiar with your anatomy?

it surprised me that it doesnt matter how long th row of soldiers is because that will cancel out. anyways the answer is 4 minutes.

Now the same soldiers march side by side. The same dog runs behind the soldiers as they march. How much time will it take for the same dog to run from one end of the column to the other if they march for 3 minutes?

[size=8]FIRST SOLDIER PROBLEM
L = length of column
Vd = dogs speed
Vm = men's speed

We want L/Vm.

L = (Vd - Vm) / 2
L = (Vd + Vm)

Vd/L - Vm/L = 1/2
Vd/L + Vm/L = 1

Therefore,
Vd/L = 3/4
Vm/L = 1/4

Since Vm/L = 1/4, L/Vm = 4 minutes.[/size]

[size=8]SECOND SOLDIER PROBLEM
Assumptions:
All speeds are the same.
The width of the row is the same as the length of the column.

From the first problem:
L = 4/3 * Vd (It takes the dog 4/3 minutes to run from one end to the other if the row is stationary).
Vd = 3 * Vm

If you consider the rectangle formed by the starting and ending positions of the row, the dog runs along a diagonal.

The ratio of the diagonal to the base (L) is 3:2*sqrt(2).
So, it will take the dog 3/(2*sqrt(2)) * 4/3 minutes to reach the other end.
That is sqrt(2) minutes. The dog can make the return trip with time to spare.[/size]

correct about the dog.... i should have formulated my question differently and i see the other of my ways

ok another problem. i hope i remember the solution correctly:

a dog sees a squirrel directly north of himself, 96 meters away. at the same time, the squirrel sees the dog. the squirrel starts running directly west at 6 meters/second. the dog starts chasing the squirrel at 10 meters/second, but keeps adjusting his path so he always faces the squirrel. when will the dog catch the squirrel?