quote:Just below its boiling point, liquid helium behaves much like other liquids with low boiling points. In this state, it is known as helium I. However, when chilled several degrees below its boiling point, liquid helium enters a second state and is known as helium II. Helium II has some very unusual properties, among which are the lowest viscosity (resistance to flowing) of any liquid; the ability to flow uphill; and the capability of penetrating tiny passages where no other liquid can flow.
if you are bicycling and come to a stop light that senses vehicles, you can trip the sensor by laying your bike down over the sensor area (usually visible as a rectangle in the pavement) because they use electrical inductance, which falls off with distance squared ...
Perhaps the sensor was turned down (to prevent the door opening for dogs etc), and it probably used an average over a set time period usually taken to walk along the sensor area (to eliminate people walking across the sensor area tripping the door).
That's my guess, as it would eliminate light weight and short impacts of momentum.
A third of the time, he opens a goat and you chose the car. (Sit.A)
You win if you don't change and lose if you do.
A third of the time, he opens a goat and you chose a goat (Sit.B)
You lose if you don't change and win if you do.
A third of the time, he opens a car and you chose a goat (Sit.C)
You lose if you don't change and lose if you do (which may seem pointless unless you a filled with boundless optimism ...).
∑ 4 losing situations and 2 winning situations.
Since the presentator opens a goat, you know you are in situation A or B, and you know each situation is just as likely as the other (33% each), then you also know that since you are as likely to be in sit.A then B, then you are just as likely to have the car behind your door then behind the other door. Hence a 50/50 chance
But you are forgetting to include the results of sit C in your calcs. Just because you see the car, it doesn't mean you don't lose.
I'm not forgetting to include sit.C, it's just that nothing worth of interest comes out of doing so.
But it is of interest to me: it means there is a lose-lose case that cannot be omitted from the calculations of all the possibilities. Once the door is opened on the car, there is zero possibility of you winning. This is what changes the original results.
Probability calculations are (properly) based on evaluation of the relative success of each and every one of the possibilities (this is why probability calculations cannot be made without knowing all the possibilities), so let's look at the possibilities (your initial pick is door 1, as per the original game analysis, but here the host randomly picks door 2 or door 3):
Case
door 1
door 2
door 3
host
stay
switch
Sit A1
car
goat
goat
door 2
win
lose
Sit A2
car
goat
goat
door 3
win
lose
Sit C1
goat
car
goat
door 2
lose
lose
Sit B1
goat
car
goat
door 3
lose
win
Sit B2
goat
goat
car
door 2
lose
win
Sit C2
goat
goat
car
door 3
lose
lose
It appears that you have a 1/3 possibility of winning whether you switch or not (unless you want to win the goat). This is the way the game appears at first.
Now it could be argued that in the original game the host eliminates Sit C1 and Sit C2, and this leaves you with a 50:50 chance ... what am I missing?
One involves 3 plain cards with faces coloured red/red, red/green & green/green. Now if one card is chosen at random and placed on a table so that you can see one side (red say), what it the probability that the other side is also red?
2/3 ?
Isn't this similar to your last version of the Monty game? Once you have a card on the table (with any one card being chosen 1/3rd of the time), there are only two cards that can have the face that is up (ie it cannot be the green/green card), and one of them the same on the bottom and the other is not: 50:50?
The thing to remember is that we are only interested in answering the question: "Should I swap or stay?" If Monty chooses the car, then the question is moot; the answer is neither "swap" nor "stay".
Curiously, you do not walk away with the car. You still have the option to swap or stay, whether you have a possibility of winning or not.
RAZD, I assume you are referring to a yacht tacking. I hate to be pedantic but that isn't "directly" into the wind. However, it is pretty mindblowing that a wind-powered vessel can travel into the wind, directly or indirectly.
Nope.
A vessel with a propeller that is spun by the wind powers a propeller that drives the vessel directly upwind. Same principal as the vehicle going downwind faster than the wind.
quote:The 36 foot catamaran, Revelation II, is powered by 3 20-foot long carbon fiber propellers on a 30 foot rotating mast. The windmill transmits power to a 6 blade propeller underwater, with the net result that the boat can make way even directly into the wind.
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