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Flash
15687.  Sat Feb 26, 2005 9:46 am Reply with quote

Q: You have a bullet in one hand and a gun in the other. You fire the gun horizontally and drop the bullet simultaneously. Which bullet hits the ground first?

A: The gun propels the bullet horizontally, and the vertical force (gravity) acting on both bullets is identical, which will make them land simultaneously. However, if the bullet from the gun travels far enough the Earth will curve away from it, so it'll land second (though I suppose this effect would be comfortably dominated by irregularities in the terrain, over any distance you like, so I suppose we need to specify a salt flats location).

If you could achieve a velocity of 5 miles per second (which you couldn't) the bullet would go into orbit and not land at all.

I have to say that this sounds a bit rough'n'ready to me, so put me right. My info, uncorroborated, from http://www.straightdope.com/classics/a5_201a.html.

 
Gray
15728.  Mon Feb 28, 2005 11:43 am Reply with quote

It's true that the gun-fired bullet would land second, for the reasons given - the Earth would curve away from it slightly during its flight. See this page for a brief run-down.

As for the speed at which you'd have to fire a bullet for it never to hit the Earth, it can be easily calculated by equating the acceleration of the bullet towards the centre of the Earth (mv^2 / R) with the force of attraction between the Earth and the bullet (GMm/R^2). The solution (which is then independent of the mass of the bullet - Galileo was right) is simply got by taking the square root of [G (universal grav constant) x Mass of Earth divided by Radius of Earth], which does indeed come out at almost exactly 5 miles per second (thank goodness for a calculator that can convert metres-per-second into miles-per-hour!)

Of course, air-resistance would stop it from maintaining that air-speed, so it would slow down and start to drop. That's why satellites have to be out of Earth's atmosphere to stay up there - no air friction.

----------

Returning briefly to the Monty Hall problem, I've just come across a lenthy section in Paul Hoffman's excellent book on the mathematician Paul Erdös (The Man Who Loved Only Numbers, pp 234-239). An explanation she gave (after three columns devoted to her communications on the subject) was as follows:
Quote:
Imagine that just after the host opened the door, revealing a goat [the booby prize], a UFO lands on the game-show stage and a little green woman emerges. Without knowing what door you originally chose, she is asked to choose one of the unopened doors. The odds that she'll randomly choose the car are 50-50. But that's because she lacks the advantage that the original contestant had - the help of the host.

Suppose you've picked Door 1 to start with. If the prize is behind Door 2, the host shows you Door 3. If the prize is behind Door 3, the host shows you Door 2. So if you switch, you win if the prize is behind Door 2 OR Door 3. If you stick, then you ONLY win if the prize is behind Door 1.


The situation was programmed into a computer by a mathematician friend of Erdös's, and it was played a hundred times, with the computer choosing randomly whether to switch each time it played (this is known as the 'Monte Carlo method). Sure enough, the outcome favoured switching two to one (i.e. it won twice as often when employing the switching tactic).

The reason why people get worked up about it all is that we tend to imagine that probabilities are 'fused' with the object in question, and therefore when the odds change (as they do when you receive new information), we don't update our ideas of what the probabilities are.

 
knightmare
1064146.  Tue Mar 18, 2014 11:44 am Reply with quote

Flash wrote:
Is this trick question very old hat? I hadn't encountered it until today. It's apparently called the Monty Hall Dilemma


Isn't the old hat (based on) a Joseph Bertrand Paradox? We don't always need a Monty Hall quiz show nor the Mythbusters to visualize a problem.

There are three boxes, containing 6 items: AA, AB or BB. You look at one of the items. It's an A. The odds that the second item in the same box is also an A is 2 / 3, because you can exclude the BB box.

We had A-A-A-B-B-B. We don't have BB, so the new collection is A-A-A-B. We already have one of the A's in our hand, so the remaining collection is A-A-B. So the odds are 2 out of 3 that the second item is an A. Switching, part of Hall's dilemma, reduces this to 1 - (2 / 3) = 1 / 3.

The right answer isn't 1 / 3 (1 out of 3 boxes), because we already can exclude the box containing BB. The right answer isn't 1 /2 (1 out of 2 remaining boxes, AA or AB), because we know for sure that we have at least 1 A, there are more remaining A's (2 out of 3 remaing items than B's (1 out of 3 remaining items).

 
Spud McLaren
1272116.  Fri Jan 26, 2018 6:45 pm Reply with quote

A thought has just occurred to me: in the original Monty Hall Dilemma, behind the three doors are two goats and a car (not a car and two empty garages). The question as to the correct course of action to give you the best chance of netting the prize is as outlined above, elsewhere on this site, and yet elsewhere on the internet.

However, that presupposes that you want a car out of the deal. Is the procedure any different if you wanted a goat?

 
monzac
1272131.  Sat Jan 27, 2018 2:11 am Reply with quote

You'd sit rather than switch, unless the host wanted to direct you away from the goat. You'd need two cars in that scenario though.

 
Spud McLaren
1272154.  Sat Jan 27, 2018 5:52 am Reply with quote

No, I meant if there were still two goats (ie, the main prize was the car but the contestant was unexpectedly "alternative").

 
monzac
1272186.  Sat Jan 27, 2018 11:03 am Reply with quote

That's what I thought you meant. You'd sit, wouldn't you?

 
Spud McLaren
1272192.  Sat Jan 27, 2018 1:04 pm Reply with quote

I think so, but since the first answer is at first sight counter-intuitive, I wondered whether this is too. I don't think so, but then I'm no probability theorist.

However, I despaired at my colleagues' ability to answer the other day's Puzzle for the Day.

 
monzac
1272202.  Sat Jan 27, 2018 3:10 pm Reply with quote

Not seeing a puzzle there. What number is it?

Edit: No 149. Can't view the puzzle at Puzzles for Today either. I can see the previous ones :/

 
Baryonyx
1272391.  Mon Jan 29, 2018 6:26 am Reply with quote

Yes I think the opposite is the case; switching gives a 2/3 chance you WON'T get the goat you so desire.

2/3 chance you select a goat initially, switching is 100% a car
1/3 chance you select a car initially, switching is 100% a goat

 
dr.bob
1272408.  Mon Jan 29, 2018 9:36 am Reply with quote

 
'yorz
1272415.  Mon Jan 29, 2018 10:07 am Reply with quote

So that's what you're doing as a sideline, dr.bob! :-p

 
suze
1272427.  Mon Jan 29, 2018 12:29 pm Reply with quote

Do the rules of the problem state that the open door is out of the game, and you can't switch to it? If they don't, then if you actually want a goat you obviously switch to the door which has already been shown to hold one, as in the cartoon above.


As for Puzzle #149 mentioned by Spud, to me that is trivially easy. Am I missing something? (Husband adds that he wasn't much luckier at the pre-game coin spin than Alec Stewart was. At least, he imagines this is why the team often referred to him as a "useless tosser" ...)

 
Spud McLaren
1272436.  Mon Jan 29, 2018 1:40 pm Reply with quote

suze wrote:
As for Puzzle #149 mentioned by Spud, to me that is trivially easy. Am I missing something?
You are not.

 

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