Nope. If he doesn't know where the prize is, then you're right. But since he will open only a 'bad' door and not a 'good' door, it's really no different than had he not openned any door, and had instead said, Do you want the one door you chose, or one of these two that you didn't. If the door that he opens is random, then 1/3 of the time the game ends there. Since that 1/3 probability never happens, it screws up the odds. Personally, i wouldn't trust Monty, but statistically, you should switch.
We've been through this one here before. I didn't want to believe it either, but bnb is right (hell----that is if I remember correctly). So what about our current issue: 64=65?
I still don't think it makes much sense. After looking at it again, I understand that Monty will take away a bad door that you didn't pick, regardless of whether you currently have the right door or not. But I don't see how changing your pick will necessarily improve your odds. It's still two doors left. I think that you're saying that in order to improve your odds, you should repick after a bad door gets taken away. But I contend that picking the same door that you already had is a valid pick and isn't anymore likely to be correct than the other pick. Edit: I searched for the problem and came across this website explaining it: http://mathworld.wolfram.com/MontyHallProblem.html I still don't get it, but will just go along with Jeff and LegendZ3 on this one. Edit #2: I found another page that goes into the problem even more indepth. And this one is a bit more thorough with it's pros and cons. http://jimvb.home.mindspring.com/monthall.htm After skimming through this page, I can only conclude that, dees guys are nurdz. And at least I'm not spending my free time, debating math problems on the internet. . .
I saw something similiar to this before. Like mentioned, it isn't what it appears. The one I saw looked like it had the same pieces, but a line changed evert so slightly, but just enough to give an extra piece. Unless you were looking for it, you wouldn't notice it. Lucky for me, I find this site that broke it down, stopping and zooming in on the pieces and then highlighting them to make them more visible. Too bad I can't remember where that was or I'd post a link. (we still have that privilege, right? ) I once saw this one gif that's kind of similiar to this one. It has a group of 12 people, and then it exchanges top left and right sides, lining up bodies with heads again, revealing 13 people. It then asks how they go an extra person. I remember going over this before as well, but I forgot how they did it again. I think there is some shoe or hair or something that helps. Can't remember exactly:
This is the simplest explanation that can be given of the Monty Hall problem: Car is behind Door #1. You choose Door #1. You win by staying. Car is behind Door #1. You choose Door #2. You win by switching. Car is behind Door #1. You choose Door #3. You win by switching. Car is behind Door #2. You choose Door #1. You win by switching. Car is behind Door #2. You choose Door #2. You win by staying. Car is behind Door #2. You choose Door #3. You win by switching. Car is behind Door #3. You choose Door #1. You win by switching. Car is behind Door #3. You choose Door #2. You win by switching. Car is behind Door #3. You choose Door #3. You win by staying. In 2/3 of the possible situations, switching is favorable.
And to elaborate further on the facts: * 2/3 of the time you pick a door, it will be an incorrect choice. * 100% of the time where you originally pick the incorrect door and then switch, you will pick the door with the car. Therefore, switching is favorable 2/3 of the time, and those results agree with the results shown above.
for 64=65, if you calculate the slope of the lines that are touching each other, they are different, so its not a real rectangle. the slope of the diagonal line for the green triangle is 3/8. the slope of the diagonal line for the blue figure is 2/5. since they have two different slopes, connected all the figures by those lines would result in a figure that isnt a rectangle.
Okay, that actually makes sense to me. Buuuut, you'd still be massively pissed if you originally picked the right door but ended up switching.
To finish that thought, when drawn to scale, the rectangle would have a parallelogram missing from the middle of it. Dividing the parallelogram down the diagonal of the rectangle would create two triangles whose three sides (according to the Pythagorean formula) are the square roots of 194, 73, and 29 respectively. Heron's formula will tell us the area of one of these triangles: area = sqrt{s(s-a)(s-b)(s-c)} s is the semiperimeter of the triangle, (a+b+c)/2. Trying to add up all those square roots results in a horrific mess unless you use approximate values. However, multiplying out Heron's formula results in something more manageable: area = sqrt{2(a^2b^2 + a^2c^2 + b^2c^2) - (a^4 + b^4 + c^4)} / 4 Plugging in the values for a, b, and c yields: area = sqrt{2*(14162+5626+2117) - (37636+5329+841)} / 4 which (IMHO, almost miraculously) equals sqrt(4)/4, which is 1/2. So, the two triangles that make up the parallelogram have a total area of 1, which, when added to the area of the original square (64), gives the area of the new rectangle (65), q.e.d. (Stuff like this is why I love math -- why should a triangle with such exotic side lengths have such a simple area?) P.S. The missing parallelogram has angles of approximately 178.75° and 1.25°. No wonder it's so hard to see!
This one stumped Isabel also, but I, her husband Ferdinand, have encountered this puzzler before and know the solution. Your explanation is partially correct; you're right in saying that zooming in on the pieces will help reveal the solution, but it's not quite what you think it is. As the animation shows, one can indeed cut those four pieces of an 8x8 square and re-arrange them so that one can construct a rectangle of dimension 5x13. It's a true rectangle; all four vertices contain within them a right angle, so the definition of a rectangle (a quadrangle containing within it four right angles) is fulfilled, and it is in fact of dimension 5x13. The rectangle therefore encloses 65 square units, whereas the square encloses 64 square units. How could the rectangle "add" an extra unit of area? After all, you have to conserve the area marked in square units. The secret is that, in fact, the rectangle encloses 65 square units of area, but contains only 64 square units of material. What does that mean? This is where a close-up view of the rectangle helps (it might help even more if you made these pieces yourself using a large scale and then putting them together as the rectangle). The rectangle appears to be made of two right triangles, one on the upper left and the other on the lower right, and then joined along their hypotenuses to form a rectangle. They do indeed form a rectangle, but in fact they're not triangles. If you do the trigonometry, you'll find that the two "triangles" are actually quadrangles. Three of the four vertices of these quadrangles are obvious; the fourth vertex is obscured because the angle that subtends it is very close to 180 degrees, thus making it appear (at large scale ) to be a straight line (in fact, that angle is about 178 3/4 degrees). The rectangle that is formed then is from two quadrangles positioned so that the two sides that almost make a straight line are against each other. However, since they're not actually straight lines, they don't fit flush against each other, and there's some empty space between them. That empty space is in the shape of an extremely elongated quadrangle that happens to have an area of exactly one square unit. You can think of the rectangle this way: it contains 64 square units of material marked in lines to form squares and 1 square unit of empty space. Therefore, the area marked in square units is indeed conserved; the rectangle formed from the four pieces simply incorporates within it a void of one square unit. The reason this illusion works is because the mind is tricked by the pattern of straight lines. The void between the two quadrangles that make up the rectangle is so small relative to the size of the rectangle itself that the mind makes the lines run continuously across the area of the rectangle instead of being interrupted by the void space as they truly are. This makes the mind perceive the rectangle as being composed of 65 squares of 1 square unit each, when in reality it's composed of 49 squares of 1 square unit each and 15 rectangles (which are very nearly squares) that together occupy 16 square units of area. Ferdinand Proud Houston Rockets fan since 1971
The problem is not with the math - you can divide 10 by 3. The answer is 3 1/3. The problem is trying to express a fraction as a decimal.
The 8 x 8 square at the beginning is NOT a square. The sides are not equal which means one side is longer and the actual area is 65 (not 64) . So when the 'sqaure' is turned into a rectangle the size didn't change, it was simply redistributed.
Ferdinand is correct. I've seen this before. There is no magic to the puzzle - it's an optical illusion. The 2 "right" triangles that result to form the final rectangle actually aren't true triangles - their hypotenuses are slightly curved. The curvature of the hypotenuses leaves an "open" area within the rectangle extending from the upper right to the bottom left. The area left "open" is 1 sq. unit. Thus, the rectangle contains 65 sq. units of area using 64 sq. units of "stuff". Solution
