Oh, and the answer to your riddle, puglsy, is that it's impossible for him to accomplish that feat time = distance/rate In order to average 240 dm/hr over a 4 km course, you would have to travel the entire course in .0167 hours. If you travel 2 km at 120 km/hr, you've already traveled for .0167 hours....To me, that would seem to make it impossible... I think?
Midget. Dude can reach the 6 button, but can't reach the 10. Only takes elevator to 10 when another person is in the elevator to push the button for him.
Monty Hall problem: If you always switch, you'll lose when you originally pick the correct door and win when you don't originally pick the correct door. What's the probability of originally picking the correct door? 1/3. So, your probability of winning if you always switch is 2/3. (Note: This assumes that Monty will always open a door. It also assumes that the prize doesn't move around behing closed doors -- if it moved randomly after the first door was revealed, then it wouldn't matter if you switched or not.) Another problem: I have 2 siblings. One of them is female. What is the probability that the other one is male?
It's true. Think of it this way, when you initially choose a door, you're picking with a 1/3 chance. You also have a 2/3 chance of getting it wrong. When the host opens one of the other doors, he always opens a wrong door because he knows which one is empty. This doesn't change your probability because you chose before he opened a door. If you chose after he opened a door, you'd have a 1/2 probability of getting it right. Or think of it this way. I tell you to guess what number I'm thinking from 1 to 10. When you pick a number you have 1/10 chance of winning. Then I eliminate 8 numbers. The probability you had of getting it right on your first guess is 1/10. SWITCHING to the other number is basically saying that you guessed the wrong number which has a probability of 9/10. Since 8 numbers are eliminated, the remaining choice has 9/10 probability of being correct. SWITCHING IS EQUIVALENT TO THE PROBABILITY OF YOUR FIRST GUESS BEING WRONG.
A-train is correct. It's impossible to average 240km/h. Wait, maybe you can if you can travel 186,000 miles/sec.
That's silly. What if there are two of us picking. You pick door A, I pick door B. The host reveals door C, which turns out to be empty. Are your chances now 67% of winning if you switch to door B? And my odds are 67% if I switch as well? The odds of one door having the prize are exactly the same as the other door having the prize: 50%.
The monty hall problem is very famous because of Marilyn vos Savant . Basically, she got it right but hordes of professional mathematicians swore she was wrong for years. Switching is the correct move. EDIT: This is an even better link: http://www.willamette.edu/cla/math/articles/marilyn.htm
This is exactly what I thought, but everyone seems to insist that the other answer is correct. I wish the other one could be put in a simple mathematical form so you could see it easily. Good job A-Train, that's the correct answer to the "math problem." Pugs
A (possibly futile) attempt at simplifying the Monty Hall problem: Let's call the door I originally choose A, the leftmost door that I didn't choose B, and the rightmost door that I didn't choose C. The prize is either behind door A, B, or C with equal probability. Let's assume that I always switch. If the prize is behind door A, Monty will open either door B or door C. I'll change to either door C or B, respectively, and lose. So, the probability that I win when the prize is behind door A is 0. If the prize is behind door B, Monty will open door C. I'll switch to door B and win. So, the probability that I win when the prize is behind door B is 1. If the prize is behind door C, Monty will open door B. I'll switch to door C and win. So, the probability that I win when the prize is behind door C is 1. Multiply the probability of my willing in each case by the probability of that case occurring ... 0 * 1/3 + 1 * 1/3 + 1 * 1/3 ... and you get the probability of me winning in all cases ... 2/3. [EDIT: I was trying to use Google Calculator to show that your average speed would only be 239.999996 km/hr. It didn't work, so the joke kinda falls flat. Plus, as you mentioned, if you assume the time frame of the racer, then the speed of light is a reasonable answer.]
Fine, I'll try to show you guys in a different way. Suppose I ask you to guess what number I'm thinking from 1 to 10. You guess a number. NOW, instead of doing all the weird stuff, I ask you whether you want to bet that you guessed the correct number or that you guessed the wrong number. Obviously you'll bet that you got the wrong number because you have a 9/10 chance of getting it wrong while only a 1/10 chance of getting it right. Out of the remaining 9 numbers, 8 of them are wrong at the very least. Even if I told you that those 8 numbers aren't the numbers I'm thinking, It doesn't change anything. The probability you guessed the correct number is 1/10, and the probability that you got it wrong is 9/10. By eliminating 8 of the chioces, the remaining choice has a 9/10 probability of being right.
Good explanation. However, I do believe I am correct in saying 186,000 miles/sec is good enough It is the speed of light, and if you travel at that speed, time stops. Hence, it just may work
On the door question you have to make a couple of assumptions: 1) he always opens a door 2) he knows where the prize is 3) he alway opens the door to an empty room. In this case, if your original guess is wrong, he will always open the only other 'wrong' door, so the leftover door is right. Essentially, by choosing the leftover door, you've got a 2/3 chance of being correct. Your only chance of being wrong is if your original choice was right (1/3). This doesn't hold true if the host has the choice of offering or not offering a switch...or if the host doesn't know which door is right. If the host doesn't know which door is right, then 1/3 of the time he'll open the winning door, and the games over -- meaning you'll be offered a choice only 2/3 of the time -- so 1/3 you'll be right originally, 1/3 the game will end when he opens a door, and 1/3 you'll be correct to switch (or a 50/50 once you're offered the choice).
Still 50/50. If one thing (like your sibling being female or male, a coin coming up heads or tails, or a prize being behind a door) is an either/or thing and you have a 50/50 shot at it, it seems to me like no outside factors will change that. As for the door problem: regardless of math, you picked a door and, with what you know now (after opening the one wrong door), you have a 50/50 chance that your door is right. I heard it explained that the reason you should still switch anyway is that: after you picked your door, the game show host had to choose one that was definitely wrong to show you. Kimble14 explained it fairly well below. But the reasoning here is that there is a human element involved - Monty - who has to choose something to show you. (and I read that somewhere... not smart enough to come up with it on my own )
I can see the math, but what if that same offer was given to a guy who hadn't made an initial pick, an audience member perhaps. In other words, door C (let's say) was revealed to be empty, and he was then given the choice to pick A or B. He's only got a 50% chance of being correct if he chooses B, right? But your guy making that same choice (switch to B) has a 2/3 chance?
To add the probability of Monty opening the winning door given he doesn't know the answer to kimble's formula (or of a random audience member who doesn't know the answer and will thus end the game 1/3 of the time), you get: 0 x 1/3 + 2/3 x (1 x 1/3 + 1 x 1/3) = 1/3. So your chance is always 1/3 of being right...or 50/50 of making the switch. If you throw in the bit of Monty having a choice of making the offer at all, then it comes down to how much you trust someone in a plaid jacket.
The thing confusing me is.....after the door is eliminated, how is there a better chance the prize is in the other door you didn't choose over the one you did choose? I just don't see how eliminating one door makes one remaining door more likely than the other. Pugs
lol. you think by adding a second person, it is an extension of the problem when in fact it changes the dynamics of it entirely. you assume probability to be constant, when in fact it changes depending on your perspective. For example, yesterday Rox had a 50% chance of winning game 5 but today they have 0% chance of winning game 5. For example, I tell person A that I have a rock in one hand. I tell person B that I have a rock in one hand and it's not in the left hand. A has a 50% chance of getting it right while B has a 100% chance of getting it right even though it's the same situation BECAUSE the INFORMATION each has is different. In your situation, the contestant and the audience member has a different amount of information and time frame in which to choose, that plays a crucial role in the percentages.
argh! You must see that the host revealing an empty door does nothing to the problem. Think of the problem and instead of the host showing the empty door, he says, "one of these two doors does not contain the prize." THAT is essentially what the host is saying when he opens up a door he knows to be empty. That statement is always true. It does not change the fact that you have a 1/3 chance of picking the right door. and the fact that the other two doors have a 2/3 chance of being the right door.
