I don't think this is the right explanation. They each gave $10, and were refunded $1. So its $9 from each man spent. The question at the end is wrong. $27 was spent by the 3 men, $25 of which went to the restaurant and $2 to the waiter. There is no extra dollar in the equation. Breaking it down further: Code: Customers Restaurant Waiter $30 $0 $0 $0 $30 $0 $0 $25 $5 $3 $25 $2 So, customers originally spent $30, but were refunded $3 so ultimately they spent $27, which equals the amount gained by the restaurant and the waiter.
No no no no no. This is one of my favorites, has been for years. People almost have it right. The ROOM was $25, not $30. The guys paid $25 for the ROOM. They basically also paid $2 to the bellhop as a tip. They DID in fact pay $27 total - 25 +2 = 27. So they did in fact pay $9 each in reality for everything. They paid $30 originally, the bellhop DID give them each $1 back, so they DID pay $27 total out of pocket between them. The trick is to SUBTRACT the $2 from $27, (to reach $25, the price of the room), not to try to ADD the $2 to the $27, which is what the question tries to push you into thinking you should do.
Impossible. There is always some part of one of the runners that is in front of the other. Down to the micro, pico level there is at least one quark in front
<br> I don't understand this. Don't basic probability rules state that the gender of your children is exclusive and does not affect one or the other? Wouldn't it just be 1/2 chance it's a girl?
No, if one's a boy, then there are 3 possibilities: boy - girl boy - boy girl - boy So, the probability that "the other" is a girl is 2 out of 3.
I'm interested in a statisticians formal answer as well, but the math that gets you 2/3 is as follows. There are 2 kids. each kids can either be a boy or a girl. So your options are: Boy Boy Girl Girl With the 2 kids, its obviously 50/50. No you know what one of them is, a Boy. So take away one boy, and you still have. Boy Girl Girl 2 out of 3 chance that the second one is a girl. While each child is independent 50/50, on the combined basis you know what your choices are, so it has to be 2/3... ...not sure if right, but that's my logic
<br> It's been a while, but i've taken a few logic classes and i'm pretty sure you're wrong about this. <br> I could be speaking from ignorance here, but I think there's a difference between something like "what is the sex of my next child" and "what is the probability i pick a yellow and blue card from a deck that contains only those colors" <br> In the first instance, the sex of one child has no bearing on the sex of another. It does not matter that one is a boy or a girl. The chance that the next one will be a girl is still 50%. You can't say the possibilities are boy-girl, boy-boy, girl-boy. But if I'm speaking about drawing a yellow and blue out of a deck that contains only those colors, my possibilities can be: Yellow-blue blue-yellow with the other possibilities being: yellow-yellow blue-blue <br> In the first instance, order does not matter, but in the second instance it does.
No, the probability that the two children = 1 boy and 1 girl is twice the probability that its 2 boys. So that needs to be counted twice. Hence, 2 out of 3, not 1 out of 2.
There are two children. We know that there's a 1/4 chance that they're both boys, a 1/4 change that they're both girls, and a 1/2 chance that its one boy and one girl. Agreed? Then, if you're told that one is a boy, that eliminates the case where they are both girls. So, we're left with 2 cases -- both boys, and one boy and one girl, but the latter is twice as likely. So the answer can't be 1/2. It is 2/3.
<br> I think this is a true paradox, because I just cannot agree to that. <br> I just do not think order matters in this instance at all. Each child has to be treated as an independent/exclusive probability. It is 1/2 every time. Apparently, this has caused great debate over the years. Just found some wikipedia info on it http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
Let me rephrase the question in terms of flipping a coin, so you can verify this for yourself. I flip a coin twice. One of the flips turned up Heads. What's the probability that the other flip turned up Tails? You can verify this yourself experimentally. Take a coin, and flip it twice. Record what the result of the two flips is. Now, repeat that N times. Then out of the N+1 trials, pick out all the times in which at least one flip was Heads. Now, from that resulting sample, what percentage is the other flip Tails? You'll find that as N increases, the answer should converge to 2/3, not 1/2.
No one ever asks these questions .... all they ever want to do is create crazy riddles that never really happen. In the meantime we got whackos running around hanging people on country roads. Wait, I think I just contradicted myself.
