bonanova

Cna yuo raed tihs? Olny 55 plepoe out of 100 can. i cdnuolt blveiee taht I cluod aulaclty uesdnatnrd waht I was rdanieg. The phaonmneal pweor of the hmuan mnid, aoccdrnig to a rscheearch at Cmabrigde Uinervtisy, it dseno't mtaetr in waht oerdr the ltteres in a wrod are, the olny iproamtnt tihng is taht the frsit and lsat ltteer be in the rghit pclae. The rset can be a taotl mses and you can sitll raed it whotuit a pboerlm. Tihs is bcuseae the huamn mnid deos not raed ervey lteter by istlef, but the wrod as a wlohe. Azanmig huh? yaeh and I awlyas tghuhot slpeling was ipmorantt! if you can raed tihs be aifrad - be vrey arifad!

You're right ... they are told there are two color groups. The OP says ... among other things ...

SPG - yeah ... that's it. Kudos also to joaopsr and scuttill And to Itachi-san for the best out of the box answer.

Hi Lightz, Your post was long enough and detailed enough to expose where you went wrong. Don't mean to call you out, but you were pretty dismissive of the correct analysis and those who have explained it many times. Your logic and your arithmetic are absolutely correct. Yea! The only problem is your premise is incorrect. Oops! Take a look: The OP does not permit ruling out the Boy, Girl case. [only Boy, Boy.] Quite simply then, there are three cases, not two; and the probability is 1/3, not 1/2. OK? There are a couple rather obvious points you might have thought of before saying that: [1] the OP states that you can assume the chance of a boy or girl birth is 50%. What kind of a riddle would it be if the question then posed was simply: what are the odds that the 2nd child is a girl? [2] what kind of people do you think come to this site to post and answer riddles who would not immediately say it's 50% if that were the question being asked? Hint: when it seems to you that a whole lot of fairly smart people have got an answer wrong, it might be time to consider whether you got the question right. Either way, right or wrong, respect for another's opinion is always appropriate.

Do NOT mess with this lady ... she is good.

Why don't you ever see the headline "Psychic Wins Lottery!"? Why is "abbreviated" such a long word? Why is a boxing ring square? Why is it called lipstick if you can still move your lips after you use it? Why is it necessary to nail down the lid of a coffin? Why is what doctors do called "practice?" Why is lemon juice made with artificial flavor, and dishwashing liquid made with real lemons? Why is the third hand on a watch called a second hand? Anyone who has an answer for any of these please do not post it. Life's more fun just as it is...

Posted and answered many times. Locked.

Nice puzzle. Camel version here. Locked.

A woman weed-whacking her yard accidentally cut off the tail of her cat. She picked up the tail and the cat and drove to Wal-Mart.

It's true that there are two cases: [1] both are girls and [2] there is one of each. One of those cases [the first] is the favorable one. The OP asks, what is the probability of [1]? You answer that it must be 1 out of 2 because 1 case out of 2 cases is favorable. What I want to help you to see is that counting the number of cases gets you into all kinds of trouble unless you do one more thing. To see what that is, try this thought experiment: You buy a lottery ticket. How many outcomes are there? The answer is two: [1] it's the winning ticket, or [2] it's a losing ticket. What is the probability your ticket is the winning ticket? Count the cases: 2. Count the favorable cases: 1. Therefore your probability of winning the lottery is 1 out of 2. Ooops. When I asked you if you believed that mixed-gender 2-child families and 2-girl 2-child families had equal likelihood, I wasn't concerned about wording or how you described the boy-girl or girl-boy order. I was trying to help you remind yourself that those probabilities are not equal: p[girl-girl] = 1/2 p[girl-boy, boy-girl]. The lottery ticket experiment is an easy way to see that counting up cases that aren't equally likely leads you into error. The reason that it's helpful to consider the boy-girl and girl-boy cases separately is that then you can count up all the cases: [1] GG, [2] GB [3] BG. Why? Because these three cases have equal probabilities. And since they are the only possible cases, their probabilities add up to 1. p[GG] + p[GB] + p[bG] = 1. Now do the math, and find p[GG]. Does that help?

More like a mad scramble ... read posts > note the incorrect guesss > the solvers > update OP > read more posts > iterate. None of the previous ones were finished under an hour, but other than that, didnt keep previous times.

Nikyma closes it out with [12] Modern day record ... all words found in less than one hour. Great job all!

Itachi-san has [1] One to go ... [12].

Two more have been found - [7] -congrats to Enlightenment and wakejon and [4] - Congrats to Nikyma Just two to go.... [1] and [12]

Going fast - great job!!! Ten of the fourteen have been found: Nayana has [2], [3], [5], [6] and [8] Itachi-san has [5] , [8] and [11] Enlightened has [9] Nikyma has [8], [10] [13] and [14] <- the "difficult" ones Going to update the OP.

New to the game? Rules and previous chapters are here. Enjoy! Puzzle completed in 47 minutes! Kudos to the solvers [in bold below]! [1] cast _ _ _ toss - Itachi-san [2] cast _ _ _ couch - Nayana [3] cast _ _ _ _ opt - Nayana, Itachi-san [4] cast _ _ _ _ chop - Nikyma [5] cast _ _ _ _ musty - Nayana, Itachi-san, wakejon, Nikyma [6] cast _ _ _ _ fashion - Nayana, Nikyma [7] cast _ _ _ _ parcels - wakejon, Enlightened, Nikyma [8] cast _ _ _ _ _ oust - Itachi-san, Nikyma [9] cast _ _ _ _ _ afghan - Enlightened, Nikyma, Nayana [10] cast _ _ _ _ _ _ elite - Nikyma, wakejon [11] cast _ _ _ _ _ _ impute - Itachi-san [12] cast _ _ _ _ _ _ company - Nikyma [13] cast _ _ _ _ _ _ _ _ _ _ arrival - Nikyma [14] cast _ _ _ _ _ _ _ _ _ _ handcuff - Nikyma, Itachi-san

Not so ... Chuck Norris = Chronic Rusk, for one; Urchins Rock for another.

What she said. ... <_<

Not really. The OP is a physics problem pure and simple. OK I'll meet you half-way. It's a physics riddle.

It goes back to my first answer: the total weight of the system comprising the woman and two balls is 120 lbs. If that entire system attempts to cross the bridge together, it will fall into the chasm unless the bridge keeps it from falling. To do that, the bridge must exert an upward force [on average during the crossing] of 120 lbs. So says Sir Isaac Newton. So if there is ever a moment when the exerted upward force is less than 120 lbs, [a ball is in the air] there must be another moment when it's greater than 120 lbs - the average must be 120. And if at any moment it exceeds 112 lbs, the bridge will fail. So the entire system cannot cross at once. But, 100 feet is not that far. Throw one of the balls across the chasm; then carry the other one across. There are other analogies. Soldiers break cadence when marching across bridges to keep their instantaneous weight within the capacity of the bridge. Bouncing up and down in an elevator by a near-limit group of passengers can create a memorable result. And so on. The weight that needs to be supported can be reduced momentarily, but only at the price of increasing it later on.

Hi Billybob, You're thinking along the right lines, but it's either [1] a little more complicated, or [2] it's a little simpler. Here's what I mean by that. [1] The 6 in your equation probably means to you the length of the hole. If so, your formula has the units of length. To express volume, you need an expression with units of length cubed: length x length x length. Doing it this way, you'd need the volume of the original sphere [4pi/3] R3 where R is the sphere's radius, and then subtract the volume removed by the drill. That's the complicated part. If you can visualize it, there's a cylinder of length 6" and two spherical "caps": The volume of the cylinder is easy; the caps are a little harder. Find them here. Then it's just arithmetic. [2] The easier way is to recognize that the problem did not say how big the sphere was. So you can infer it doesn't matter how big it is. So you choose an easy size to work with: a diameter of 6"! Then if you drilled a hole with the smallest imaginable diameter, you'd have a 6" hole that removed no volume at all! So the remaining volume would simply be the original volume = [4pi/3] 33 = 36pi cubic inches. If a picture would help, look at this post or simply http://brainden.com/forum/uploads/monthly_02_2008/post-3375-1203670267.png. Hope that helps. This puzzle has challenged a lot of smart people. - bn

It's a clever solution - not original with me - and very satisfying to see how it works. Sometimes I can explain things better than I can analyze them.