The solution given there differs slightly from the one <john barleycorn> posted. In his notation: ...>

Yes, variations are possible as long as the first move is not Harry bringing over his 800 :-).

Still, anxiously awaiting the alternate solution of *the mental giant* <morfishine> based on where the wording in the problem is not clear.

<al wazir>

Thanks for that explanation! It would seem <ughaibu> is right then, that the brothers could still carry the full weight while juggling (or carry more weight, as you've described, depending on the height).

Would there ever be a point at which juggling could possibly solve the issue? Suppose one of the brothers had SEVEN bars of gold and was ONE bar over weight. Since he's only got TWO hands, could he possibly exert the same force on the bridge while juggling as his twin brother who has THREE bars of gold and was ONE bar overweight?

Juggling SEVEN balls in slow motion
https://www.youtube.com/watch?v=ioZ...

Here's the set-up I'm thinking about:

< Two brothers approach a bridge. The older brother has THREE bars of gold, and the younger brother has SEVEN bars of gold. It just so happens that the younger brother's weight plus the weight of his SEVEN bars of gold is equal to his older brother's weight plus the weight of his brother's THREE bars of gold. They would like to carry their gold across the bridge. Unfortunately the bridge has a weight requirement, and it can't carry both of them at the same time. To add to their dilemma, it turns out that each brother is precisely ONE bar of gold overweight. They decide on a plan together, and the little brother crossed the bridge first with all of his gold in tact. But when it was the older brother's turn to cross with all of his gold, the bridge collapsed, even though he had followed the same plan. Why? >

They should seek help from Tim.

Esmeralda arrives at a river bank bringing with her a goat, a wolf, and a head of cabbage. The only available means of transportation is a little boat which will hold Esmeralda and no more than one of her possessions. The wolf will eat the goat if they are left alone together, but won't touch the cabbage. The goat, however, will eat the cabbage if it gets a chance. How can Esmeralda cross the river with all her possessions intact?>

Dang, that must be one big head of cabbage!

The goat, the wolf, and Esmeralda herself each weigh exactly 80 pounds. Just as the bridge in <zborris8>'s problem will collapse if it's overloaded by even a couple of ounces, the boat in Alcuin's problem will sink if it is loaded with even an ounce more than 160 pounds.

If you weren't bothered by the sensitivity of that bridge, you shouldn't worry about the size of the cabbage.>

This may well be so, however I didn't even read about the bridge.

I only ready about the cabbage/goat/wolf/woman puzzle.

But a couple people already had it solved, so I couldn't very well jump into it so much later with a solution.

I was left with the option of making a joke.

Now I don't know how other people go about solving this, but it occurred to me that if you outline your boundary conditions first, then they force you to conclude something that is required in the solution.

Since the wolf and the goat can never be left alone, *and* the cabbage and the goat can never be left alone, it is obvious that the goat must go first. And last! So if the goat must go first, and then go last, the goat must also come back.

I wonder if the folks who posted solutions used a different method?>

https://www.youtube.com/watch?time_...

I figured they couldn't all make it, so one must be sacrificed to the zombies while the others escaped. Maybe the janitor (easily replaced) or the professor (has already lived a long time and should have retired years ago anyway).

The 4 possible outcomes are

1. B B
2. B G
3. G B
4. G G

and they are equally probable.

We know #4 is not an option because one of the children is a B.

So the chance that the other child is a boy is included in #1, and the other is a girl is option #2 or #3.

So P(G) = 2/3.

There are 3 doors. Behind one of the doors is a banana peel. Behind another door is a dead rat. Behind yet another door is $5000.

You choose door #3. The probability of the $5000 being behind #3 is P = 1/3

Door #2 is opened to reveal the banana peel. Time to switch your pick to door #1, where P ($5000) = 2/3>

Dang. You guys have heard them all.

Suppose that you are worried that you might have a rare disease. You decide to get tested, and suppose that the testing methods for this disease are correct 99 percent of the time (in other words, if you have the disease, it shows that you do with 99 percent probability, and if you don't have the disease, it shows that you do not with 99 percent probability). Suppose this disease is actually quite rare, occurring randomly in the general population in only one of every 10,000 people.

If your test results come back positive, what are your chances that you actually have the disease?