A test of your common sense

Taz writes:Unless I'm mistaken, a quick run through in my head says the ball costs 5 cents. But I have a feeling this is not the answer because there's a trick I'm not catching.

The surprising answer is that, if you call I have the Ace of Spades, it is more likely that you have another Ace than if you just state I have an Ace!
If you state you have the Ace of Spades there is an 11686 / 20825 chance you have (at least) another Ace, which is 56.12%. If you state you have An Ace then the chance you have (at least) another Ace is 5359 / 14498 (which is 36.27%). It’s over 50% less likely!How come? Well, this puzzle is more about a confidence problem than a probability problem. It’s about the disclosure of information.

Tangle writes:Clear? I'm not.

First a few basics:Total_Number_Hands All possible ways to combine the cards to make a Bridge hand. There are 52 cards in a deck, and 13 cards in a hand = b{52,13}No_Aces The number of hands that have no Aces. There are 48 non-Ace cards, and 13 in a hand = b{48,13}At_Least_One_Ace The number of hands that have at least one Ace = Total_Number_Hands - No_AcesExcatly_One_Ace The number of hands that have just one Ace. (Chose any Ace, then choose 12 non-aces) = b{4,1} x b{48,12}At_Least_Two_Aces Start with the total number of hands, then subtract hands with either no Aces, or one Ace. You are left with the number of hands with two or more Aces = Total_Number_Hands - (No_Aces + Exactly_One_Ace)We now have enough to calculate our first ratio (chances of having a second Ace if you state you have an Ace):At_Least_Two_Aces / At_Least_One_Ace = 36.27%To calculate the ratio where the person claims to have the Ace of Spades, we need some additional terms:With_Ace_Spades Number of hands with Ace of Spades. (We choose the Ace of Spades, and then have to choose an additional 12 cards from the remaining 51) = b{51,12}No_Other_Aces Number of hands with no other Aces = b{48,12}At_Least_Two_Aces_(Including_Ace_Spade) Number of hands with at least two Aces, one of which is the Ace of Spades = With_Ace_Spades - No_Other_AcesWe now have enough to calculate our second ratio (chances of having a second Ace if you state you have the Ace of Spades):At_Least_Two_Aces_(Including_Ace_Spade) / With_Ace_Spades = 56.12%

We have a long thing made from a hard material that is designed to only withstand 500 pounds. The long thing is held up at the ends by two 10-96's. 2 insanely overweight women, each weighing 500 pounds, decide to sit on two points L distance from the ends and the women are also L distance from each other.Where on the long thing will the crack or breaking occur?

Taz writes:
We have a long thing made from a hard material that is designed to only withstand 500 pounds.

xongsmith writes:P.S. - what's a 10-96? Is that some kind of cinder block?

It's cop talk.

Just to jump in here.
I'm a structural/stress engineer and deal with this kind of problem regularly.
For clarity (and correct me if I'm wrong here Taz)
The triangle at point A represents a "Simple Support" [edit: a simple support allows rotation but not translation] and the circle at point B represent a support that allows a horizontal translation and prevents vertical translation [edit: and allows rotation].

Message 39: He has perfectly shown how people with little or no knowledge of the subject at hand still feel they can wade in and point out what they see to be deficiencies in the diagram and therefore dismiss the question as nonsense.
When in fact to someone who knows the subject matter, the diagram is clear, consise and as has been shown in this thread, makes the point that Taz wanted to make quite well.

RAZD writes:Indeed. This is why learning the technical "language" of a field is critical if you are going to discuss it intelligently.

And no, I do not deal with such things when I walk across a bridge.