Hailstorm

http://info.med.yale.edu/genetics/ward/tavi/Trblesht.html Good luck. Also, what's the GC content you're attempting to clone? Are your primers good?

Something else, a drink manufacturer "Fuze" (I believe) in the US has HFCS with the red "no" sign through (like on a traffic sign), but one of the main ingredients is "crystalline fructose." Go figure. Link: http://www.fuzebev.com/refresh.asp also: http://www.bevnet.com/reviews/fuze/facts.asp

Another easy way to look at it is as follows: There are infinity +2 doors, there is one car behind one door, and one goat behind each additional door. If you choose any number, and then the infinity goats are removed, you're clearly better switching. Your chance of getting a car on your first guess is zero (since this is the math forum it is [math] 1/ (infinity+2)[/Math). Whereas the chance of you getting a car should you switch your choice is [math] (Infinity +1)/(Infinity+2).

Visual, four trials of each to make it easy: ||_| = Door one = Car = Trial one selection |_| = Door two = Goat = Trial two selection |_| Door three = Goat = Trial three selection |_| = Goat released In trial one: The door with the car behind it - First set of four lines In trial two: The door with a goat behind it - Second set four lines In trial three: The door with a goat behind it - Third set four lines Please note: your door selection is bolded ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins Hence: Switch wins 8/12 (2/3) and Stay wins 4/12 (1/3) The logical flaw seems to come in with the first selection. I felt for some reason that there should be four of each goat door opened (instead of two of each), but that would have left me choosing the first car door twice as often. My original question of whether or not he can remove a goat from the door you chose would only further bolster the viability of switching (staying on a door with a goat in a game-show setting would be simply stupid, whereas when scripting a program, it seems more reasonable). Viewing it as a 50/50 chance would require: ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins But when looking at it, you'd have to choose the door with the car in it (in this case door #1) far more often than random chance allows (100% more likely). In order for a true 50/50 chance, you would have to be able to see the car... or perhaps smell the goats with fairly good accuracy on the first selection only, but not the second. Perhaps the release of the first goat would skew goat-sense.