unreality

one two three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen twenty twenty-one twenty-two (...) twenty-nine thirty (...) forty (...) fifty (I'm guessing one thousand right about now...) sixty,seventy,eighty,ninety one hundred Aha! one-hundred-and-one

Assuming there are 52 cards, four of which are aces, and that the cards are shuffled randomly and not in any order that we know about: drawing one ace: 4/52 If he did draw an ace, he takes it away, now there are 51 cards- and 3 aces drawing another ace: 3/51 (in other words, Alex is correct) In order to get both, it is AND, and 'and' is multiplication: (4*3) / (51*52) or 12/2652 or 1/221 Assuming the deck isnt touched, is still random, has 52 cards, etc: To keep drawing and draw two CONSECUTIVE aces, you WILL draw one ace. The chances to draw an ace on the next card is n/# where n is the number of aces left in the deck and # is the number of cards left in the deck. So now we have to figure out the average probability. Say Card 1 is an ace. We have a 3/51 chance. If it is not an ace, we have 3/50 for the next draw, etc. This method is nice for looking and understand, but not for getting the answer. We have to look at the overall positioning of the deck, saying its random, since nothing is moved or replaced between drawings. Now it is clear that the cards being randomly distributed is important. The chance of one particular ace being a particular spot: 1/52 The chance of another partciular ace of being in a particular spot (above or below): 1/52 Thus one would think that there are 1/(52^2) possibilities. But they would be forgetting that there are FOUR aces and TWO spots- above and below the first card. First the above and below: First card: 1/52 Second card: 2/51 (twice chance of first) however now we factor in that there are FOUR and then it would be down to THREE if one is drawn: First card: 4/52 Second card: 6/51 (3*2) thus 24/2652 or 2/221 Remember the first bet was 1/221 This only twice as likely... but neither are very likely. If he did go for the bet, to make it even and if he pays a penny ($0.01) to do it: 2/221 = 1/110.5 He should win $1.10 or $1.11 if he wins, depending on how you round am i close to right? lol ;p

Most Liar/Truthteller riddles with two (or three) directions/doors/paths/etc to go are based on this princriple, ie, asking one man what the other would say (in this case, the "other" man is the liar or truthteller of yesterday because he was different yesterday)

lol let me just break this down for anyone still clueless: every combination of ten binary flags is 2^10 or 1024. Thus ANY combination of ten 0s or 1s (or tails/heads) has a 1 in 1024 chance of being realized, or flipped, in other words. Thus 1111111111 isn't any more unlikely than 011010111. They have equal probabilities. Of course if you get the second one, you'll be like "okay" but if you get all 1's youll be like "whoa, crazy, what are the odds!" Just because it is a spectacular result. In this case, it was the goal which means it is desired to get.... and suddenly the odds of getting a non-all-heads are 1023/1024 and the odds of getting all heads are 1/1024 I have a question for you... Jonny has flipped: 101101 (6 flips so far) Albert has flipped: 0001000 (7 flips so far) What is the probability that Albert will get MORE 1's IN A ROW than Jonny?

So using 2 legs to pinch one's nose counts as not having them?

what page Martini?

your Trojan horse hint definitely pointed to the last answer given about painting the lead golden... so he shaved off some of the gold maybe... and the scales in those days were sucky, and couldnt measure micrograms? lol can I ask you a question: is the gold he returns PURE 100% gold? I.e. its not Pyrite, like i suggested, or the lead painted with gold leaf or paint, etc? Would lead 'expand' in the making of gold? Nobody knows since you cant do it and it would be really ****** to make up your own imaginary chemical outcomes and such. So it probably isnt something chemical like lead making more gold than its weight... is it?

the play on words, ya mean?

I dont know about moles and that (i'm just starting chemistry, gimme a break ) but maybe the gold was pyrite?

oh i thought it was the MOST in a row... and actually it IS a word

I know spanish really good but i'm not bilingual yet... and once I do become, (hopefully soon) i wouldnt have enough time to translate... sorry

(if reality is an illlusion...UNREALITY IS REAL!!!!!! MUHAHAHAHHAHAHA!!!!!!!!!!!!!!!!) The one unquestionable fact happened suddenly because something desired

Nope. Number 8 is actually: subbookkeeper

THE ANT-LEG-EATING DUNG MONSTER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! (with a belly only big enough for two legs)

any of you geniuses care to divulge? lol

it is a star shape... am i right?

lol are you being serious or are you just joking?

The one unquestionable fact happened suddenly because

the... uh... rune E!

I have already posted this... see Six Cups somewhere here in New Riddles in a page

yeah i was wondering where you went welcome back! (both you and martini had exactly 155 posts when i read that- w00t!)

all three are incorrect thus: box 1 - S or BOTH box 2 - N or BOTH box 3 - N or S now lets pick S for box 1 (this color means that is not possible to have 1 of each): box 1 - S box 2 - N or BOTH box 3 - N or S thus it must be S, BOTH and N if box 1 is S but say box 1 is BOTH: box 1 - BOTH box 2 - N or BOTH box 3 - N or S thus is must be BOTH-N-S there are two possibilities. So somone please correct me if i am wrong, but it could be either of these two... another way to come at that conclusion is that the only way to stay differnet is to keep the same order, just shift it by 1 or 2 to get different variations BOTH = B current variation: NSB shift one: BNS (matches my second solution in the above proof) shift two: SBN (matches my first solution in the above proof)

his second solution is the best by far IMHO... good use of the equilateral triangle... lol

yep

They would just give you one coin, the least expensive one