Aside from the bots, it seems offtopic has been a bit too on-topic lately. Therefore, I present a puzzle, possibly to be followed by more if someone gets this one.

You are flipping an unbiased coin and looking for a certain sequence of results. For example, if you were looking for tails-tails-tails, you would keep flipping until you got three tails in a row.

On average, does it take more flips of the coin if you are looking for the sequence heads-tails-heads or for the sequence heads-tails-tails or does it matter?

Explain.

considering flipping a coin is a 50 50 chance, i feel like it shouldnt matter and that on average it would take the same number of flips (considering you performed the experiment numerous times). but that seems to easy to be the answer.

It is easier to get heads-tails-tails, just cause

you're correct.
Excellent explanation, sir.

It would take more flips to get heads-tails-heads than heads-tails-tails

to fail and have to start looking for the sequence again, both have the same condition for the second flip: it must flip a heads.

However, on the last flip, the first sequence fails by flipping a tails, while the second fails while flipping a heads. When the first fails, you need to start ALL over again and start with a new heads. When the second one fails, you start over and already have the first value for your sequence. Therefore, the second sequence should take less flips.


BOOOOOM

in short yeah rhk is totally right

I feel stupid.

Here's a riddle:
A man stops at a hotel. He then loses all his money. How did this happen?

(And no google you cheaters)

thats easy: monopoly. and probably park place.

Jerk. Why do you have to be good at this?

Ha, I got one. What's nokills name? I win!

well um i'd tell you but then he'd get mad.

(if you'd listen to the podcast you'd know I found out)


EDIT://
Argentina >.>

Hehe I know it too.


Here's one: Three brothers share a family sport. An endless marathon. They race every day and last the whole day long, the first of them is short and stout, the second tall and thin, the third is tiny, short and thin but he is sure to win. What are they?

They're Mexican!!!!!!

There are n wizards held in a prison, and they are collectively offered a game for their individual freedom or death. Before the game begins, they are told the rules and allowed to agree on a strategy. The next day, the game begins. They are arranged in a line with each wizard wearing a hat. Each hat is either black or white, and the hats are distributed randomly. There can be any number of each type of hat; their numbers aren't necessarily equal. Each wizard can see all the other wizards and their hats, but not his own. Once the wizards are in the line and have their hats, they are not allowed to rearrange themselves. One by one, the wizards are asked the color of their own hats. They respond by saying either "black" or "white", and all the other wizards can hear this response. The response carries exactly one binary bit of information; nothing else can be communicated by trickery such as intonation. If the wizard is correct, he leaves the line for freedom. If not, he is killed. Each wizard, being a wizard of great power, can examine all the others and come to a decision arbitrarily quickly. What strategy should they agree on so that the fewest, as a worst case, die, and how many is this?

The first guy is the only one with the risk. The strategy is this: the first guy will answer based on what he sees. If he sees an odd number of black hats, he answers black. If he sees an even number of black hats, he answers white. Say that he answers black. Based on whether or not the first guy gets eaten for his guess, the next guy knows that there are either odd or even black hats. He sees 8 black hats. So obviously that means that there is one more blàck hat out there, so it must be his. So he answers black. And so on, and so on. So everyone can get out alive, except for maybe the first guy.

I might run out sooner than I imagined... have you guys seen these before?


You are given two uniform glass balls and a 100 story building. These balls are identical and possibly much harder than normal, and they land the same way each time. Your goal is to determine the highest floor of the building from which they can be dropped without breaking. Once a ball breaks, it can not be used again for any more drops. Neither ball is required to survive the search. What is the search algorithm with the fewest drops as the worst case, and how many drops is this?

The correct answer is PIE!