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## Comments

16,1510016,151From what I can see from the first riddle...

02,164It is easy to show this is true for n=1:

1*2/2 = 1

Assume it is true for n=k; that is:

1+2+3+...+k = k*(k+1)/2.

Then it must also be true for n=(k+1):

1+2+3+...+k+(k+1)

= k*(k+1)/2 + (k+1)

= k*(k+1)/2 + 2*(k+1)/2

= (k^2 + 3*k + 2)/2

= (k+1)*(k+2)/2 which is the similar formula for (k+1)

QEDSo if it is true for k, then it is also true for (k+1) and therefore for (k+2), etc. But since it is true for 1, it must be true for 2, and for 3, and for 4, and ... up to infinite.

1,050Well, if we ignore variables placed onto it all by the throw... We are really left with 3 possibilities of which one is quite rare... It can land on Heads, Tails or on it's side, though the side will require it to be leaning against something... If we assume it's a flat surface that can't really happen...

For chances of getting heads, it's 1 out of 2, so you for one heads you would need 2 flips...

I don't think there's really a reliable method of calculating how many it would take to get x amounts of heads in a row.

180Is my name Booker DeWitt?

/scoffchortle I use wit to disguise a mental shortcoming.

4,681Now suppose the equation is true for some value m. If we can prove that the equation holds for m+1, we can prove that it is true for all positive integers because we've just proved that it is true for the starting case of m=1.

Original: 1 + ... + m-1 + m = m(m+1)/2

Add m+1 to each side: 1 + ... + m-1 + m + m+1 = m(m+1)/2 + m+1

Make right side a single fraction: 1 + ... + m-1 + m + m+1 = [m(m+1)+2m+2]/2

Multiply amd simplify: 1 + ... + m-1 + m + m+1 = [m^2+3m+2]/2

Factor: 1 + ... + m-1 + m + m+1 = (m+1)(m+2)/2

This is also the result you get if you simply substitute m+1 for m in the first equation. In other words, if the equation is true for m, it's true for m+1. Since we know it works for n=1, it must also work for n=2, and therefore for n=3, and so on.[/spoiler][spoiler=Second riddle]Let's say the expected number of flips is x. If we flip once and get a tail (0.5 chance), we have x+1 flips left since the tail doesn't put us any closer to two heads. Now let's say we flip a head (0.5 chance). On our second flip, we can get either a head or a tail. If heads, we're done (2 flips); if tails, we have to start over (x+2 flips).

Add up the probabilities and solve for x:

x = 0.5(x+1) + 0.25(2) + 0.25(x+2)

x = 6

So the expected number of flips is 6.

For 3 in a row, we do a similar thing.

- Tail on first flip: 0.5 chance, x+1 since have to start over

- Head on first flip:

--Tail on second flip: 0.25 chance, x+2

--Head on second flip:

---Tail on third flip: 0.125 chance, x+3

---Head on third flip: 0.125 chance, 3 (done!)

x = 0.5(x+1) + 0.25(x+2) + 0.125(x+3) + 0.375

x = 14

For 4 in a row... eeeh.

x = 0.5(x+1) + 0.25(x+2) + 0.125(x+3) + 0.0625(x+4) + 0.0625(4)

x = 30

In order to calculate expected flips for n in a row, you have to solve for x = [the sum of (0.5^m)(x+m) for m=1 to n] + (0.5^n)(n). The pattern appears to be 2(1-0.5^n)/0.5^n.

- You get the numerator by summing m*0.5^m for m=1 to n and then adding m*0.5^m again (for the term without an x).

Original: (0.5 + ... + n*0.5^n) + n*0.5^n = 2(1-0.5^n)

Initial case (n=1) : 1 = 1

Extend left side to n+1 by subtracting n*0.5^n and adding 2*(n+1)*0.5^(n+1), then do the same to the right side to balance the equation: (0.5 + ... + n*0.5^n) + (n+1)*0.5(n+1) = 2(1-0.5^n) - n*0.5^n + 2*(n+1)*0.5^(n+1)

Distribute the second 2: (0.5 + ... + n*0.5^n) + (n+1)*0.5(n+1) = 2(1-0.5^n) - n*0.5^n + (n+1)*0.5^n

Combine terms on right side: (0.5 + ... + n*0.5^n) + (n+1)*0.5(n+1) = 2(1-0.5^n) + 0.5^n

Keep combining: (0.5 + ... + n*0.5^n) + (n+1)*0.5(n+1) = 2-0.5^n

Now pull out the 2: (0.5 + ... + n*0.5^n) + (n+1)*0.5(n+1) = 2(1-0.5^(n+1)

Hey, it works for n+1! So this must be the numerator.

- You get the denominator putting x on one side of the equation; you divide by its coefficient.[/spoiler]

4,1222. What is the only number whose letters are written in alphabetical order (for example 2 = two: t is before w in the alphabet, but w is after o, so two doen't work)

3.

0 to the power n is always 0 for any n

n to the power 0 is always 1 for any n

0 to the power 0 has no solution, but if it did have a solution, would it be closer to 0 or 1?

2,1643.

I pass on 2 for now.

4,6810002,64302,64353534th guy is able to see 2nd and 3rd and if their hats same he would know his own hat because of there is only 2 black and 2 white hat. and if he is not able to talk, 2nd and 3rd guy should have different hats. so 3 guy could talk after 4 minute and 59 seconds as a last resort.

the thing is 3rd guy can see 2nd guys hat and if he can trust that 4th guy is not a moron then he can talk!

bu there is the possibility to hearing 4rd guy saying "hodor, hodor hodor" and they are all dead

1,6673 will know his hat is different from 2's if 4 doesn't answer.

6,4561,050And I wonder... What kind of hats are they? As even though they can't move their heads, they can still move their eyes and look up to see their hat unless it's a tiny thing on the top of their heads they can't see...

6,4561,667Also, it's got to be up to the person asking the question to say whether an answer is correct. I'm not really interested in seeing the other answers until I know if I'm right or not. So instead of waiting to see if @zurathan is correct, I posted my own answer.

161As for the second:

Assuming a fair coin, the number of flips needed for a single heads is a random variable with a Geometric Distribution (with p = 0.5), so the expected number of flips needed is 2 (EV of a Geometric Distr. is 1/p).

For two consecutive heads, we can start with the same Geometric random variable and flip an additional coin. If it's heads, we're done. If it's tails, we need to repeat the process. Interestingly, the number of times we need to repeat the process is itself a Geometric random variable (with p = 0.5). The expected number of repetitions we need is therefore, again, 2. The expected number of flips 'inside' each repetition is 3: 2 from the 'inner' Geometric variable, plus the additional flip. Those 'inner' Geometric variables are independent of each other and of the 'outer' Geometric variable, so the expected number of flips = 3*2 = 6.

For three consecutive heads, we can simply repeat the same principle: get to two consecutive heads, then flip an additional coin: again, the number of repetitions is a Geometric random variable, EV = 2; the 'inner' variables have EV = 6+1, so the expected number of flips needed to get 3 consecutive heads is 14.

This progresses quite predictably:

4 heads: 30 flips

5 heads: 62 flips

6 heads: 126 flips

This, perhaps predictably, is a (partial) geometric series, so in general we have

N heads: 2^(N+1) - 2

There's probably a more mathy way to prove this, but it's been a while since I've worked with Geometric distributions (Normal distribution FTW!), so I'm a bit rusty.

00053we have four numbers and any calculator operation like +, -, squareroot, power,..... the goal is founding 100

and there were two example too; (5+5)x(5+5) and (99+9/9)

oops forgot to say the four numbers we given in this puzle was 7

this puzzle has an other version too; this time we have five numbers 1,7,7,7,7 and limited operations to +,-,/,x and ofcourse ( ) parenthesis.