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New Premium Module: Tyrants of the Moonsea! Read More

Attention, new and old users! Please read the new rules of conduct for the forums, and we hope you enjoy your stay!

JonSnowIsAlive
Member Posts: **196**

Since the wait for the legal dispute to be resolved is so maddening, I thought starting a new thread about riddles/problems might be fun (fun fun fun actually)!

Some of these riddles will naturally be more mathematically challenging, so I would like to state beforehand that there is absolutely no shame in not being able to solve some or all of them. In fact there is no shame in not caring at all!

**First riddle:**
**Answer to the first riddle:**

Here's the easiest way I know to show this:

First of all, let's denote the total sum by S, that is

S = 1 + 2 + 3 + ... + n - 2 + n - 1 + n.

We can rearrange the terms in backward order without altering the sum's value:

S = n + n - 1 + n - 2 + ... + 3 + 2 + 1.

Now, if we were to add those two we would obtain:

Thus, we can isolate S:

S = n(n + 1) / 2

And that's it!

**Second riddle:**
**Answer to the second riddle:**

Non-mathematical solution coming...

**Third riddle:**
**Answer to the third riddle:**

The north pole.

It would also be fun if people added riddles of their own as well. Will add more myself if the thread actually piques people's curiosity, if not... well... that's still 5 less minutes of waiting for me!

P.S. Sorry for the grammar...

EDIT: to add answers, to add riddles, and to better format the post...

Some of these riddles will naturally be more mathematically challenging, so I would like to state beforehand that there is absolutely no shame in not being able to solve some or all of them. In fact there is no shame in not caring at all!

What is the sum of the firstnintegers (starting from 1, not 0), i.e. show that

Here's the easiest way I know to show this:

First of all, let's denote the total sum by S, that is

S = 1 + 2 + 3 + ... + n - 2 + n - 1 + n.

We can rearrange the terms in backward order without altering the sum's value:

S = n + n - 1 + n - 2 + ... + 3 + 2 + 1.

Now, if we were to add those two we would obtain:

S + S | = | (n + 1) + (n - 1 + 2) + ... + (2 + n - 1) + (1 + n) |

2S | = | (n + 1) + (n + 1) + ... + (n - 1) + (n - 1) |

= | n * (n + 1) |

Thus, we can isolate S:

S = n(n + 1) / 2

And that's it!

What is the expected number of coin flips required to obtain your first 2 consecutive heads? for 3? for 4? Is there a patern?

Non-mathematical solution coming...

You're standing in a rectangular cabin which has one window per wall. You stare through all 4 of them and realize you've been looking south every time. Where are you?

The north pole.

It would also be fun if people added riddles of their own as well. Will add more myself if the thread actually piques people's curiosity, if not... well... that's still 5 less minutes of waiting for me!

P.S. Sorry for the grammar...

EDIT: to add answers, to add riddles, and to better format the post...

Post edited by JonSnowIsAlive on

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## Comments

15,98119619615,981From what I can see from the first riddle...

196Basically, if

n = 3, you'd have a total sum of1 + 2 + 3 = 6. Ifn = 4, you'd have a total sum of1 + 2 + 3 + 4 = 10. Ifn = 5, well, you get it. Show that for an arbitraryn > 0, the pattern is the one given in the OP.Keep the good work!

2,139It 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?

2,1393.

I pass on 2 for now.

4,681196@ajwz Here's a very complete answer to your third riddle:

Will now add answers to my original riddles and then add some more!

196Fourth riddle:P.S. This is a logic challenge. The context is irrelevant (e.g. sand could be replaced with mud, etc.).196@Jalily Your answer to #1 is correct. Your answer to #2 is also very insightful, though I will have to look back on your generalization. Cheers!

2,643196They can speak, but they can't turn their heads. Only one of them can speak, and only one. What he says (the color of his own hat) has to be the correct answer or they all get killed.

2,64353534th 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.

196196196You answer is not only right, it is very insightful. Damn this thread is cool! Cheers!

Also, to be completely honest, I don't like the normal distribution that much, its tail is way too thin. Try the Gumbel distribution for a change!

Jon

53we 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.