[Is this number divisible?]

16 and 17?

[April Fool's Day and the typewriter]

I think we can go a little higher 

4x3x5x7x11x13 = 60,060 should work - as long as the individual cycles are co-prime it won't repeat until the product of the cycle lengths. 

For instance with 7 keys 2 3 4 1 6 7 5 shouldn't repeat for 12 cycles, indeed

2341675
3412756
4123567
1234675
2341756
3412567
4123675
1234756
2341567
3412675
4123756
1234567

So we start taking the first however many primes whose sum is <=n then with whatever keys we have left over try and replace primes with the most cost-efficient prime powers but I didn't go very far into that

 

 

[Three Gods]

What happens when you ask the truth teller or the liar a question they can't answer?

[Which treatment]

I give my funding to treatment B as long as they promise to stop testing such a disproportionately small number of women.

[A simple Monty Hall variant]

Let's say Monty's playing fairly and you stay with one of your doors from the first round. If Monty picks one of the two unpicked doors (7/11 chance) there's a 3/7 chance the car's behind the other unpicked door, and a 2/7 chance it's behind your door or the other door from the first round. Say he opens the other door from the first round (4/11) then there's a 3/8 chance it's behind either of the unpicked doors and a 2/8 chance it's behind your door. So at best a 3/7 chance of winning, at worst 3/8.

This is the same if you switch in the second round except you have a 4/11 shot at the 3/7 chance of winning and a 7/11 shot at the 3/8, so it's still the best play to stay with one of your doors.

You would have the upper hand if you knew Monty always tries to go for the other picked door from the first round, not sure if there's any way Monty can intentionally make the play harder though.

[Golden Ticket]

15,000

How?

I figured you have a 5/100000 chance of getting the ticket with each candy bar, so after buying x candy bars the probability you won't get a ticket is .99995^x.

[Golden Ticket]

27,726?

[Russian roulette]

Assuming, since one is left handed and the other right, that they spun the cylinder in opposite directions the one who spun the cylinder in the same direction as it normally rotates fires his bullet on the second shot, the one who spun it counter to how it rotates would fire on the sixth shot if he wasn't already dead.

[How Many Intersections?]

Using the notation in the link:

1/[d₁d₂csc(θ₁)]+1/[d₁d₃csc(θ₂₎]+1/[d₂d₃csc(θ₂-θ₁)]

or

sin(θ₁)/(d₁d₂) + sin(θ₂)/(d₁d₂) ₊ sin(θ₂ - θ₁)/(d₂d₃)

As the area of the plane approaches infinity one intersection corresponds to one 'block' or 'section' of the grid made by two intersecting sets of parallel lines. So calculating how many of those 'blocks' from each pair of intersecting sets fit into one unit of area I got the above equation.

[Analyze this: A simple card game]

Here's what I got (through exhaustion using what little I know of java) to be the number of starting positions from which you reach the end in the maximum number of steps for T1-8, and the total number of starting positions that will reach the end regardless of how many steps for T1-6:

T1: 1 1

T2: 1 1

T3: 1 1

T4: 3 7

T5: 16 42

T6: 65 239

T7: 293 ??

T8: 1267 ??

I considered a starting position to be one that couldn't be the result of another turn, for example [1, 1, 2, 2, 3] is a starting position but [1, 1, 2, 5] is not.

The starting positions that will get you to the end in the maximum number of steps also aren't all with the minimum amount of decks.

[Three intersecting cylinders]

Is the three cylinder problem also doable without using calculus?

[Elevators problem]

Have on elevator positioned between floors 3 and 4, the other between floors 9 and 10. As soon as one elevator gets a call the other moves into position between 6 and 7 until the first elevator is done.

[Bag and balls revisited]

Probability of the added ball being red would be 2r/(3b+2r) where r and b are the initial probabilities of the ball being either red or blue.

Forgot exactly how I got there but here goes.

Let n be the number of balls in the bag, then the possibility of selecting any one ball after adding a red ball is r*1/(n+1), similarly the possibility of selecting any one ball after adding a blue ball is b*1/(n+1). If you add a blue ball there are 3 possibile scenarios where you draw a blue ball, with red there are two so 3b/(n+1) + 2r/(n+1) is the probability you'll draw a blue ball at all.

Therefore the ratio of the probability of drawing blue after red to drawing blue at all is 2r/(n+1) : 3b/(n+1)+2r/(n+1), assuming a blue ball is drawn by setting the latter half of the ratio to equal one and simplifying we get 2r/(3b+2r) : 1.

[Bag and balls revisited]

Taking a hint from benjer. There are five possible scenarios where we draw a blue ball: A red ball is added and you draw blue1 or blue2, or a blue ball is added and you draw blue1, blue2, or the newly added ball. So a red ball is added in 2/5 scenarios where a blue ball is drawn, therefore the chance is 2/5.

[Nice puzzle !]

1

2 1

2 9 1

2 9

2 1 9

1 9

1 5 9

6 1 5 9

6 1 2 5 9

1 2 5 9

1 2 8 5 9

1 2 8 9

1 2 5 8 9

6 1 2 5 8 9

6 1 2 3 5 8 9

1 2 3 5 8 9

1 2 3 7 5 8 9

1 2 3 7 8 9

1 2 3 6 7 8 9

1 2 3 6 7 8 9 5

1 2 3 6 4 7 8 9 5

1 2 3 4 7 8 9 5

1 2 3 4 6 7 8 9 5

1 2 3 4 6 7 8 9

1 2 3 4 5 6 7 8 9