[Friday Riddler] Meet for lunch

[Mike]

This week's riddlers: https://fivethirtyeight.com/features/wh ... for-lunch/

Riddler Express
On a lovely spring day, you and I agree to meet for a lunch picnic at the fountain in the center of our favorite park. We agree that we’ll each arrive sometime from noon and 1 p.m., and that whoever arrives first will wait up to 15 minutes for the other. If the other person doesn’t show by then, the first person will abandon the plans and spend the day with a more punctual friend. If we both arrive at the fountain at an independently random time between noon and 1, what are the chances our picnic actually happens?

Riddler Classic
On a standard chessboard, what is the largest number of each piece (work it out first for kings only, then for knights only, bishops, rooks and queens) that can be placed on the board so that none of the pieces attack each other? (You don’t have to do pawns!)

What is the smallest number of each piece that can be placed on the board such that every empty square is under attack?

Extra credit: For each question above, how many possible correct arrangements are there?

[Mike]

Riddler Express: the answer is .

I just graphed it out. Laid out the probabilities on a graph and calculated the area under the curve.

[Eliahad]

What were your axes, if you don't mind my asking?

[Mike]

The chess one is harder.

Starting with the first question, for all the pieces, I have a minimum... that is, I have a number that I can arrange to make it work. There may be a way to squeeze more in there, but I don't know yet.

Kings -- 16 minimum
Knights -- 24 minimum
Bishops -- 14 minimum
Rooks -- 8... I'm pretty sure the answer is just 8.
Queens -- 8... Once again, final answer is 8.

[Mike]

[Mike]

Alternately, I suppose one axis is the time I arrive (from noon to 1pm), and the other axis is the time YOU arrive (from noon to 1pm), and then you color in all the parts of the graph that would make a successful meet-up and calculate the area of that. Different graph, same area.

[Eliahad]

Okay, so like, if you arrive at Noon, there's a 25 percent chance that the friend would arrive in the window. But at 12:01 that chance goes up slightly, because what if the friend showed up at Noon? or 12:16?

[Mike]

Correct. The shortcut I took is that I know that if I arrive at noon, chances of success are .25. I know that if I show up at 12:15, chances are .5. Chances remain a constant .5 from 12:15 to 12:45. Chances then decrease after that to .25 again at 1pm.

[poorpete]

Yeah it's at 50% chance or 30/60 chance from 12:14 to 12:45.

As in at 12:14 you'll catch your friend if they arrived between 12:00 and 12:29.

So for for half of the time there's half a chance. Then it decreases before and after that interval until you are arrive at 12:00 or 12:59, which gives you only a 25% or 15/60 chance.

Update: what Mike said.

I always have issues with putting the probabilities together. Because it's not addition nor mulitiplication right? I/2 time to do this or 1/2 time to do that is still 1/2 time. Not 1 nor 1/4. Do I average?

[poorpete]

30 minutes have 30/60

2 minutes have 29/60
2 minutes have 28/60
...
2 minutes have 15/60

Something like:
((15! / (60 x 15)) + 1/4) + 1/2) x 1/2 ?

Been a while.

[Mike]

Technically, it's the average of all those probabilities. But since there are an infinite number of points between noon and 1pm, taking the average of an infinite number of points is a pain in the ass. That's why I plot it 2-dimensionally and measure it as an area of space instead of trying to add up a bunch of increasingly small increments.

Spoiler warning: Here's my original graph...

[poorpete]

So average of 8/16 and 4/16 is 6/16 and average of 6/16 and 8/16 is 7/16?

[Mike]

[Phoebe]

Agreed on the chess answers except I think you can get thirty two knights on there.

[Mike]

I have two ways of arranging knights that both come up to 24. I'm stumped on this one.

[Phoebe]

Knights can only threaten a square of the opposite color from the one they're sitting on.

[Mike]

[Eliahad]