# Riddler Express, August 28 2020

I didn’t add last week’s Classic to this blog, but I did produce a pretty graph in Jupyter from which you can read the right answer (in either graph, the point on the x-axis where the orange four-post line first exceeds the blue three-post line.)

This week’s Express is not difficult, but I did find the answer a little surprising.

# Riddler Express, July 24 2020

Got the Classic from two weeks ago correct, although I hadn’t got the closed-form solution; I needed to spend more time playing with toy examples, but I’m not sure I would have discerned the pattern anyway. It’s neat though.

This week’s Express is not too difficult (please don’t let me get it wrong!). The thing that surprised me on thinking about it is that the total number of combinations of shires is only 1024. That’s  210, since there are 10 shires, each of which can be in or out of any particular subset.

Half of the possibles combinations of shires don’t have enough votes to win, and of those combinations that do, many of them could lose a shire and still win, so they can be ignored as well.

My 33 lines of code will print out an answer in about a millisecond. I assume in Haskell it could be done in about four lines! I should take a look at that language sometime.

# Riddler Express, June 26 2020

Simple question for this one:

In Riddler City, the city streets follow a grid layout, running north-south and east-west. You’re driving north when you decide to play a little game. Every time you reach an intersection, you randomly turn left or right, each with a 50 percent chance.

After driving through 10 intersections, what is the probability that you are still driving north?

# Riddler, May 15 2020

Since I’m still learning my way through Jupyter, I tried the Classic as a notebook.

I’m also not sure what the best way to present a notebook is, if I have a WordPress site? The include I used last time wasn’t that great.

For the Express, the best score I can manage is 6. In this grid, there’s no legal place to put a 1 in the top row.

# Riddler Express, July 13 2018

I got a shoutout for last week’s solution and “nice explanation”. Definitely an improvement on the week before.

This week’s puzzle is easily stated:

This year’s World Cup has been chock full of exciting penalty shootouts. Historically, about 75 percent of soccer penalty kicks are successful. Given that number, what are the chances that a shootout goes past its fifth kick for each team and into the even more exciting sudden-death portion of the penalty-kick period?

I see two ways of solving this one. The first is to calculate it directly, by summing the probability of each possible draw. The second is to generate every possible outcome, and find the draws among them.

# Riddler Express, July 6 2018

So, last week’s effort was, err, not quite 100%. I feel better about this week’s Express puzzle.

You are a delivery person for the finest peanut butter and jelly sandwich restaurant in Riddler City. City streets are laid out in a grid, and your restaurant is on the corner of 20th Street and Avenue F. The city has 61 east-west streets, numbered 1st to 61st, and 21 north-south avenues, named A to U.

While traveling on a given street or avenue, you can drive at 20 mph, and the blocks are all 0.1 miles long. The exception is Avenue U, also known as the Ultra-Speed Trafficway, upon which you can drive at 200 mph. (You don’t need to worry about slowing down for traffic or turns.)

What are the parts of the map for which it’s helpful to use the Ultra for your deliveries, assuming you always start at 20th and F?

This is the most Excel-friendly Classic I can remember, and I pretty much think in Excel, so that’s how I tackled this one.

# Riddler Express, June 29 2018

EDIT2: Even my second attempt was wrong. See the comments.

EDIT: I’ve got this wrong – see the bottom of the post.

This one is titled, “What Are The Odds World Cup Teams Play Each Other Twice?”. The question is simple:

Assuming we don’t know anything about the strengths of the teams in the tournament, what are the chances that any pair of teams in a 32-team World Cup plays each other twice?

I’m pretty confident I’ve got this one, so here goes!

The chance of any two teams meeting each other is the product of a lot of independent events. These events being independent means that

What are the events? Or to put it another way, what needs to happen for the two teams to meet each other twice?

Firstly, to meet in the first round, they must be in the same group.

Then, to have any chance of meeting again, the two teams must finish first and second in the group.

Then, both teams must win their second round matches, and their quarter-final matches.

Lastly, both teams must achieve the same result in the semi-final match: either both must win their semi-finals, or both must lose. (If one team loses and the other wins, then one team will go to the final, and the other to the third-place play-off.)

If those are the things that need to happen, what are the chances of each of them happening?

Being in the same group: there are 32 spots in the World Cup, in eight groups of four. Whichever group the first team is in, there are then 31 spots left, and three of them are in the same group as the first team. So the chances of this are 331.

Finishing first and second in the group: this is related to the question, “how many ways can you choose two things out of a set of four different things?”, also called four choose two, or 4C2. There are two related ways of thinking about this:

1. Call the teams in the first round group A, B, C, and D. We want our pair of teams to win. We can say that A and B are the teams we care about, but it doesn’t matter which of the four letters we pick. How many different possible winning pairs are there? The pairs are easy to list: AB, AC, AD, BC, BD, CD. The teams A and B only appear together once in that list of six, so the chances of them coming first and second are 16.
2. Let’s say again, that the teams we care about are A and B. How many different possible results are there for the whole group? We can count them, starting with first place: there are four possibilities for the team in first, then there are three possible teams left that can come second, and two left that can come third. If we know the teams in first, second, and third place, then there’s no question about which team is fourth – it must be the team that is left. So the number of possibilities are 4 × 3 × 2 = 24.  What are those combinations?

All 24 combinations are here, and you can see that the four combinations in which A and B are the top two teams have been underlined. Four possible results out of 24 = 16. That’s the same as the answer that we got through the first method, thankfully.

Both teams win their second round and quarter-final matches: this is a total of four matches which all need to be won. Since we don’t know the strength of any of the teams, we can say that each of our teams has a 50% chance of winning any match it is in. 50% is a probability of 12, so the chance of all four matches being won is 12 × 12 × 12 × 12 = (12)4 = 116.

Both teams get the same result in their semi-finals: there are four possible results in the semi-finals (team A wins or loses, at the same time as team B independently wins or loses). Two of these four results are good enough: two out of four = 24 = 12.

Putting it all together: the combined chance of all of these events happening is just the chance of each of them happening individually multiplied together:

331 × 16 × 116 × 12 = 131 × 12 × 116 × 12 = 11984

So, not very likely!

That’s the Riddler Express for this week. Riddler Classic is still stumping me.

Edit: a friend pointed out that I have misinterpreted the question:

…what are the chances that any pair of teams in a 32-team World Cup plays each other twice?

I have considered “any pair” to mean “any given pair” – answering the question, “If you pick two qualified teams before the draw for first round groups even happens, what are the chances that your pair of teams will play each other twice?” But “any pair” means “any pair at all”, not “any pair you pick before the tournament starts.” So the question actually means, “After the end of the tournament, what is the chance that at least one pair of teams will have played each other twice?”

I think I can answer that as well:

At the end of the first round, there are 16 teams left in the competition; eight pairs, where each pair played each other in the first round, and the two teams came first and second in their group.

Each pair has eight teams in the top half of the bracket, and eight teams in the bottom half of the bracket. Regardless of what happens in the bottom half of the bracket, four teams go out in the top half of the bracket after the second round matches, and two more go out after the quarter-finals. Two  teams are left in the top half of the bracket, after the semi-final: one in the final, and one in the play-off.

Now in the bottom half of the bracket, there are the other halves of all eight pairs. Only two of those eight still have a paired team in the top half of the bracket – for the other six, their paired team is already out.

So two teams in the bottom of the bracket have a chance to play again against a team they played in the first round. To do that, either of them would have to win their second round and quarter-final match (50% × 50% = 25% chance), and then get the same result in the semi-final that their paired team in the top half of the bracket did (another 50% chance, for a total of 50% * 25% = 12·5% = 18 for each of the two teams).

What’s the chance that at least one of those two teams makes it to play their other pair in the final or play-off? The chance that a particular one of them makes it is 18, so the chance that a particular team doesn’t make it is 78. The chance that neither team makes it is 78 × 78 = (78)2 = 4964. So the chance that at least one team plays another team twice is 1 – 4964 = 1564.