# Tag: Puzzles

# Riddler Classic, January 24 2020

EDIT: Nope! Not far wrong, but definitely wrong.

My strategy of starting with the easy cases and building out to more complex cases was used successfully by other solvers. My analysis of the easy cases was correct, but there was a mistake in the way I built out. My diagram probably didn’t help; the one drawn by the winner didn’t distinguish between the larger and smaller piles, and I think that’s where my error crept in.

For example, when I considered (9, 5), I thought it was a losing position because I didn’t see that taking six coins from the larger pile produced (5, 3), since the larger pile became the smaller pile in that move. So the “simplification” I introduced, of specifying one pile as larger, while not wrong, made it easier for me to make a mistake. It helped to break my mental model by suggesting an identity for the piles (“larger” or “smaller”) that isn’t persistent, but depends on the progress of the game. Will try not to do that again!

EDIT 2: Revised the original spread sheet showing the rows that I had calculated wrongly, and producing a correct game state table.

Another week of being *fairly* confident about this week’s Riddler Classic. (Bolstered by getting exactly the right answer for the last one I did, if by a slightly inefficient method.)

The puzzle is a coin-picking game, with the usual aim of taking the last coins, but the unusual condition of having two coin stacks:

# Riddler Classic, January 3 2020

Initially I thought this week’s Classic puzzle might be too tough to solve without some serious insights, but the more I thought about it, the more it seemed like I might be able to break it down to manageable pieces.

# Riddler Classic, November 22 2019

Most of the Classics look too difficult for me to be able to solve, but this week’s looked like I could approach it. No code required, either.

Here’s the question:

Five friends … are playing the … Lottery, in which each must choose exactly five numbers from 1 to 70. After they all picked their numbers, the first friend notices that no number was selected by two or more friends. Unimpressed, the second friend observes that all 25 selected numbers are composite (i.e., not prime). Not to be outdone, the third friend points out that each selected number has at least two distinct prime factors. After some more thinking, the fourth friend excitedly remarks that the product of selected numbers on each ticket is exactly the same. …

What is the product of the selected numbers on each ticket?

There might be a neat, elegant way of solving this, but I chipped away at it bit by bit.

# Riddler Express, November 15 2019

# Riddler Express, August 17 2018

This week’s seems a little more involved than the Expresses usually are (although I’ll admit July 27^{th}‘s really stumped me!):

Take a standard deck of cards, and pull out the numbered cards from one suit (the cards 2 through 10). Shuffle them, and then lay them face down in a row. Flip over the first card. Now guess whether the next card in the row is bigger or smaller. If you’re right, keep going.

If you play this game optimally, what’s the probability that you can get to the end without making any mistakes?

I got nowhere when I tried visualising this as a decision tree. Too wide and deep for me to understand it. Then I did the sensible thing, and broke it down into simpler problems. I also tried staying away from Excel for a while, a new one for me!

# Riddler Classic, July 13 2018

EDIT: I have a better answer, but had too much help to claim it as my own! See the first comment.

I have a sort-of answer to this week’s Riddler Classic. Probably not the exact answer sought, but it’s further than I usually get!

Say you have an “L” shape formed by two rectangles touching each other. These two rectangles could have any dimensions and they don’t have to be equal to each other in any way. (A few examples are shown below.)

Using only a straightedge and a pencil (no rulers, protractors or compasses), how can you draw a single straight line that cuts the L into two halves of exactly equal area, no matter what the dimensions of the L are? You can draw as many lines as you want to get to the solution, but the bisector itself can only be one single straight line.

# 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 excitingsudden-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 ^{3}⁄_{31}.

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 _{4}C_{2}. There are two related ways of thinking about this:

- 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
^{1}⁄_{6}. - 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?

ABCD ABDC ACBD ACDB ADBC ADCB BACD BADC BCAD BCDA BDAC BDCA CABD CADB CBAD CBDA CDAB CDBA DABC DACB DBAC DBCA DCAB DCBA 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 =

^{1}⁄_{6}. 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 ^{1}⁄_{2}, so the chance of all four matches being won is ^{1}⁄_{2} × ^{1}⁄_{2} × ^{1}⁄_{2} × ^{1}⁄_{2} = (^{1}⁄_{2})^{4} = ^{1}⁄_{16}.

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 = ^{2}⁄_{4} = ^{1}⁄_{2}.

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:

^{3}⁄_{31} × ^{1}⁄_{6} × ^{1}⁄_{16} × ^{1}⁄_{2} = ^{1}⁄_{31} × ^{1}⁄_{2} × ^{1}⁄_{16} × ^{1}⁄_{2} = ^{1}⁄_{1984}

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% = ^{1}⁄_{8} 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 ^{1}⁄_{8}, so the chance that a particular team *doesn’t* make it is ^{7}⁄_{8}. The chance that neither team makes it is ^{7}⁄_{8} × ^{7}⁄_{8} = (^{7}⁄_{8})^{2} = ^{49}⁄_{64}. So the chance that *at least one team plays another team twice* is 1 – ^{49}⁄_{64} = ^{15}⁄_{64}.