Easy math puzzles

Consider a deck with only 2 cards (2, 3). How 'bout a 3 card (2, 3, 4) deck?

2-10: 775 / 4,536 (1 / 5.8529)
2-11: 115,991 / 907,200 (1 / 7.8213)
2-12: 950,171 / 9,979,200 (1 / 10.5025)

Anything larger than that took too long to calculate with brute force, but I do have some simulation-based results:
2-20: 1 / 119
2-30: around 1 / 2750
Pretty much anything 50 or higher had too few "wins" to simulate with any degree of accuracy without running for hours

I couldn't find a way to generate the pattern 1, 5/6, 16/24, 62/120,...; it is listed in the Online Encyclopedia of Integer Sequences, but the only listed generating formula is recursive

It’s about 17 percent.

The optimal strategy is fairly simple: You should say “bigger” if there are more bigger cards remaining and “smaller” if there are more smaller cards remaining. (If there is an equal number of bigger and smaller cards, you can just guess randomly — for simplicity, let’s say you always guess “bigger” in this case.)

Getting to the probability of winning is a little trickier. Because we are dealing with cards 2 through 10, we are dealing with nine cards. Therefore, there are 9! — or 362,880 — ways that these cards might be arranged. What we need to do next is figure out how many of those arrangements will lead us to a win. If we divide the number of winning arrangements by the total number of arrangements, we’ll get the probability that we win the game.

To ease into these calculations, let’s first imagine a smaller game — just two cards, numbered 2 and 3. In this case, you’re guaranteed to win because you’ll know for sure after seeing the first card whether the second and final card is bigger or smaller. Now, let’s move to a slightly larger game — three cards, numbered 2, 3 and 4. Things get a little more interesting here. There are six (3!) decks you might face. Five of these — every arrangement except 324 — leads to a win. In a game with four cards, there are 24 (4!) possible decks, 16 of which lead to victory.

Solver Keith Hudson developed a clever way to find and visualize the underlying pattern here, which we can extend to solve our nine-card game. Imagine numbers in a pyramid, where each row going down represents a game of a certain size: one card, two cards, etc. The numbers in each row of the pyramid add up to the number of ways you can win that game. From left to right, the individual numbers in each row are the number of ways you can win that game if you happen to start with the lowest card, then the second-lowest card, and so on to the highest card. For example, in a one-card game, there is only one card you can start with and one way to win. In a two-card game, there are two cards you can start with, and one way to win in each of those cases. In a three-card game, there are three cards you can start with, and two, one and two ways, respectively, to win in those scenarios. The first six rows look like this:



We can build this pyramid with some pretty simple rules. If, for example, we start by drawing either the smallest or biggest card in our deck, we are essentially then playing the game as if we had started with one fewer card, so the outer numbers (for example, 62 and 62 in the bottom row above) in the rows are the sums of the whole row that comes above them.

You’ll notice that you want to draw the biggest or smallest cards — the closer you get to the middle, the fewer ways you have to win the game. For example, say you’re playing with six cards (the last row in our pyramid above). If you draw the biggest or smallest card (the outer numbers of the row), you have 62 ways to win. But if you draw the third-biggest or third-smallest, you have only 35 ways to win.

You’ll also notice that the numbers have a relationship with each other. If a number is on the interior of the pyramid and on the left side, it is the sum of everything above it and to the right. Same goes for a number on the interior of the pyramid and to the right — it’s the sum of everything above it to the left.

Why is this? Because if we draw the biggest or smallest cards (the outer numbers), it’s like we haven’t added a new card at all — we have the same number of ways to win as we did in total in the smaller game (the row above). But as you move closer to the middle of a row, you’re adding more and more complications, so your options to win are fewer. Remarkably, though, the past games help us calculate the number of winning paths in the new ones.

Continuing in this way, the sum of the numbers in the ninth row of this pyramid is exactly 62,000. So our probability of winning is 62,000/362,880, or about 17 percent.


Puzzle by Freddie Simmons (at Riddler forum)

I only have had a quick look at this and saw some recursion as a valid approach. So to get from N cards left to N-1 you have to get the next one correct.
Consider N cards in a row, if you pick the smallest one then there's (N-1) / (N-1) chances of winning (as all the others are larger). If you pick the next one then the chances are (N-2)/(N-1). Continue this and you get (N-1) (N-2) (N-3) etc and either get one or two occurrences in the middle.
With 2 cards left your chances are 1/1 1/1 (as you'll always be correct) which add up to 2/1 so average out to 2/2. (Note 2 = 2*1).
With 3 cards left your chances are 2/2 1/2 2/2 which add up to 5/2 so average out to 5/6. (Note 6 = 3*2, 5 is 3 more than 2)
With 4 cards 3/3 2/3 2/3 3/3 which add up to 10/3 so average out to 10/12 or 5/6. (Note 12 = 4*3, 10 is 5 more than 5)
With 5 cards 4/4 3/4 2/4 3/4 4/4: 16/4, 16/20 or 4/5. (Note 20 = 5*4, 16 is 6 more than 10)
With 6 cards 5/5 4/5 3/5 3/5 4/5 5/5, 24/5, 24/30 or 5/6. (Note 30 = 6*5, 24 is 8 more than 16)
With 7 cards 6/6 5/6 4/6 3/6 4/6 5/6 6/6, 33/6, 33/42 or 11/14. (note 42 = 7*6, 33 is 9 more than 24)
Thus you can see the pattern emerging.

I get p = ( 3 - sqrt(5) ) / 2, which is about 38.2%.

Let x = probability of making a touchdown with a possession.

Going for a 2-point conversion, Red can win by either making the 2 points and then another touchdown, or missing the 2-point conversion, making another touchdown with 2-point conversion and then winning in overtime.

The probability of Red's winning with the above strategy is: p * x + (1 - p) * x * p * (1 / 2)

Forgoing 2-point conversion attempts, Red can win by making one more touchdown and then winning in overtime.

The probability of Red's winning with this strategy is: x * (1 / 2).

Equating the two probabilities and simplifying yields: p² - 3p + 1 = 0

Solving for p yields: p = (3 - sqrt(5)) / 2.

~38.2% for p, when you go for it on the first touch down.

Agree with above. Incidentally, the answer of ~38.2% is 2 minus the golden ratio