ETA: Mine eyes have been opened by Tom Keith.

This week the question is: Can You Win at Non-Traditional Blackjack?

You’re playing a modified version of blackjack, where the deck consists of exactly 10 cards numbered 1 through 10. Unlike traditional blackjack, in which the ace can count as 1 or 11, the 1 here always has a value of 1.

You shuffle the deck so the order of the cards is completely random, after which you draw one card at a time. You keep drawing until the sum of your drawn cards is at least 21. If the sum is exactly 21, you win! But if the sum is greater than 21, you “bust,” or lose.

What are your chances of winning, that is, of drawing a sum that is exactly 21?

And for extra credit:

You’re playing the same modified version of blackjack again, but this time, whenever there’s even the slightest chance you could bust on the next card, you quit the round and start over. On average, how many rounds should you expect to start until you finally win?

Highlight to reveal solution:

In short: I find all the possible card combinations, and add together their probabilities. One possible combination is 1+2+3+4+5+6. It’s the only possible one with 6 cards. So its probability is 1/(10*9*8*7*6*5). (I can draw these 6 cards in 10*9*8*7*6*5 different ways, depending on their order.) Another possible combination is 1+2+3+5+10. It turns out, there are 9 different combinations with 5 cards. Their combined probability is 9/(10*9*8*7*6*5). Notice that these 2 cases (5 cards and 6 cards) aren’t overlapping. Like, one of the 5 card combinations can’t be combined with 1 more card into a 6 card combination, because we already hit the right sum with 5 cards. So we can safely add these probabilities together. If I keep going in this way, the resulting probability is 1996/151200 or approximately 0.013. Assumption: that I listed all the possible combinations.

My program solution gets a wildly different result. Sigh.

And for extra credit:

If there’s the slightest chance of busting, I start over. This translates to:

1. The sum is 10 or less, and I can’t reach 21 with 1 more card, but I keep going.
2. The sum is 11, and I can reach 21, if I draw a 10.
3. The sum is 12 or more, and I stop.

Case 1 turns into either case 2 or 3, depending on the next card. My only chance of winning is case 2, drawing a 10. This case reduces to: What is my chance of getting the sum 11, and then drawing a 10? After some calculations, the result is 193/30240, or approximately 1/157. On average, I have to play 157 games to win.

And again, my program says otherwise. Hmf.

Addendum: In both puzzles I forgot that, e.g., the card combination (2,9) can also be (9,2), so I need to multiply by 2! in this case. Also in extra credit I forgot the valid combinations (before drawing the final card) (2,4,5) and (2,10), the latter followed by a 9. Sigh. Updated calculations for puzzle and extra credit .

This week the question is: When Is a Triangle Like a Circle?

… let’s define the term “differential radius.” The differential radius r of a shape with area A and perimeter P (both functions of r) has the property that dA/dr = P. (Note that A always scales with r2 and P always scales with r.)

For example, consider a square with side length s. Its differential radius is r = s/2. The square’s area is s2, or 4r2, and its perimeter is 4s, or 8r. Sure enough, dA/dr = d(4r2)/dr = 8r = P.

What is the differential radius of an equilateral triangle with side length s?

Highlight to reveal solution:

I pretty much just worked through equations for equilateral triangles. I also assumed r = a * s. In other words, a simple relationship. After some calculations, I got r = s / (2 * √3).

This week the question is: Can You Beat the Heats?

There are 25 sprinters at a meet, and there is a well-defined order from fastest to slowest. That is, the fastest sprinter will always beat the second-fastest, who will in turn always beat the third-fastest, and so on. However, this order is not known to you in advance.

To reveal this order, you can choose any 10 sprinters at a time to run in a heat. For each heat, you only know the ordinal results, so which runner finishes first, second, third, and so on up to 10th. You do not know the specific finishing time for any runner, making it somewhat difficult to compare performances across heats.

Your goal is to determine with absolute certainty which of the 25 sprinters is fastest, which is second-fastest, and which is third-fastest. What is the fewest number of heats needed to guarantee you can identify these three sprinters?

And for extra credit:

At a different meet, suppose there are six sprinters that can race head-to-head. (In other words, there are only two sprinters per heat.) Again, they finish races in a consistent order that is not known to you in advance.

This time, your goal is to determine the entire order, from the fastest to the slowest and everywhere in between. What is the fewest number of head-to-head races needed to guarantee you can identify this ordering?

Read more: #ThisWeeksFiddler, 20240503

Highlight to reveal solution:

After a heat, I can eliminate the 7 slowest sprinters. I need to narrow the field to 3 sprinters. 25 – 3 = 22 = 3 * 7 and a bit. So I need 4 heats, just to find the top 3.

(Assume there is no faster way to do this, like choosing the sprinters for the 3rd heat in a clever way. I can’t think of such a clever procedure.)

Luckily that’s also enough to determine their order. Spend 3 heats eliminating 21 sprinters (always choosing sprinters that haven’t been eliminated yet), and there’s 4 left. The 4th heat will clearly show their internal order.

We need 4 heats.

And for extra credit:

With 2 sprinters, it’s easy. A and B race, we’re done.

With 3 sprinters, we need a little more. A and B race, turns out (WLOG) that A is the fastest. We want to check whether C fits in between them. So we race C against (WLOG) A. If C wins, we’re done. If not, we also need to race C against B. At most 3 races.

Order	Race	Winner	Order
	(1) A B	A	A B
	(1) A B	B	B A
A B	(2) A C	A	A (B or C)
A B	(2) A C	C	C A B
B A	(2) B C	B	B (A or C)
B A	(2) B C	C	C B A
A (B or C)	(3) B C	B	A B C
A (B or C)	(3) B C	C	A C B
B (A or C)	(3) A C	A	B A C
B (A or C)	(3) A C	C	B C A

(Going forward, I assume it’s possible to just add the nth sprinter to n-1 ordered sprinters, and that this is efficient. That I can’t do something clever like, I don’t know, shuffling 2 ordered groups into each other.)

With 4 sprinters, the complexity grows again. Assume we need 3 heats to order A, B and C. Turns out (WLOG) that the order is A B C. First we race B and D. If D wins, race A and D. If D wins again, it’s D A B C. If D loses, it’s A D B C. If on the other hand D loses to B, race D and C. If D wins, it’s A B D C. If D loses, it’s A B C D. 3 + 2 heats.

Notice D didn’t have to race everybody. Racing against B eliminates some options.

With 5 sprinters, there’s a little leap again. We use 5 races to obtain (WLOG) the order A B C D. Race E against (WLOG) B. Worst case, B loses and loses to C and will have to race D. 5 + 3 races.

With 6 sprinters, it’s more of the same. I want to add F to (WLOG) A B C D E, obtained after at most 8 races. Race against C, then against A or D, and worst case against B or E. 8 + 3 = 11 races.

Note: Binary Insertion Sort. There’s something going on with log₂ here. I need 1 race to add a 2nd sprinter, 2 races to add a 3rd sprinter, 3 races to add a 5th sprinter. log₂( x – 1) + 1 races to add the xth sprinter?

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