#ThisWeeksFiddler, 20250926

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This week the #puzzle is: Can You Take a “Risk”? #ExpectedValue #probabilities #permutations #MonteCarlo

Regarding Risk. At the end of each turn in the game in which you conquer at least one enemy territory on the board, you are dealt a card.
There are 42 territory cards in the deck—14 that depict an infantry unit, 14 that depict a cavalry unit, and 14 that depict an artillery unit. Once you have three cards that either (1) all depict the same kind of unit, or (2) all depict different kinds of units, you can trade them in at the beginning of your next turn in exchange for some bonus units to be placed on the board.
If you are randomly dealt three cards from the 42, what is the probability that you can trade them in?

And for extra credit:

The full deck of Risk cards also contains two wildcards, which can be used as any of the three types of cards (infantry, cavalry, and artillery) upon trading them in. Thus, the full deck consists of 44 cards.
You must have at least three cards to have any shot at trading them in. Meanwhile, having five cards guarantees that you have three you can trade in.
If you are randomly dealt cards from a complete deck of 44 one at a time, how many cards would you need, on average, until you can trade in three? (Your answer should be somewhere between three and five. And no, it’s not four.)

Can You Take a “Risk”?

Highlight to reveal (possibly incorrect) solution:

Program

Let’s split the situation up into cases:

  • I get 3 cards of the same kind. This happens with probability 1 * 13/41 * 12/40 = 156/1640 ≈ 0.095122.
  • I get 3 cards of different kinds. This happens with probability 1 * 28/41 * 14/40 = 392/1640 ≈ 0.23902.
  • Added together this is about 0.33415.

I also wrote a monte carlo program to confirm this.

And for extra credit:

Image

Again, let’s split up the situations:

  • AAA: I have 3 cards of the same kind. This happens with probability 42/44 * 13/43 * 12/42 = 6552/79464 ≈ 0.082452. Stop here.
  • ABC: I have 3 cards of different kinds. This happens with probability 42/44 * 28/43 * 14/42 = 16464/79464 ≈ 0.20719. Stop here.
  • *AA, A*A, AA*, *AB, A*B, AB*: I have 1 wild card. This happens with probability 3 * 2/44 * 42/43 * 41/42 = 10332/79464 ≈ 0.13002. Stop here.
  • **A, *A*, A**: I have 2 wild cards. This happens with probability 3 * 2/44 * 1/43 * 1 = 6/1892 ≈ 0.0031712. Stop here.
  • Added up this is a probability of about 0.42283 of stopping after 3 cards.
  • Or I can’t trade with only 3 cards. This happens with probability about 1 – 0.42283 = 0.57717.
  • AABA, ABAA, BAAA: I have 3 cards of the same kind and 1 that’s different. The last card drawn is of the former kind. This happens with probability 3 * 42/44 * 13/43 * 28/42 * 12/41 = 550368/3258024 ≈ 0.16893. Stop here.
  • AABC, ABAC, BAAC: I have 3 cards of different kinds and 1 extra. The last card drawn was a new kind. This happens with probability 3 * 42/44 * 13/43 * 28/42 * 14/41 = 642096/3258024 ≈ 0.19708. Stop here.
  • AAB*, ABA*, BAA*: I have 2 cards of 1 kind and 1 card of another kind, and the last card drawn is a wild card. This happens with probability 3 * 42/44 * 13/43 * 28/42 * 2/41 = 91728/3258024 ≈ 0.028154. Stop here.
  • Added up this is a probability of about 0.39416 of stopping after 4 cards.
  • Or I can’t trade with either 3 or 4 cards. This happens with probability about 1 – 0.42283 – 0.39416 = 0.18301.
  • I can trade after drawing the 5th card. With probability 0.18301.
  • In all I expect to draw 3 * 0.42283 + 4 * 0.39416 + 5 * 0.18301 = 3.7602.

Again, my program confirms this.

Output from program:

Wild cards? 0
   1000000 loops in all:
    334503 trades with 3 cards (33.45030%).
    443999 trades with 4 cards (44.39990%).
    221498 trades with 5 cards (22.14980%).
Expected no. of cards: 3.88700

Wild cards? 2
   1000000 loops in all:
    423209 trades with 3 cards (42.32090%).
    393234 trades with 4 cards (39.32340%).
    183557 trades with 5 cards (18.35570%).
Expected no. of cards: 3.76035

Also see the image of the situations with 0-9 wild cards.

2017: Flammen

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I juli var vi på ferie i #Viborg. Her besøgte vi bl.a. Flammen.

En større jysk by med respekt for sig selv har naturligvis en Flammen . Lige den her inkarnation er ikke meget for at indrømme det, men de er fra 2017 . Den store fordel ved dem er, at de har buffet med masser af kød, salat osv. Alle bør kunne finde noget, de kan lide. Vi skulle mødes med en håndfuld venner, og stedet passede fint. Og jeg fik prøvet at spise kænguru.

2017: Nytorv 11

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I juli var vi på ferie i #Viborg. Her besøgte vi bl.a. Nytorv 11.

Nytorv 11 , åbnet i 2017 , er såmænd et sted, hvor man kan spise. Og det gjorde vi så. (På billedet er det nr. 11 til højre.) Jeg mener, vi fik “PASTA TENERA (OKSEKØD EL. KYLLING): Skært oksekød eller mør kylling, portobello svampe i en fløde-trøffelsauce smagt til med trøffelolie, toppet med friskrevet parmesan”.

2010: For enden af gaden

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I juli var vi på ferie i #Viborg. Her besøgte vi bl.a. For enden af gaden.

For enden af gaden lå meget tæt på hotellet, så selvfølgelig fik vi både noget at spise og drikke der. (Et af de der firmaer, det er svært at finde et årstal på, men til sidst fandt jeg frem til 2010 .) Med retter komponeret af så mange slags ting, så var noget mad genialt og noget ikke. Og så havde vi glæde af en stor papegøje, der også er et springvand, og som udendørsdelen af restauranten bare er bygget op rundt om.

Fish & Chips deluxe — Deluxe serveres med kuller, Black tiger i fillodej og dagens friskfanget fisk, alt pakket i sprød panko og toppet på vores grove fritter med jysk krydret sauce tatar, chili mayo og grillet citron
Små og store oste — 3 skæve oste fra vores oste mand i Århus med rugbrøds tuilles, salt og sødt

1972: Interferenser

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I juli var vi på ferie i #Viborg. Her kom vi bl.a. forbi kunstværket Interferenser.

I toget kom vi bl.a. gennem Fredericia, der har kunst på stationen. Sådan, noget stor kunst. 2 m høj (?), 5 m lang (?), og i 3-4 forskellige versioner. Og jeg var ved at blive drevet til vanvid af, at det var svært at finde en hjemmeside, der kunne fortælle mig noget om dimserne, såsom navn og kunstner. Vha. netværket lykkedes det at få nogle data, men processen var stadig indviklet. (Senere fandt jeg en lille smule mere kunstbeskrivelse .) Så jeg besluttede selv at tage nogle billeder. Pga. regnvejr blev det ikke så godt. Men I skal have dem alligevel! Her er 3 af dem altså, Interferenser fra 1972 af Gunnar Aagaard Andersen.

1937: Skovgaard Museet

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I juli var vi på ferie i #Viborg. Her besøgte vi bl.a. Skovgaard Museet.

Jeg har allerede nævnt domkirken, og der er en direkte forbindelse over til Skovgaard Museet . Stiftet i 1937 og flyttet ind i den nuværende bygning, det gamle rådhus, i 1981, så er der her et museum, der nærmest kun eksisterer for at huse diverse udstillinger, der har med Joakim Skovgaard at gøre, deriblandt hans skitser til vægmalerierne i domkirken.

Da vi var der, så var der også en særudstilling om Niels Larsen Stevns, en ven og kollega.

Ved den udtørrede Damhus Sø. 1908. Olie på lærred. 56,5 x 72,5 cm.
Kornet køres hjem. Dyreborg. 1933. Olie på lærred. 66 x 104 cm.
Skovbillede. 1941. Olie på lærred. 54,5 x 68,5 cm.

Og så alt det andet, bl.a. skabt af andre medlemmer af familien.

Peter Christian Skovgaard (1817-1875). Sommereftermiddag på malkepladsen ved Knabstrup, 1875.
P.C. Skovgaard. Udsigt over havet fra Møens Klint, 1850
Ebbe Skovgaard (1895-1982). Figur i form af et afrikansk vildsvin.
Niels Larsen Stevns. Kristus helbreder en spedalsk. 1913

#ThisWeeksFiddler, 20250919

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This week the #puzzle is: Can Your Team Self-Organize? #permutations #LongestIncreasingSubsequence

In a recent team building exercise at work, a group of people (including myself) was asked to quantify themselves in various ways. For example: “What outdoor temperature do you prefer?” No one reveals their answer at first. Instead, each person places a card with their name on an unmarked line, one at a time. In this example, folks who prefer higher temperatures would place their names farther right; folks who prefer lower temperatures would place their names farther left. However, the line is unmarked and doesn’t have any units.
Once all the names are placed on the line, their values are revealed. For example, a group of six might have generated the following numbers, in order from left to right on the line: 60, 67, 65, 74, 70, 80.
The team’s score is the length of the “longest increasing subsequence.” In other words, it’s the maximum number of elements in the list you can keep such that they form an increasing subsequence. In the example above, you can remove the 67 and the 74 to get the following increasing subsequence with four numbers: 60, 65, 70, 80. There are a few other ways to get an increasing subsequence of length four, but there’s no way to get a sequence of length five or more, so the team’s score is four.
Suppose a total of four people are participating in this team building exercise. They all write down different numbers, and then independently place their names at random positions on the line.
On average, what do you expect the team’s score to be?

And for extra credit:

Instead of four people, now suppose there are 10 people participating in the team building exercise. As before, they all write down different numbers, and then independently place their names at random positions on the line.
On average, what do you expect the team’s score to be?

Can Your Team Self-Organize?

Highlight to reveal (possibly incorrect) solution:

Spreadsheet Program Image

Method 1: Write out all the permutations, measure the longest increasing subsequence for each, calculate the average. This is what I did in the spreadsheet. Result: 2.41667.

Method 2: Create an image, a graph, with all the permutations and how one can move between them moving just 1 number. This gives me the number of permutations with subsequences of length 1, 3 and 4 (1, 9 and 1). For length 2 it gets messy. For them the number is “the rest” (24 – 1 – 9 – 1 = 13). Do the same calculation as above.

Method 3: Write a program to go through all the permutations. This replicates method 1. Same result.

And for extra credit:

Theorem

Do method 3, just with 10 members instead of 4. Result: 4.33496.

Apparently the Baik–Deift–Johansson theorem says something about this situation as well.

1935: Palads Hotel

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I juli var vi på ferie i #Viborg. Her boede vi på Palads Hotel.

Palads Hotel er svært at komme udenom. Etableret i 1935 på en enkelt adresse breder det sig i dag i flere nabobygninger også, og vi skulle (fra værelse 150) krydse gågaden for at komme til morgenmad. Det var dog også lidt en udfordring på første dag. Bookingen snakkede om Centrum 24-7. Kortet snakkede om Best Western. Og da vi så stod der, så sagde baldakinen Palads. Det var sådan set rigtigt alt sammen, men en kende forvirrende.

Badeværelset havde 5 brusere! Og sauna! Selve værelset havde et trin, der desværre lå et mørkt sted, når man lige var kommet ind, så det var muligt at falde. Men der var god plads, lækkert.

Apropos morgenmad! (Ja, et par af billederne er rystede. Men du ved jo godt, hvordan 2 slags juice ser ud, ikke?)

Og så spillede døren en lille melodi.