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.