Can You Take a "Risk"?

Fiddler

Earlier this week, I was playing the board game Risk with one of my kids. 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?

Solution

There are only three groups of cards you can receive :

1. All depict the same kind of unit.
2. Two depict the same kind of unit, the third is different.
3. All depict different kinds of units.

2. is the scenario where you cannot trade them in.

1. ${14 \choose 2}$ ways to have two cards depicting the same unit.
2. ${28 \choose 1}$ ways to have one card depicting a different unit.
3. ${3 \choose 1}$ ways to choose one out of the three units for the first two cards.

Answer

Using complementary counting,

\begin{align*} P &= 1 - \frac{ {14 \choose 2}\cdot 28 \cdot 3}{{42 \choose 3}} \\ &= 1 - \frac{14 \cdot 13 \cdot 14 \cdot 3 \cdot 3 \cdot 2 \cdot 1}{42 \cdot 41 \cdot 40}\\ &= 1 - \frac{7 \cdot 13 \cdot 3}{41 \cdot 10} \\ &= \boxed{\frac{137}{410} \approx 33.41\%}\\ \end{align*}

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.)

Solution

Let $A$, $C$, $I$, and $W$ denote an artillery, cavalry, infantry, and wild card, respectively. Let $X$ denote any card.

Might as well tree it out...

• $\dfrac{42}{44}$ you will receive $A$, $C$, or $I$.
  • $\dfrac{13}{43}$ you will receive the same unit as your first card.
    • $\dfrac{12}{42}$ you will receive the same unit as your first and second cards.
      • $AAA$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{13}{43} \cdot \frac{12}{42} \cdot 3}$
    • $\dfrac{28}{42}$ you will receive a different unit than your first and second cards.
      • $\dfrac{12}{41}$ you will receive the same unit as your first and second cards.
        • $AACA$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{13}{43} \cdot \frac{28}{42} \cdot \frac{12}{41} \cdot 4}$
      • $\dfrac{13}{41}$ you will receive the same unit as your third card.
        • You can trade with any additional card.
        • $AACCX$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{13}{43} \cdot \frac{28}{42} \cdot \frac{13}{41} \cdot \frac{40}{40} \cdot5}$
      • $\dfrac{14}{41}$ you will receive the unit different from your first/second and third cards.
        • $AACI$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{13}{43} \cdot \frac{28}{42} \cdot \frac{14}{41} \cdot 4}$
      • $\dfrac{2}{41}$ you will receive a $W$.
        • $AACW$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{13}{43} \cdot \frac{28}{42} \cdot \frac{2}{41} \cdot 4}$
    • $\dfrac{2}{42}$ you will receive a $W$.
      • $AAW$, etc. $\rightarrow$$\boxed{\frac{42}{44} \cdot \frac{13}{43} \cdot \frac{2}{42} \cdot 3}$
  • $\dfrac{28}{43}$ you will receive a different unit than your first card.
    • $\dfrac{13}{42}$ you will receive the same unit as your first card.
      • $\dfrac{12}{41}$ you will receive the same unit as your first card.
        • $ACAA$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{28}{43} \cdot \frac{13}{42} \cdot \frac{12}{41} \cdot 4}$
      • $\dfrac{13}{41}$ you will receive the same unit as your second card.
        • You can trade with any additional card.
        • $ACACX$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{28}{43} \cdot \frac{13}{42} \cdot \frac{13}{41} \cdot \frac{40}{40} \cdot 5}$
      • $\dfrac{14}{41}$ you will receive the unit different from your first or second cards.
        • $ACAI$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{28}{43} \cdot \frac{13}{42} \cdot \frac{14}{41} \cdot 4}$
      • $\dfrac{2}{41}$ you will receive a $W$.
        • $ACAW$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{28}{43} \cdot \frac{13}{42} \cdot \frac{2}{41} \cdot 4}$
    • $\dfrac{13}{42}$ you will receive the same unit as your second card.
      • $\dfrac{13}{41}$ you will receive the same unit as your first card.
        • You can trade with any additional card.
        • $ACCAX$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{28}{43} \cdot \frac{13}{42} \cdot \frac{13}{41} \cdot \frac{40}{40} \cdot 5}$
      • $\dfrac{13}{41}$ you will receive the same unit as your second card.
        • $ACCC$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{28}{43} \cdot \frac{13}{42} \cdot \frac{12}{41} \cdot 4}$
      • $\dfrac{14}{41}$ you will receive the unit different from your first or second cards.
        • $ACCI$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{28}{43} \cdot \frac{13}{42} \cdot \frac{14}{41} \cdot 4}$
      • $\dfrac{2}{41}$ you will receive a $W$.
        • $ACCW$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{28}{43} \cdot \frac{13}{42} \cdot \frac{2}{41} \cdot 4}$
    • $\dfrac{14}{42}$ you will receive a different unit than your first or second cards.
      • $ACI$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{28}{43} \cdot \frac{14}{42} \cdot 3}$
    • $\dfrac{2}{42}$ you will receive a $W$.
      • $ACW$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{28}{43} \cdot \frac{2}{42} \cdot 3}$
  • $\dfrac{2}{43}$ you will receive a $W$.
    • You can trade with any additional card.
    • $AWX$, etc. $\rightarrow$ $\boxed{\frac{42}{44} \cdot \frac{2}{43} \cdot \frac{42}{42} \cdot 3}$
• $\dfrac{2}{44}$ you will receive a $W$.
  • You can trade with any two additional cards.
  • $WXX$ $\rightarrow$ $\boxed{\frac{2}{44} \cdot \frac{43}{43} \cdot \frac{42}{42} \cdot 3}$

Answer

$$\boxed{\frac{490,029,120}{130,320,960} \approx 3.7602}$$

Rohan Lewis

2025.09.29