Question

suppose you rolled a twelve - sided number cube instead of a standard six - sided one. how would the probability of these events change? the theoretical probability of landing on a star space on your first roll. the theoretical probability of not landing on a question mark on your first roll. the theoretical probability of reaching the end space on your first roll

Answer:

To solve this, we first analyze the game board (from the image):

• Star spaces: Let's count. From the board, there are 4 star spaces (yellow with star).
• Question mark spaces: 3 (blue with ?).
• Total spaces: Let's count all: Start is 1, then green (cat town, etc.), star, question, etc. Wait, actually, when rolling a number cube, the number of "outcomes" is the number of spaces? Wait, no—wait, the problem is about a twelve - sided cube vs six - sided. Wait, maybe the original game (with six - sided) has a certain number of spaces, but now we use a twelve - sided cube (so 12 possible outcomes, each roll gives a number 1 - 12, and we map to spaces? Wait, maybe the game board has a certain number of spaces, but the key is: when we use a 12 - sided cube, the total number of possible outcomes is 12, whereas with a 6 - sided, it's 6. Let's assume the game board has, say, for the first roll:

1. Probability of landing on a star space

• Let's assume in the original (6 - sided) game, the number of star spaces is \( n_{star} \), and total spaces (outcomes) is 6. With 12 - sided, total outcomes is 12. But from the board, let's count the star spaces: looking at the image, there are 4 star spaces (yellow with star: top, left, bottom left, right). Question marks: 3 (blue with ?). Green spaces: let's count: left green (2), cat town column (3), right green (3), plus start? Wait, maybe the total number of spaces (the "length" of the path) is such that with a 6 - sided cube, each roll moves 1 - 6 spaces, but the problem is about "landing on a star space on your first roll"—so maybe the star spaces are at positions that correspond to numbers on the cube. Wait, maybe a better approach: Probability \( P = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \).

• For a 6 - sided cube: total outcomes = 6. For a 12 - sided cube: total outcomes = 12.

• Let's assume the number of star spaces (favorable for star) is \( S \), question marks \( Q \), etc. From the image, let's count:

• Star spaces: 4 (yellow with star: top, left, bottom left, right).
• Question marks: 3 (blue with ?).
• Green spaces: Let's count: left (2), cat town column (3: cat town, above, below), right (3: before end, two before star, one after star? Wait, maybe total spaces: Let's list all:

Start (1), then green (2), star (1), question (2), star (1), green (3: cat town area), question (1), star (1), green (3), star (1), end? Wait, maybe the total number of "spaces" (the possible landing spots) is such that when rolling a cube, you move that many spaces. But the problem is about "landing on a star space on your first roll"—so the number of star spaces is 4, question marks 3, and other spaces (green, end, start? Wait, start is a starting point, not a landing space? Maybe the total number of possible landing spaces (excluding start) is, say, 15? No, the cube is 12 - sided, so the number of possible outcomes (rolls) is 12. Wait, maybe the game board has a path where the number of spaces you can land on in one roll (with 12 - sided) is 12, and we need to see how many are star, question, etc.

Wait, maybe the original problem (with 6 - sided) has:

• Star spaces: Let's say in 6 - sided, the probability of star is \( \frac{S}{6} \), with 12 - sided, it's \( \frac{S}{12} \). But from the image, let's count the star spaces: 4. So:

• For 6 - sided: \( P_{star,6} = \frac{4}{6} = \frac{2}{3} \)? No, that can't be. Wait, maybe I mis - count. Wait, the image shows:

• Top: star (1)
• Left: star (1)
• Bottom left: star (1)
• Right: star (1)

So 4 stars.

Question marks: top left (1), top right (1), bottom (1) → 3.

Green spaces: left (2: two green), cat town column (3: cat town, above, below), right (3: three green before end, one after star? Wait, no, let's count all non - star, non - question, non - start, non - end:

Start (1), green (2), star (1), question (1), question (1), star (1), green (1), cat town (1), green (1), green (1), question (1), star (1), green (1), green (1), green (1), star (1), end (1). Wait, this is confusing. Maybe the key is: when using a 12 - sided cube, the total number of possible outcomes (rolls) is 12, so the denominator for probability is 12, whereas for 6 - sided, it's 6.

Let's re - frame:

1. Probability of landing on a star space

• Let \( S \) = number of star spaces. From the image, \( S = 4 \).
• For 6 - sided cube: \( P_{6} = \frac{S}{6} = \frac{4}{6} = \frac{2}{3} \) (if 6 spaces, but that's not right). Wait, no—maybe the game board has a path where the number of spaces you can land on in one roll is equal to the number of sides of the cube. So with 6 - sided, you can land on 6 spaces; with 12 - sided, 12 spaces. But the star spaces are 4, so:

• 6 - sided: \( P_{star,6} = \frac{4}{6} = \frac{2}{3} \)
• 12 - sided: \( P_{star,12} = \frac{4}{12} = \frac{1}{3} \)

So the probability decreases (since \( \frac{1}{3} < \frac{2}{3} \)).

2. Probability of not landing on a question mark

• Number of question marks \( Q = 3 \).
• For 6 - sided: Number of non - question = \( 6 - 3 = 3 \), so \( P_{6} = \frac{3}{6} = \frac{1}{2} \)
• For 12 - sided: Number of non - question = \( 12 - 3 = 9 \), so \( P_{12} = \frac{9}{12} = \frac{3}{4} \)

So the probability increases (since \( \frac{3}{4} > \frac{1}{2} \)).

3. Probability of reaching the end space on the first roll

• The end space is 1 (only one end).
• For 6 - sided: \( P_{6} = \frac{1}{6} \) (if end is 1 space out of 6)
• For 12 - sided: \( P_{12} = \frac{1}{12} \)

So the probability decreases (since \( \frac{1}{12} < \frac{1}{6} \)).

Final Answers (assuming the dropdowns are for "increase", "decrease", "stay the same"):

1. The theoretical probability of landing on a star space on your first roll: decreases
2. The theoretical probability of not landing on a question mark on your first roll: increases
3. The theoretical probability of reaching the end space on your first roll: decreases

(Note: The exact counts of star, question, and end spaces depend on the game board. If the game board has, for example, 4 star spaces, 3 question marks, and 1 end space, the above logic holds. The key is that with a larger number of sides (12 vs 6), the probability of a specific outcome (star, end) decreases (since denominator is larger), and the probability of "not question" increases (since non - question count increases more than denominator).)