Question

game 3 - monopoly

1. what are the chances of every person in your group landing on beverly railroad on their first turn? (hint: you might want to reference the sum chart on your addition rule notes - mutually exclusive and inclusive events.)
2. what are the odds of landing on go (square with the red arrow)?
3. when drawing a chance card, what is the probability of drawing the get out of jail free card?
4. which space on the board did you land on the most after 20 trials? to do this, start at go (square with the red arrow) each time and roll the dice and record where on the board you would land. do this 20 times and record your data.

roll # space roll # space roll # space
1 2 3
4 5 6
7 8 9
10 11 12
13 14 15
16 17 18
19 20
which space on the board did you land on the most after 20 trials?

1. which space should you land on the most? (theoretical answer. remember the sample space of rolling 2 dice and which sum has the highest probability)

is your answer to question #4 the same as the theoretical answer?

Explanation:

Step1: Determine sample - space for landing on Beverly Railroad

In Monopoly, when rolling two dice, the sample - space of dice rolls is \(n(S)=6\times6 = 36\) possible outcomes. To land on a particular space, we need to consider the number of ways to get a certain sum of the dice rolls that will move us to that space. However, without knowing the distance from the starting point to Beverly Railroad, we assume a general approach. If we assume that each space on the board can be reached by a unique sum of two - dice rolls.

Step2: Calculate probability for landing on GO

The board has a finite number of spaces. Assuming a standard Monopoly board with 40 spaces, if we start from GO and roll the dice, the probability of landing on GO again depends on the number of ways we can roll the dice to move exactly 0 or a multiple of 40 spaces. When rolling two dice, the minimum sum is 2 and the maximum is 12. The probability of landing on GO on the first roll from GO is 0 (since we need to roll a sum that moves us back to GO). In a full - turn around the board, if we consider the cycle of 40 spaces, we need to find the number of ways to roll a sum that is a multiple of 40. Since the sum of two dice is between 2 and 12, the probability of landing on GO on a single roll is \(P=\frac{1}{40}\) (assuming all spaces are equally likely to be landed on in a non - special move situation).

Step3: Calculate probability of drawing Get Out Of Jail Free card

In a standard Monopoly game, there are 16 Chance cards. There is 1 Get Out Of Jail Free card among the Chance cards. So the probability \(P=\frac{1}{16}\) of drawing the Get Out Of Jail Free card when drawing a Chance card.

Step4: Conduct 20 trials (simulation)

This requires actual rolling of dice and recording. Since we can't perform the physical experiment here, in a real - life situation, you would roll the dice 20 times starting from GO, record the landed spaces, and then count the frequency of each space to find the most - landed - on space.

Step5: Theoretical most - landed - on space

When rolling two dice, the sample space \(n(S) = 36\). The sums and their frequencies are:

Sum	Frequency
3	2
4	3
5	4
6	5
7	6
8	5
9	4
10	3
11	2
12	1

The sum with the highest probability is 7. So, theoretically, the space that is 7 spaces away from the starting point (GO) should be landed on the most (assuming no special cards or other game mechanics interfere).

Answer:

1. Depends on distance to Beverly Railroad from start. General approach: Consider sum of two - dice rolls and sample space of 36.
2. \(\frac{1}{40}\)
3. \(\frac{1}{16}\)
4. Requires actual 20 - trial simulation.
5. Space 7 spaces away from GO.