Probability Game: Find Chance of Landing on Each Spot 1-10

[Data&Stuff]

I need help with finding the probability of the following game, I am making game for my data management class. The game consits of one die, so rolling anything between 1 and 6, there is 10 spots from the beginning to end and every 3 spots there is a hurdle where the player has to flip coin to determine if he crosses or not.
First question :
I want to know how can I find the probability of landing on each different spot from one to ten ? The probability of the coin is pretty simple, but I need help . please help me :(

 

[micromass]

The rules of the game are not quite clear to me... For example: what happens if I roll a 6?

 

[Data&Stuff]

well there is just one dice, and there is 10 cases. so if you roll a 6 you land on the sixth spot ?
I just want to know how to find the probability of finishing the game and rolls.

 

[micromass]

Well, with the information you gave me, the transition matrix of your game is

$$\left(\begin{array}{cccccccccc} 0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0 & 0 & 0\\ 0 & 0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0 & 0\\ 0 & 0 & 0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0\\ 0 & 0 & 0 & 0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6}\\ 0 & 0 & 0 & 0 & 0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{3}\\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{2}\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{6} & \frac{1}{6} & \frac{2}{3}\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{6} & \frac{5}{6}\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ \end{array}\right)$$

With that matrix, you can easily calculate that the probability that you finish the game is 1. But that is probably not the answer that you want... Maybe you should ask the question: what is the probability that I finish the game in 4 turns (or something likely).

The probability of finishing the game in 1 turn is 0
The probability of finishing the game in 2 turns is 0.27
The probability of finishing the game in 3 turns is 0.74
The probability of finishing the game in 4 turns is 0.94
The probability of finishing the game in 5 turns is 0.99
The probability of finishing the game in 6 turns is 0.99
The probability of finishing the game in 7 turns is 0.99
The probability of finishing the game in 8 turns is 0.99
The probability of finishing the game in 9 turns is 1


Of course, this probabilities are without the "hurdles" every 3 places. I did not factor them in because you did not yet explain what they do...

 

[blue_raver22]

I would approach this problem by first defining the variables involved. In this case, we have a single die with 6 possible outcomes, and a coin with 2 possible outcomes. The game also has a specific set up with 10 spots and hurdles every 3 spots.

To find the probability of landing on each spot, we need to consider the possible combinations of outcomes from rolling the die and flipping the coin. For example, if the player rolls a 1 on the die and flips a heads on the coin, they would advance 1 spot on the game board. If they roll a 2 and flip a tails, they would advance 2 spots.

We can create a table to show all the possible outcomes and their associated probabilities:

Die Roll | Coin Flip | Result | Probability
1 | Heads | Advance 1 spot | 1/6 * 1/2 = 1/12
1 | Tails | Stay on same spot | 1/6 * 1/2 = 1/12
2 | Heads | Advance 2 spots | 1/6 * 1/2 = 1/12
2 | Tails | Stay on same spot | 1/6 * 1/2 = 1/12
3 | Heads | Advance 3 spots | 1/6 * 1/2 = 1/12
3 | Tails | Stay on same spot | 1/6 * 1/2 = 1/12
4 | Heads | Advance 1 spot | 1/6 * 1/2 = 1/12
4 | Tails | Stay on same spot | 1/6 * 1/2 = 1/12
5 | Heads | Advance 2 spots | 1/6 * 1/2 = 1/12
5 | Tails | Stay on same spot | 1/6 * 1/2 = 1/12
6 | Heads | Advance 3 spots | 1/6 * 1/2 = 1/12
6 | Tails | Stay on same spot | 1/6 * 1/2 = 1/12

From this table, we can see that the probability of landing on each spot is not