The Median number of rolls required to complete a game is With a median of 29, this means that there are the same number of games that are completed in less moves than 29 as there are games that take more than 29 moves to complete. To complete the average trifecta, we'll calculate the Mean. The arithmetic mean is simply the sum of all the rolls of the die divided by the number of games played. During the billion simulated games played, the die was rolled a total of 36,,, times which results gives an average of the number of die rolls per game of approx There is a subtle difference.
If the same snake was encounterd three times in a single game, the count for this snake would be increased by 3, not just 1 "Used or not in this game".
Think of it like a "Toll" charge for using the ladder or snake. The least frequently used ladder, not surprisingly, is ladder 1 which can only be used if a player rolls a 1 on their initial roll. If a 1 is not rolled on the first roll, it is impossible to come back to this square.
Refreshingly, looking at the count of the number of times this ladder was used in the entire simulation run results in a percentage of The next least used ladder is ladder 2. Again, not a surprise since, like ladder 1, once passed, there is no way to return to take it again.
Unlike ladder 1, however, there is more than one way of getting onto this ladder over a series of early rolls, so the percentage this ladder is used is higher than ladder 1. Objective approaches often work extremely well, but there are some limitations. Sometimes, for instance, it is simply not possible to repeat an experiment multiple times. Sometimes you only get one shot. The work around for this leads us neatly to an alternative mechanism for calculating the distribution of expected game lengths.
Games like Chutes and Ladders are ideal candidates for Markov Chain analysis because, at any time, the probability of events that will happen in the future are agnostic about what happened in the past. If a player is on grid square 18 of the board, the probability of what will happen on the next roll is independent on how the player got to square It is this memorylessness that enables Markov chain analysis to work.
At the heart of Markov Chain analysis is the concept of a Stochastic Process. This is just a fancy word to say that, from a given state, there are a series of possibilities that could happen next, defined by a probability distribution. Implied in the definition is that all the probabilities add up to 1. In a game of Chutes and Ladders , a player can be on a particular square. We don't care how they got there, we just know that they roll the die again and act based on the results of the roll. If a player is at grid square G when he rolls again, one of six things could happen with equal probability , and based on these probabilities the player would advance to one of the next squares.
These probabilities can be represented as a sparse matrix which records the probability of moving from GridSpace i to GridSpace j by the entry in row- i and column- j. We'll call this the Transition Matrix. A vanilla snippet of this matrix can be seen below. Things get a little more interesting when we add Snakes and Ladders into the mix. Now there is a chance that a roll will land a player onto the business-end of one of these special entities and they will get 'teleported' to a new location. An example of the what the transition matrix would look like in this ficticious location is shown below.
An example of this can be seen on row A roll of 3 will take the player to square 53 , but a roll of 6 will also land the player on square 53 because landing on square 56 is the head of a snake which slides the player back to This is shown in the matrix snippet below. The second condition to care about is the boundary scenario when the player is close to the finish.
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Here, according to our house rules, as an exact roll is not needed, there are multiple ways to get to square And watch out for the snake on square 98 which slides you back to square 78! Tokens start off-board, then the first roll lands the player on the board. Well the simple explanation is that it is impossible for a player to rest on the head of a snake or the bottom of a ladder. You can see this in row 98 above. Interestingly removing these redundant rows makes a non-trivial difference to calculation speed.
Matrix multiplication which as we will see below is used for this calculation is O n 3 so reducing the size of a square matrix from to 82 doubles the speed!
The transition matrix encapsulates the probability of moving from any square to any other square. Now, all we need to do is provide it with an input.
A player starts the game off board, and nowhere else, so we create a column vector with 1. Next we multiply our column vector by the transition matrix, and the vector produced at the output is the probability distribution at the end of roll 1. Each row value in the output vector is the probability that the player token will be in that square at the end of that roll. On the grid, non-zero probabilities are painted in red. The stronger the probability, the more intense the colour. After one iteration multiplication , the results are pretty palpable.
There are six shaded squares, each with equal probability representing the squares that would have been achieved with each distinct roll of the die. You can see that two of the rolls resulted in the use of ladders. We can use the output of the first roll as the input for the second roll.
The output of the first roll shows the probability density of the grid. If we multiply probability density by the transition matrix again our output will be the probability density after two rolls A superposition of all the probabilities from rolling the die again from every position on the grid.
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The dark shading of some of the cells especially on the lower row , highlight the superposition of probabilities and show how these spaces are more likely to be occupied after two rolls because of the multiple ways to get there. Here we can see that or maybe not, it's pretty faint! Seven rolls is the least number of rolls required to complete the game, and it is the first time that our probability cloud reaches this square.
Continuing on, here is a picture of the board after 10 rolls.
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You can see the colour in square getting deeper, as more and more games reach completion and the probability that a game would be completed increases. Here we are after 20 rolls. The pure-white squares show the start cells of the ladders and snakes. It is impossible zero probability for player to be on one of these squares.
The very light cells are the ones with almost no chance of the player residing in them. To calculate the probability of completing a game after n -turns, we can examine the value in row of the output matrix after the initial identity input column vector has been multiplied by the TransitionMatrix n. Notice the curve? Actually, it's very close indeed.
The only difference is that, not surprisingly, the Monte-Carlo is slightly less smooth. The curves are so close that when I plot them both the same graph, the lines obscure each other. To differentiate them, I've made the Markov generated line a little thicker and show a zoomed in portion of the curve in the image below. The fact that the curves are so similar despite being generated in two very different ways cooroborates that our code is working correctly. It also confirms that the number of simmulations we ran for the objective analysis was an appropriate number.
Now that I have a test-frame for running simulations it's easy to tweak the game and see the changes to the expected results. Generally, adding ladders to the game shortens the average number of moves, and adding addition snakes lengthens the average number of moves to complete. But that is not always the case! Initially this sounds paradoxical and counter intuitive. How can adding a snake reduce the number of steps required to complete?
After all, if you don't land on the new snake, it makes no difference to the number of moves, but if you do land on it, it sends you backwards! What's going on? Well, if the snake in question happens to send you back to before the start of a really long ladder that you previously missed, it gives you another chance to take this upwards boost.
Similarly, if a new short ladder is added that makes the user bypass a much longer ladder, then the small boost gained by taking the short ladder is outweighed by the chance of taking the big one. Adding too many new snakes or ladders before a significantly longer one also makes a noticable difference to the average moves. With a long clean run-up to an important entity there are multiple combinations of rolls that will land the player on that square.
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