Chute Ladder Solution Game Examples: Operation Management

Question

Chutes and Ladders is a popular children’s game where players compete to move along a game board the fastest. It involves both hazard squares that can set you back a few spaces and some helpful squares that can move you forward a few spaces. 

A player starts at the red “Start” square and finishes at the green “Finish” square. A player advances along the board by spinning a spinner that results in moving forward one to three squares.

If a player ends his turn on a Ladder (depicted by the blue ladder icon) he immediately moves upwards along the ladder. For example, if a player ends his turn on square 6, he would immediately move to square 11.

If a player ends his turn on a Chute (depicted by the orange slide) he immediately moves down the slide. For example, if a player ends his turn on square 7, he would automatically move to square 2.

A player wins the game when he lands on or passes the Finish square.

Question: What is the average number of turns needed to finish the game?

Solution

We can solve the model using Markov chain principle as explained by our operations management homework help experts.

I will show you below the method

as the spinner is having 3 numbers so each number is having an equal probability of 1/3

Transition state probability

From the diagram, you can just think that we cannot be in square 4, 6, 7, and 9 as they have ladder or chute attached to them so we can be in other remaining squares in a state i.e. [1,2,3,5,8,10,11,12]

Explanation of transition state

Suppose in square1

next step

we cannot be in square 1 in our next step because we have to spin either 1, 2 or 3 so we can be in state 2, 3 or 5

each having an equal probability of 1/3

Suppose in square 2

if we spin 1 we can be in square 3 so that probability is1/3

if we spin 2 we will end up in 5 because 4 is having a ladder to take us to 5 and also if we spin 3 we will be in 5 so total probability will be 1/3+1/3

For square 3 we can end in 5 with spin 1 or 2 so both add up to 1/3+1/3 and also we can go to square 11 if we spin 3 that having another probability of 1/3

for square 5 we can end to 2 if we spin 3 having probability 1/3 similarly we can go to 8 with spin 3 and can go to 11 with spin 1 each having probability of 1/3

for square 8 we can remain in 8 if we spin 1, can go to 10 and 11 if we spin 2 and 3 respectively each having probability 1/3

for square 10 we can remain in 10 if we spin 3 as we can not go anywhere and we can go to 11 and 12 if we spin 1 and 2 respectively each probability 1/3

similarly, for 11 if we spin 2 or 3 we have to be in 11 as the game won't end and we can not go anywhere so total probability1/3+1/3 and to go to 12 probability is 1/3

For 12 we can only go to start i.e square 1 with a probability 1 as the game ends there

The transition matrix is shown in excel below

Transition Matrix
	1	2	3	5	8	10	11	12	Total
1	0	0.333333	0.333333	0.333333	0	0	0	0	1
2	0	0	0.333333	0.666667	0	0	0	0	1
3	0	0	0	0.666667	0	0	0.333333	0	1
5	0	0.333333	0	0	0.333333	0	0.333333	0	1
8	0	0	0	0	0.333333	0.333333	0.333333	0	1
10	0	0	0	0	0	0.333333	0.333333	0.333333	1
11	0	0	0	0	0	0	0.666667	0.333333	1
12	1	0	0	0	0	0	0	0	1

Please notice total probability if each row is 1

the rows show the current state and the column shows the future state

the explanation I have given is just transformed into the matrix

Now to calculate the expected no of steps we have to calculate the expected passage time equation and to solve it

from the transition matrix, it is prepared

it is denoted as 

so if we are in 1 and to go to 12 we should multiply the probability of each transition state and its passage time to end at 12

for example

like this, we have to calculate up to 11th square

this is shown in excel below

Expected passage time  equations
mu(1,12)=	1+0.33333mu(2,12)+0.3333mu(3,12)+0.33333mu(5,12)
mu(2,12)=	1+0.3333mu(3,12)+0.6667mu(5,12)
mu(3,12)=	1+0.6667mu(5,12)+0.33333mu(11,12)
mu(5,12)=	1+0.3333mu(2,12)+0.33333mu(8,12)+0.33333mu(11,12)
mu(8,12)=	1+0.3333mu(8,12)+0.3333mu(10.12)+0.33333mu(11,12)
mu(10,12)=	1+0.3333mu(10,12)+0.3333mu(11,12)
mu(11,12)=	1+0.6667mu(11,12)

We can solve the equation in solver by taking all the notations to LHS and keeping 1 in RHS

Shown below

mu(1,12)	mu(2,12)	mu(3,12)	mu(5,12)	mu(8,12)	mu(10,12)	mu(11,12)	LHS	RHS
1	-0.33333	-0.3333	-0.3333	0	0	0	1	1
0	1	-0.3333	-0.6667	0	0	0	1	1
0	0	1	-0.6667	0	0	-0.3333	1	1
0	-0.3333	0	1	-0.3333	0	-0.3333	1	1
0	0	0	0	0.6667	-0.3333	-0.3333	1	1
0	0	0	0	0	0.6667	-0.3333	1	1
0	0	0	0	0	0	0.3333	1	1
mu(1,12)	mu(2,12)	mu(3,12)	mu(5,12)	mu(8,12)	mu(10,12)	mu(11,12)	Total steps
7.13064	6.7889	5.841818	5.762439	4.49955	2.99985	3.0003	36.02349519

Formulas used in excel

mu(1,12)	mu(2,12)	mu(3,12)	mu(5,12)	mu(8,12)	mu(10,12)	mu(11,12)	LHS	RHS
1	-0.33333	-0.3333	-0.3333	0	0	0	=SUMPRODUCT(J14:P14,$J$23:$P$23)	1
0	1	-0.3333	-0.6667	0	0	0	=SUMPRODUCT(J15:P15,$J$23:$P$23)	1
0	0	1	-0.6667	0	0	-0.3333	=SUMPRODUCT(J16:P16,$J$23:$P$23)	1
0	-0.3333	0	1	-0.3333	0	-0.3333	=SUMPRODUCT(J17:P17,$J$23:$P$23)	1
0	0	0	0	0.6667	-0.3333	-0.3333	=SUMPRODUCT(J18:P18,$J$23:$P$23)	1
0	0	0	0	0	0.6667	-0.3333	=SUMPRODUCT(J19:P19,$J$23:$P$23)	1
0	0	0	0	0	0	0.3333	=SUMPRODUCT(J20:P20,$J$23:$P$23)	1
mu(1,12)	mu(2,12)	mu(3,12)	mu(5,12)	mu(8,12)	mu(10,12)	mu(11,12)	Total steps
7.13064170040816	6.78889614755953	5.84181815612355	5.76243911222972	4.49955003374775	2.99985000749963	3.000300030003	=SUM(J23:P23)