## Update: Chutes and Ladders is long, but not *that* long

It occurred to me, as I was failing to pay attention in a class this past week, that I neglected an important detail in my recent post analyzing the expected number of turns to complete the game Chutes and Ladders: namely, that one generally does not play the game alone.

Recall from the previous post that we can express the expected value $x_i$ of the number of die rolls (or spins of the spinner) needed for a player to reach the final square 100, starting from square $i$, “recursively” as

$x_i = 1 + \frac{1}{6} \sum_{j=1}^6 x_{f(i,j)}, i < 100$

$x_{100} = 0$

where $f(i,j)$ is the number of the square reached from square $i$ by rolling $j$ (thus encoding the configuration of chutes and ladders on the board).  Solving this system of 100 equations yields the value $x_0$ of about 39.5984 turns on average for our hypothetical player to finish the game.

However, it is a bit misleading to stop there, since the expected total number of turns in a game with multiple players does not simply scale directly as the number of players.  That is, for example, in a game with two players, we should not expect nearly 80 total rolls of the die, 40 for each player.  As simulation confirms, the game typically ends much more quickly than that, with an actual average of about 52.5188 total rolls, or only about 26 turns for each player.

(Why is this?  This is a good example of the common situation where we can gain insight by considering “extreme” aspects or versions of the problem.  In this case, first note that the shortest possible path from the start to the final square 100 is just seven moves.  It is very unlikely that any single player will actually take this short route, with a probability of only 73/46656, or about 1 in 640.  But now suppose that instead of just one or even two players, there are one million players.  It is now a near certainty that some one of those million players will happen to win the lottery and take that short route… and so we should expect that the average number of total turns should be “only” about 7 million, instead of 40 million as the earlier post might suggest.)

We can still compute this two-player expected value exactly, using the same approach as before… but it starts to get expensive, because instead of just 100 equations in 100 unknowns, we now need 100^2 or 10,000 equations, to keep track of the possible positions of both players:

$x_{i,j} = 1 + \frac{1}{6} \sum_{k=1}^6 x_{j,f(i,k)}, i,j < 100$

$x_{i,100} = 0, i < 100$

Solving yields the desired value $x_{0,0} \approx 52.5188$.

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