The Mathematics of Snakes and Ladders
Many of us have at some point played the childhood game of Snakes and Ladders. Although there are no decisions to make, there is still quite a bit of mathematical structure to the game, which is useful in understanding the positional aspects of game theory, and probability theory in general.
A typical board is shown in Figure 1. Players start at square 0 (off the board), and in turn throw a 6sided die and move the given number of spaces. If they land on a snake they go down to the bottom, and if they land on a ladder, they ascend to the top. The winner is the first player to reach square 100, but this requires an exact throw.
Like much of game theory, even simple games can pose some interesting questions:
 1. What is the maximum number of throws required to reach the finish square?
 2. What is the minimum number of throws required to reach the finish square?
 3. How many throws, on average, does it take to reach the finish square?
 4. Would you rather be on square 81 or square 53? Does it matter who goes next?
 5. In a twoplayer game, what is the probability that the player who starts first wins?
 6. In a game of three players, at positions 10, 15 and 27, who act in that sequence, what’s the probability that each player wins?
 7. In a tenplayer game, what is the probability that the player who goes last wins?
 8. How many rounds does a 2player game last, on average?
 9. How many rounds does a 3player game last, on average?
 10. With three players at positions 10, 15 and 27, how many more rounds is the game expected to last?
 11. What’s the probability that you reach the finish in the minimum number of throws?
 12. What’s the probability that you need more than 100 throws to reach the finish?
 13. What is the probability of landing on square 80?
 14. What is the expected number of times you will land on the big ladder on square 28, before reaching the finish?
 15. Which are the least and most likely squares you will land on (apart from the start and finish)?
 16. Which square will you land on the most often before reaching the finish?
 17. What are the fairest starting positions for 10 players? (Assume that players don’t all have to start from the same square.)
Do try these questions yourself first, but you’ll need a computer to actually calculate the answers. There's also a bit of subtlety in the implementation which I won't go into but can be seen in the source code.
For all of these questions, precise answers can be formulated. This is much more insightful than doing a simulation to approximate the answers, although simulation is useful to validate the results.
Source code is here: http://www.calumgrant.net/projects/snakesandladders. I wrote this because I haven’t seen any other analyses of Snakes and Ladders, and it turned out to be a lot more interesting than I expected.
Answers
1Unlimited, 27, 335.5432, 481,no, 50.509027, 60.280842,0.335736, 0.383422, 70.0907991, 824.421, 920.4108, 1017.9524, 110.00173611, 120.0170009, 130.493964, 140.464519, 152,45, 1699, 1733,34,37 35,17,38,18,19,20,39.
Answers
1Unlimited, 27, 335.5432, 481,no, 50.509027, 60.280842,0.335736, 0.383422, 70.0907991, 824.421, 920.4108, 1017.9524, 110.00173611, 120.0170009, 130.493964, 140.464519, 152,45, 1699, 1733,34,37 35,17,38,18,19,20,39.
Question 1
What is the maximum number of throws required to reach the finish square?
On most boards, such as the board in Figure 1, there is no upper bound on the number of throws to finish. It’s possible to carry on playing forever.
Question 2
What is the minimum number of throws required to reach the finish square?
We can express the minimum number of throws T for each square s as min(T). This can be expressed as as 1 + the minimum number of throws from each square we can land on from s.
min(T(s)) = 1+min[1<=d<=6](min(T(move(s,d)))) (2)
min(T(100)) = 0
where move(s,d) is the square we end up on if we throw a d on square s.
min(T) can be computed in various ways. For example, we can iterate over an array, or treat the squares as a graph and use a "singlesource shortest path” algorithm.
The plot of min(T) for each square is shown in Figure 2.
The answer is min(T(0)), the minimum distance from square 0, which is 7.
Question 3
How many throws, on average, does it take to reach the finish square?
The expected number of remaining throws E(T(s)) for a square s is given by
E(T(s)) = 1 + sum[1<=d<=6]E(T(move(s,d)))/6 (3)
E(T(100)) = 0
Equation 3 comes from the sum over conditional probabilities, for each possible value of the die d, which have equal probability of 1/6.
This yields a set of linear equations which can be solved exactly using a linear equation solver. Another approach is to apply Equation 3 repeatedly on an array of E(T) initialised to 0, until the values in the array converge. This is correct because the values of E are monotonic so will stabilise by virtue of the Fixed Point Theorem.
The plot of E(T) is shown in Figure 3. The downward spikes are ladders, and the upward spikes are snakes. Squares 80 and 100 are at zero because they reach the finish.
Figure 3: The expected number of throws remaining for each square.
Calculating the expected number of throws from the start square = E(T(0)) = 35.5432.
Question 4
Would you rather be on square 81 or square 53? Does it matter who goes next?
Equation 3 gives
E(T(53)) = 23.7185
E(T(81)) = 23.6829
Thus, it seems that square 81 is better, because it is 0.0356 throws ahead  a tiny margin. The next player has one throw in hand, so it would seem to be at an advantage. But actually this isn’t true, because E(T) doesn’t tell the whole story.
The probability of the first player winning P_wins(a,b), when the players are in positions a and b, is given by
P_wins(a, b) = sum[1<=d<=6](1P_wins(b, move(a,d)))/6 (4)
P_wins(100, b) = 1
Equation 4 comes from the sum of conditional probabilities, that for each throw of the dice, the other player doesn't win. Equation 4 yields a large set of linear equations, which can be solved using a linear equation solver. This yields
P_wins(53,81) = 0.487661
P_wins(81,53) = 0.535979
Thus the player on square 81 is more likely to win, irrespective of who goes first.
Intuitively, the player with the smaller number of expected moves remaining should be more likely to win. It turns out that there is an approximately linear relationship between the delta (difference) in the number of throws remaining, and the probability of winning a 2player game. Figure 4 shows a plot of this, computed for all pairs of squares. The graph is not symmetrical because there's always a slight advantage in going first, giving an average bias of about 0.9% to the player first to act.
Figure 4 gives a quick way to estimate the probability of winning. Look up the expected values E(T) for each player in the Appendix, and P_wins is approximated by 0.018x+0.5094, or about 2% per throw.
Question 5
In a twoplayer game, what is the probability that the player who starts first wins?
Using the results from Equation 4,
P_wins(0,0) = 0.509027
I.e. the advantage of going first = 0.509027  0.5 = about 0.9%. The reason this is so low is because it’s relatively unlikely for two players to reach the finish square in exactly the same round.
Question 6
In a game of three players, at positions 10, 15 and 27, who act in that sequence, what’s the probability that each player wins?
Unfortunately Equation 4 is limited to 2 players, and calculating every state would become computationally expensive as there are 100^n game states for n players.
E(T) doesn't tell the whole story about who will win or give information about the distribution of throws remaining. We actually need to compute the random variable T(s) (representing the number of throws remaining at any any given square s).
The cumulative distribution function F(s;t) gives the probability that the finish square has been reached within t throws from square s.
F(s; t) = sum[1<=d<=6](F(move(s,d)), t1) / 6 (6a)
F(100; t) = 1
F(s; 0) = 0 for s!=100
f(t) = F(t)  F(t1)
Equation 6a uses the sum over conditional probabilities that you will reach the finish in t throws from s, if you can reach the finish in t1 throws from any square reachable by s. Since F(s;t) is defined only in terms of F(s;t1), we can iterate through t to calculate F. The probability density function f is the difference between sequential values of F.
Figure 6a shows a plot of f for various values of s, and Figure 6b shows the corresponding plots of F. The plot of f(0) is flat for the first 6 throws because it’s impossible to reach the finish in less than 7 throws.
Figure 6a: The probability distribution f of the duration of a game T, from the start (square 0), square 50 and square 90.
Figure 6b: The cumulative distribution F for the duration of a game T, from the start (square 0), square 50 and square 90.
F can be used to figure out who will win in the general case (unlike Equation 4 which is limited to the 2player case). Assuming the players are in positions p_1 ... p_2,
P(player 1 wins)
= sum_t( P(Player 1 wins in t throws) * P(no other player wins in < throws) )
= sum_t( f(p_1; t) * (1F(p_2; t1)) * … * (1F(p_n; t1) ) ) (6b)
Thus we can plug in the values 10, 15 and 27 into Equation 6b, to get
P(player 1 wins) = 0.280842
P(player 2 wins) = 0.335736
P(player 3 wins) = 0.383422
These add up to 1.
Question 7
In a tenplayer game, what is the probability that the player who goes last wins?
This is evaluated from Equation 6b, to 0.0907991.
Figure 7a shows the probability of each position winning a multiplayer game, calculated using Equation 6b, where as expected, the players who act in early position are at an advantage. The advantage to the first player is consistently around 0.9%, and the last player has the same disadvantage.
Figure 7a: The probabilities for each player winning a multiplayer game.
Why is the positional advantage consistently around 0.9%, irrespective of the square and the number of players? This actually comes from the probability that two players finish in the same round
sum[t>0](P(T(p_1)=t and T(p_2)=t)
= sum[t>0](f(p_1;t)*f(p_2;t)) (7)
where p_1 and p_2 are the respective positions of the players. Calculating Equation 7 from square 0, the probability of both players finishing in the same round = 0.0180536 ~= 1.8%. This is also the approximate advantage per throw in hand from Figure 4, as well as the difference in winning probabilities in a 2player game (see Question 5.)
Proof:
P(wins in position 1)  P(wins in position(2)
= sum[t>0](f(t).F(t))  sum[t>0](f(t).F(t1))
= sum[t>0](f(t) * (F(t)  F(t1))
= sum[t>0](f(t) * (F(t1)+f(t)  F(t1)))
= sum[t>0](f(t) * f(t))
which is the same as Equation 7.
Figure 7b shows the probability of players finishing in the same round for each square, assuming they are both on the same square. As the number of expected throws remaining decreases, so the probability that both players finish in the same round increases.
Figure 7b: Probability of two players finishing in the same round for each square.
Question 8
How many rounds does a 2player game last, on average?
The random variable T(0) defined in Question 6 gives the number of throws for one player to reach the finish. With 2 players, play stops as soon as one of the players reaches the finish, which is the minimum of two independent distributions T(0).
The cumulative distribution function (CDF) of the minimum of n identical distributions is given by
F_n(t) = 1  (1F(t))^n (8a)
The probability density function is as before
Figure 8: Probability distributions of the number of rounds in a 1 and 2player game.
The expected value of a distribution is given by
E(T) = sum_t(t*P(t=T)) = sum_t(t*f(t)), (8b)
which is computed to be 24.421. As expected, the number of rounds player by 2 players is smaller than the number of rounds played by 1 player (35.5432) because the game stops as soon as one player reaches the finish.
Question 9
How many rounds does a 3player game last, on average?
The same approach as Question 8 is used, with the minimum of 3 distributions. The new distribution is plotted in Figure 9a. The expected value of this distribution is calculated using Equation 8b, which evaluates to 20.4108. As expected, this is fewer rounds than with 2 players.
Figure 9a: Probability distributions of the number of rounds of a 1, 2, 3player game.
Figure 9b shows the expected number of rounds plotted against the number of players, which is calculated by applying Equation 8b for each number of players. It starts with the value E(T(0)) = 35.5432 and decreases asymptotically to min(T(0)) = 7, because no matter how many players there are, a game cannot be finished in fewer than 7 rounds.
Question 10
With three players at positions 10, 15 and 27, how many more rounds is the game expected to last?
A similar approach to Question 8 is required, but this time we are taking the minimum of 3 different independent distributions, because each square has its own distribution of remaining throws.
F(t) = 1  (1F(10;t))(1F(15;t))(1F(27;t))
This new distribution is plotted in Figure 10, and the expected value of this distribution (using Equation 8b) = 17.9524.
Question 11
What’s the probability that you reach the finish in the minimum number of throws?
We can answer this from the distribution for T in Equation 6a. The minimum number of throws=7, and F(7) = f(7) = 0.00173611.
Question 12
What’s the probability that you need more than 100 throws to reach the finish?
P(T>100) = 1P(T<=100) = 1F(100) = 0.0170009.
Question 13
What is the probability of landing on square 80?
For this question, we need to figure out the probability of being on a particular square s on turn t.
P(on square s on throw t) = sum[1<=d<=6](P(on square s0 on throw t1)  s=move(s0,d))/6. (13)
P(on square 0 on throw 0) = 1.
Again we are using the sum of conditional probabilities to define the probability of one state (being on square s on throw t) in terms of the probabilities of other states. This type of calculation is called a “Markov chain” where the system (i.e. the statespace of the game) is in one of many possible states (in this case, parameterised by s and t), with probabilities of transition between them = 1/6 because there are 6 sizes of a die.
Since square 80 does not have any paths back to itself (it goes straight to square 100), we know that each of states P(s,t) are mutually exclusive, thus
P(lands on square 80) = sum[t>0](P on square 80 on throw t).
We can compute Equation 13 directly by iterating over t. The answer is 0.493964. I.e. half the time you win by the ladder on square 80, and the other half if you reach the finish via the top row on the board.
Question 14
What is the expected number of times you will land on the big ladder on square 28, before reaching the finish?
Unlike in the previous question, the Markov states are not independent, so we cannot simply add up the probabilities of landing on square 28. If you land on square 28, there’s a possibility that you end up back on square 28 again.
We can however limit the Markov chain defined in Equation 13, to only permit one visit to a particular square s. This is done by simply removing all transitions in the chain away from square s, and so
P(lands on square s) = sum[t>0](P(first lands on square s on throw t)).
In fact, we can define the Markov chain from any starting square, allowing us to compute P(reaches s1 from s0) in general.
For example, P(reaches 100 from s0) is always 1, because you are guaranteed to reach the finish square if you play for long enough.
We need to figure out the distribution N_s of the number of times a particular square s is visited. We can interpret N_s as a variation of a geometric distribution, representing the number of failed attempts to reach square 100. A player either reaches s (with probability p), or reaches square 100 (with probability 1p). When they are on square s, they either reach square s again (with probability q), or they reach square 100 (with probability 1q).
P(N = 0) = 1p
P(N = k) = p.(1q).q^(k1) (14)
That is, if N=k then we reached square s from 0 (probability p), then reached ourselves k1 times (probability q^(k1)), then reached the finish (probability 1q).
Figure 14a: The probability of landing on each square at least once (blue) and the probability of landing on the square a second time (green), computed using a Markov chain.
Figure 14b shows the distribution of N for square 28, calculated using Equation 14. It falls away exponentially.
We could draw such a graph for any square on the board. Figure 14c shows the probability of landing on each square a given number of times, P(N_s=n), calculated using Equation 14.
Theorem 14: The mean of this distribution E(N) = p/(1q).
Proof:
E(N) = sum[k>=1](k.p.(1q).q^(k1))
= p.sum[k>=1]( k.(1q).q^(k1))
= p.sum[k>=1]( k.q^(k1)  k.q^k)
= p.sum[k>=0]( (k+1)q^k  k.q^k)
= p.sum[j>=0](q^j)
= p/(1q).
For square 28, we compute p = 0.37451, q= 0.193768, so by Theorem 14, E(N) = 0.38451/(10.193768) = 0.464519.
Question 15
Which are the least and most likely squares you will land on (apart from the start and finish)?
In this question we are only looking at p for each square to see if we landed on it or not. We can calculate p as P(reaches square s from 0) using the Markov chain from Question 14, to give a value of p for each square.
This gives the least likely square = 2 with p=1/6, the most likely square = 45 with p=0.590058.
Question 16
Which square will you land on the most often before reaching the finish?
This time we are looking for the square with the highest value of E(N). Calculating E(N) for each square using Theorem 14, and taking the maximum, gives the square with the most visits = 99 with 1.51811. This is because when you are on this square, you are likely with probability 5/6 to visit it again in the next throw (because you have to throw a 1 to finish.)
Figure 16 shows E(N) for each square.
Question 17
What are the fairest starting positions for 10 players?
The problem as shown by Figure 7a is that players who are in early position get an unfair advantage. So we need to find a fairer set of squares, where the probability of winning for each player is as close to 1/10 as possible. Since there are 100^10 =10^20 different possible starting positions, it’s not feasible to do an exhaustive (“bruteforce”) search. We also need to define what's meant by “fair."
These problems don't exist for the 2player case. By searching every value of P_wins (calculated in Question 4), we find that the starting positions with probabilities closest to 0.5 is squares 39 and 22, with probabilities 0.49994 and 0.50006. Of course it would look a bit weird to start from these squares!
This is an example of an optimisation problem where we need to search a large complex space for the optimum solution. The strategies I investigated were:
1) Pruning the search space, to reduce the number of positions we analyse from 10^20 to something much smaller.
2) Stoichastic methods, for example to guess solution and iterate to a better one.
In order the prune the search space, we can use two insights:
Insight 1: The positions in the solution are likely to have a similar value of E (calculated in Question 3.)
Insight 2: For a partial solution, the first n players should be "fair", irrespective of the values of the other positions.
Both of these insights provide a way to exclude large numbers of potential solutions without evaluating them all explicitly, but leave open exactly what thresholds to use.
After experimentation, the best solution I could find is: 33, 34, 37, 35, 17, 38, 18, 19, 20, 39 with standard deviation 0.000571703. More details can be found in the source code. Figure 17 shows a plot of the winning probabilities for each player, calculated using Equation 6b. Like the 2player case, there isn’t a perfect solution.
Figure 17: Probabilities of each player winning in positions 33, 34, 37, 35, 17, 38, 18, 19, 20, 39.
To evaluate each solution for fairness I used the standard deviation of the probabilities (calculated using Equation 6b)  a perfectly fair position would have a standard deviation of 0. I also disallowed starting on the top of snakes or the bottom of ladders, mainly because these squares can't be landed on in normal play.
Stoichastic methods involve random sampling of the search space (which in this case is huge), plus the ability to locally optimise a solution, for example by adjusting one position at a time. This method also found the above solution, but overall seemed much less reliable.
The following table shows the results for different numbers of players. Nevertheless this is still an open question  maybe other solutions exist?
Number of players  Fairest position 
2  39, 22 
3  35, 26, 24 
4  31, 33, 37, 38 
5  30, 32, 17, 18, 20 
6  30, 32, 33, 34, 35, 20 
7  31, 32, 33, 34, 35, 38, 20 
8  30, 32, 33, 34, 37, 35, 18, 19 
9  31, 32, 33, 34, 37, 35, 38,18, 19 
10  33, 34, 37, 35, 17, 38, 18, 19, 20, 39 
11  30, 31, 32, 33, 33, 34, 37, 35, 38, 17, 18 
12  30, 31, 32, 33, 33, 37, 34, 35, 38, 17, 18, 18 
Table 17: The fairest positions for different numbers of players.
Conclusion
That’s it for now! I’ve exhausted the mathematical possibilities of Snakes and Ladders for now, and I hope the tour has been both accessible and interesting.
It turns out that we can derive precise answers for most of the questions we can ask of a game of Snakes and Ladders. Next time you’re playing Snakes and Ladders, you can now reel off interesting facts about each square, see who’s really ahead, and estimate your odds of winning.
Appendix
The following table contains computed values for some of the figures in this article. These values are only applicable for the board shown in Figure 1.
Square  Minimum throws remaining min(T)  Expected throws remaining E(T)  Completion %  Winning probability adjustment %  Probability of finishing in the same round %  Probability of landing (p) %  Probability of landing again (q) %  Expected number of visits E(N)  Visits by Monte Carlo E(N) 
0

7

35.5432

0

32.3008

1.80536

100

0

1

1

1

6

31.1991

12.2221

40.575

1.804

16.6667

0

0.166667

0.166884

2

7

36.0292

1.36721

31.3752

1.80598

16.6667

0

0.166667

0.166236

3

6

35.4548

0.248945

32.4693

1.81191

19.4444

0

0.194444

0.194488

4

6

33.6121

5.43321

35.979

1.8131

22.6852

0

0.226852

0.227304

5

6

35.5725

0.0822114

32.2451

1.81245

22.6852

0

0.226852

0.226388

6

6

35.3918

0.426064

32.5892

1.81191

45.8211

32.6582

0.680425

0.680628

7

6

35.1843

1.00983

32.9844

1.81158

19.565

7.44933

0.211398

0.212512

8

6

34.9597

1.64174

33.4122

1.81161

22.5773

8.45734

0.246631

0.2471

9

5

32.0082

9.94588

39.0341

1.80273

24.8336

4.47104

0.259958

0.259816

10

7

35.1719

1.04472

33.0081

1.80288

19.8454

12.7869

0.227551

0.226548

11

6

34.7189

2.31934

33.871

1.80283

45.9289

37.2128

0.731501

0.732532

12

6

34.3079

3.47572

34.6538

1.80393

28.3364

18.9426

0.349584

0.349536

13

6

33.9394

4.51246

35.3557

1.80615

25.0372

14.9679

0.294444

0.29386

14

6

33.6121

5.43321

35.979

1.80935

45.208

15.5208

0.535137

0.536116

15

6

33.0615

6.98238

37.0278

1.82247

29.8098

16.3515

0.35637

0.356508

16

6

35.3918

0.426064

32.5892

1.83917

27.9983

32.6582

0.415765

0.41618

17

5

32.0006

9.96703

39.0484

1.83544

32.4128

14.2153

0.377839

0.37634

18

5

31.8418

10.414

39.351

1.83383

28.6225

10.245

0.318896

0.318352

19

5

31.7284

10.7329

39.5669

1.83349

28.1983

10.134

0.313781

0.314332

20

5

31.6485

10.9577

39.7191

1.83405

28.4871

10.1364

0.317004

0.317408

21

5

29.7577

16.2774

43.3205

1.81738

25.0151

10.8666

0.280648

0.280964

22

4

30.3402

14.6386

42.211

1.82683

20.5448

7.14363

0.221253

0.219348

23

4

30.6872

13.6624

41.5501

1.81568

23.6

8.57278

0.258129

0.259208

24

4

30.8885

13.096

41.1667

1.81523

46.2443

37.9585

0.745377

0.746348

25

4

31.0485

12.646

40.862

1.81836

26.4191

14.5726

0.309257

0.309632

26

4

31.169

12.3067

40.6323

1.8241

50.6056

40.8079

0.854939

0.85714

27

4

31.2491

12.0814

40.4798

1.83171

31.1289

21.818

0.398159

0.395384

28

3

20.999

40.9199

60.0033

1.91016

37.451

19.3768

0.464519

0.465216

29

6

32.769

7.80527

37.5849

1.91347

32.237

24.6171

0.427644

0.426368

30

5

32.0965

9.69743

38.8659

1.90021

33.5797

26.3435

0.455896

0.456448

31

5

32.0082

9.94588

39.0341

1.90612

48.8003

26.9027

0.667608

0.669336

32

5

31.8926

10.271

39.2542

1.91262

34.5031

26.1767

0.467374

0.467312

33

5

31.7295

10.7299

39.5649

1.91844

31.3426

22.1844

0.40278

0.402704

34

5

31.5231

11.3106

39.958

1.92522

31.4534

22.0582

0.40355

0.405772

35

5

31.3642

11.7576

40.2606

1.92832

35.2915

25.0406

0.470809

0.468088

36

4

28.0612

21.0504

46.5518

1.94352

36.6667

23.2919

0.478003

0.47792

37

6

31.4783

11.4365

40.0432

1.94117

31.397

21.902

0.40202

0.4042

38

6

31.1991

12.2221

40.575

1.94812

41.5718

20.7283

0.524422

0.523236

39

5

30.7509

13.4832

41.4288

1.94684

29.6123

19.3705

0.367264

0.368232

40

5

30.2847

14.7947

42.3167

1.95947

29.4499

18.4991

0.361344

0.361092

41

5

30.4111

14.4392

42.0761

1.93526

28.6024

19.2728

0.35431

0.355588

42

5

29.7577

16.2774

43.3205

1.97578

46.3619

24.6811

0.615542

0.616648

43

5

30.4665

14.2834

41.9706

1.91661

34.1804

21.8705

0.437484

0.43792

44

5

29.524

16.9351

43.7657

1.99045

35.3458

20.2836

0.443394

0.443964

45

4

28.0612

21.0504

46.5518

2.00464

59.0058

35.008

0.907893

0.907252

46

4

27.488

22.6632

47.6436

2.08662

41.0383

21.0793

0.519994

0.521984

47

4

31.169

12.3067

40.6323

2.28711

36.2262

33.7047

0.546436

0.548232

48

4

25.8376

27.3066

50.7871

2.36254

41.1458

15.5783

0.487384

0.484744

49

6

34.7189

2.31934

33.871

2.71689

32.3729

30.534

0.466025

0.465292

50

4

23.869

32.8452

54.5368

2.76238

35.4741

9.76064

0.393111

0.391412

51

3

19.2849

45.7423

63.268

2.97539

34.683

9.8511

0.38473

0.386036

52

5

24.0485

32.3401

54.1948

2.99579

21.9744

5.85699

0.233415

0.233832

53

5

23.7185

33.2687

54.8234

3.01402

31.46

29.2056

0.444385

0.440348

54

5

23.3858

34.2047

55.4571

3.03121

23.3726

10.0068

0.259716

0.258928

55

5

23.0577

35.1278

56.0821

3.04872

20.131

9.22639

0.221771

0.220616

56

5

23.7185

33.2687

54.8234

3.16363

18.9813

26.6376

0.258733

0.256224

57

4

22.2681

37.3494

57.586

3.17497

18.3697

4.92568

0.193214

0.1927

58

4

22.1426

37.7025

57.8251

3.23348

21.2575

5.69701

0.225417

0.225024

59

4

21.7383

38.8398

58.595

3.25974

21.1422

5.65066

0.224084

0.222912

60

4

21.3896

39.8208

59.2592

3.29177

30.6328

32.4877

0.453737

0.451312

61

4

21.089

40.6668

59.8319

3.33415

19.786

9.94258

0.219704

0.218804

62

3

20.7737

41.5537

60.4323

3.37495

19.7548

9.94306

0.219359

0.217904

63

3

20.4752

42.3936

61.0009

3.41927

22.8398

10.7537

0.255919

0.255048

64

4

21.3896

39.8208

59.2592

3.71724

18.6126

30.1249

0.26637

0.262968

65

3

19.3128

45.6641

63.215

3.95471

21.5882

5.64629

0.228801

0.22794

66

3

19.2975

45.7069

63.244

3.9631

21.628

5.79601

0.229587

0.228804

67

3

19.2849

45.7423

63.268

4.04499

49.6539

13.9386

0.576959

0.578504

68

2

18.8821

46.8756

64.0352

4.0272

23.5443

6.48519

0.251771

0.25064

69

2

18.6842

47.4326

64.4123

4.05157

23.9964

6.69127

0.257173

0.256788

70

2

18.6204

47.612

64.5337

4.11306

23.9811

6.82663

0.257382

0.256716

71

2

15.1073

57.4959

71.2251

4.18276

28.33

5.6542

0.300278

0.30014

72

2

19.2062

45.9639

63.418

4.4182

24.2385

7.53766

0.262145

0.261724

73

2

19.2095

45.9547

63.4117

4.82297

46.0793

30.5657

0.663639

0.662636

74

1

16.4653

53.6754

68.6386

6.0686

25.4291

9.83163

0.282018

0.281904

75

1

17.4963

50.7745

66.6747

5.41473

46.9618

30.6346

0.67702

0.677016

76

1

18.2378

48.6883

65.2624

4.99829

29.774

16.6074

0.357034

0.357108

77

1

18.7537

47.2369

64.2798

4.77477

30.3803

18.6917

0.373643

0.372936

78

1

19.0745

46.3345

63.6689

4.72228

47.9173

36.6062

0.755867

0.754336

79

1

19.2292

45.8993

63.3742

4.83408

37.8112

27.034

0.518203

0.518044

80

0

0

100

100

3.01719

49.3964

0

0.493964

0.493952

81

4

23.6829

33.3688

54.8912

2.71768

31.9694

28.4739

0.446961

0.446408

82

3

22.6868

36.1713

56.7885

2.88572

30.1589

26.1929

0.408618

0.407028

83

3

21.849

38.5284

58.3842

3.0355

30.4814

26.9409

0.417215

0.416628

84

3

20.999

40.9199

60.0033

3.23077

54.9615

38.1758

0.888997

0.889656

85

3

20.1573

43.2879

61.6064

3.48127

33.0222

26.0695

0.446666

0.44556

86

3

19.5169

45.0898

62.8263

3.71408

32.5435

25.1432

0.434743

0.4342

87

4

30.8885

13.096

41.1667

4.48449

33.2395

34.4646

0.5072

0.507272

88

2

16.71

52.9869

68.1725

4.61404

33.0701

23.5739

0.432706

0.432036

89

2

16.8223

52.6708

67.9585

4.47446

33.159

24.0729

0.436721

0.435412

90

2

15.8989

55.2689

69.7174

4.89473

34.2022

22.2628

0.439972

0.438912

91

2

15.1073

57.4959

71.2251

5.3765

51.3866

22.7749

0.665413

0.664176

92

2

15.6741

55.9014

70.1456

5.26552

31.9317

20.4873

0.401593

0.401732

93

2

19.2095

45.9547

63.4117

6.30315

29.2131

26.2421

0.396068

0.39522

94

1

11.5479

67.5104

78.0048

7.06861

34.0942

13.9182

0.396068

0.394964

95

1

17.4963

50.7745

66.6747

7.43353

29.6478

23.9725

0.389961

0.389016

96

1

10.3582

70.8576

80.2708

7.43353

28.917

39.2196

0.475761

0.476792

97

1

10.3582

70.8576

80.2708

7.43353

29.4965

54.3595

0.646278

0.643144

98

1

19.0745

46.3345

63.6689

9.09091

24.6316

23.0141

0.31995

0.319

99

1

6

83.1192

88.5718

9.09091

25.3018

83.3333

1.51811

1.52813

100

0

0

100

100

100

100

0

1

1

0 Comments:
Post a Comment
<< Home