This problem has generated quite a bit of attention. No consensus has been reached by my think tank at the Pickover Discussion Group.

You and an alien are playing a game that involves a set of tunnels. The top of the map is North. In the game, you start at the entrance to the tunnels located at the top of this diagram. As you travel, you may only travel south, east, and west, but never North. Given these permitted directions, each time you have a choice of tunnels, you are equally likely to choose either route. You win a neural stimulator that gives you ecstatic pleasure if you emerge at the tunnel marked "Win." What is the likelihood that you will win the prize? |

For this problem I want you to pretend you are a "mindless marble" always falling down or right or left (but not up). The marble keeps going until it reaches one of the four exits. This is the problem I had in mind.... What is your solution?

From: David Jones

To start the problem off, lets label each intersection (or choice) with a letter. The diagram shows points A-F, X, and Y. Second, to make things slightly easier, notice that when you get to either X or Y that you have won. You cannot lose from either of these points regardless of how you got there. Since we will be measuring probabilities, it is easier to measure them going to points X and Y rather than to the final destination. Points P, Q, and R are only used in the second problem.

Problem 1: What are the odds of winning assuming that I never backtrack. Another way to ask this is "How can I get to X or Y?" Here are the only possible ways of doing so:

1) A B C Y 2) A B C F X 3) A B D E F XNotice that if you make the wrong turn for points D or E, you have already lost. Also note that A B C F E fails because eventually you will be forced to go south on D or E (remember, no backtracking).

From here, the probability is easy. Each point only offers one of two choices. The odds of taking path one are (1/2)^3, path two is (1/2)^4, and path three is (1/2)^5. The sum of these comes to 7/32 or 21.9%.

Problem 2: What are the odds of winning assuming that I can backtrack? Again I only care about getting to points X and Y, but if I can retrace my steps the are many more ways (infinite in fact) that I can still win and by the same token still lose. Thus, path modeling is not desireable. What we need to do is compute the odds of winning from each point along the path. For example:

Assume that I am on point R. What are my odds of winning? I have a 1/2 chance of choosing point X which is a guaranteed winner. But I also have a 1/2 chance of choosing point E. Thus the odds of winning are 1/2 + (1/2)*E. Assume I am on point E. There is a 1/3 change of going to D, F, or going south and losing. Thus I get E = (1/3)*D + (1/3)*F + (1/3)*0. We will have to repeat this idea for each point. Notice a few things however. First, point C now has three choices, not two. I can go to point B, F, or P, so not all points are governed by a 1/2 probability as in the previous problem. Second, we now need to add extra points P, Q, and R. When I get to P I have two choices, I can either go south or return to C. Same ideas for Q and R. Finally, notice that point A doesn't count as a junction except as an entry point. In other words, if I am on point Q, I would not go to east to point A, I would go east to B. Stopping at point A would be just like stopping midway between P and Q; there is no purpose to it.

Applying all of these rules to every point on the map yields the following set of equations.

A = (1/2)*B + (1/2)*Q B = (1/3)*Q + (1/3)*C + (1/3)*D C = (1/3)*B + (1/3)*F + (1/3)*R D = (1/2)*E E = (1/3)*D + (1/3)*F F = (1/2)*E + (1/2) P = (1/2)*Q Q = (1/3)*P + (1/3)*B R = (1/2)*C + (1/2)The odds of winning will be whatever the value of A turns out to be since this is where the maze runner gets to make his/her first choice. Fortunately life is very easy in that we can solve this system linearly. I took the "cheap" way out and plugged the matrix into my 49G. Assuming that I did not enter a number incorrectly or create a bad equation (and I did double check), A=161/880 or 18.3% Compared to the nearly 22.9% from problem 1, in this case it appears that going backwards would be a bad idea.

Incidently, if you remove points P and R under the premise that they are not junctions and that the traveller must continue down that hallway, A comes out 1/6 making things even worse. So even when you allow backtracking, the solution to problem depends on when you allow the runner to choose a direction.

Davy

I assume that your "valid choice" mean a choice where the outcomes are unknown. Based on this assumption, whenever I can see the "win" exit a valid choice will not exist anymore even though there might still be tunnels branched out other ways.

Three possible routes leading to the exit are: E-E-E, E-E-S-S, and E-S-E-E

Therefore, I think the chance to win is (0.5)^3+(0.5)^4+(0.5)^4 = 0.25 or 25%

Cliff,

I'd say 1/4 of the time you will win (minus the time you get stuck if you go West just before winning, assuming you cannot backtrack).

Steve

> > > | > ---'-------,---------,--a---, > | | | > | | | > | | | > |--,--b---| | > | | | > | c | > | | | > | |------| > | | | > | | | > | .---d--| > | | > | > 1/8 of the time you'll go through (a) and win. 1/16 of the time you'll go through (c) and win (not going through b) 1/16 of the time you'll go through (b) and win. If you get to (d) going West you're screwed, unless you backtrack.Andy replies, "I got the same result of 1/4 but I don't think it matters whether you can backtrack or not." Andy4996

Just some trivia: there's a Greg Egan short story (I think it's called "Into The Darkness" in his collection _Axiomatic_) that has a main premise very much like this puzzle: some kind of malfunctioning time machine appears randomly and creates a thick spherical "shell" around itself in which light and matter can only move _inwards_. When the time machine departs, everything in the shell disintegrates, so it behooves people to make their way to the center. Of course, since light can only move inwards, this means you're walking into a big black wall. And if you make a wrong turn, you're dead.

Craig

We have so many different answers for this one. Here is my logic. Let me know if this logic applies, or if it only applies when the problem is worded in a particular fashion.

I can see this problem is a tough one since everyone has a different answer. I'm happy to accept different answers so long as we can agree on them given certain assumptions. (My intent is to put this in a book.)

My logic:

One way to solve this tunnel puzzle is to mark the number of possible routes you can take at each junction and add these numbers as you proceed to the four possible exits. For example, we can see there is only one way to get to the leftmost exit. (That means it's not too likely you would arrive at this exit by randomly walking through the tunnels.) The more ways she can get to an exit, the more likely it is that you will arrive there. The total number of possible routes to the exits is 1 (left-most tunnel) + 4 (left-middle tunnel) + 4 (right-middle tunnel) + 6 (right-most tunnel)= 15. Thus, the probability of you exiting at the right most exist is 6/15. This means the chance of winning the prize is about 40%.

Cliff,

What bugs me about your logic are the superfluous paths that seem to affect the outcome. Consider the modified maze I've put here

I added a detour to the leftmost leg, which increases by 1 the total number of paths and the number leading to the leftmost path. Would this change the likelihood of winning? By your logic it would change it to 6/16, slightly decreasing your odds. Somehow this doesn't seem right. What am I missing?

Steve

[Cliff says, "By the way, when I devised the problem, I was thinking that you wouldn't even try the paths that eventually lead North -- if that matters. You would only randomly choose between 'viable' paths. If you only choose between 'viable' paths, is my logic good?"']

> -----Original Message----- > From: Steve Brazzell [mailto:steve@brazzell.com] > Sent: Thursday, January 24, 2002 8:52 AM > To: CliffordPickover@yahoogroups.com > > [Cliff says, "By the way, when I devised the problem, I was > thinking that you wouldn't even try the paths that eventually > lead North -- if that matters. You would > only randomly choose between 'viable' paths. > If you only choose between > 'viable' paths, is my logic good?"'I don't think so, but perhaps I'm misunderstanding it. As I read your logic, the following (insultingly:) simple example has 4 total paths through it, 1 leading to a win, thus 1/4 chance of winning. But in this case isn't the only thing that matters whether you go East or West at the first junction (thus win=1/2)?

[ Cliff says, for your fascinating variation, I would say that

Tunnel 1 (left): 1 way of getting there Tunnel 2: 2 ways of getting there TUnnel 3: 0 ways Tunnel 4: 1 wayThus, the odds of winning are 1/4. Think of dropping a marble and trying to guess where it comes out. (We'd have to redraw this slightly so that gravity is pulling everything down properly, but you get the idea.)]

--- Steve Brazzellwrote: > Cliff, > > What bugs me about your logic are the superfluous > paths that seem to affect the outcome. Consider > the modified maze I've put here > > http://www.innovatum.com/maze.gif > > I added a detour to the leftmost leg, which > increases by 1 the total number of paths and the > number leading to the leftmost path. Would this > change the likelihood of winning? By your logic it > would change it to 6/16, slightly decreasing your > odds. Somehow this doesn't seem right. What am I > missing?

I am inclined to agree with Steve on this one.

>> One way to solve this tunnel puzzle is to mark the >> number of possible routes you can take at each >> junction and add these numbers as you proceed to >> the four possible exits. For example, we can see >> there is only one way to get to the leftmost exit. >> The more ways she can get to an exit, the more >> likely it is that you will arrive there. The total >> number of possible routes to the exits is 1 >> (left-most tunnel) + 4 (left-middle tunnel) >> + 4 (right-middle tunnel) + 6 (right-most tunnel)= >> 15. Thus, the probability of you exiting at the >> right most exist is 6/15. This means the chance >> of winning the prize is about 40%.

My objection to this is that the solution assumes that each path is as equally likely as the other. Unfortunately this is not the case. A path that required three choices, regardless of the final destination, is still less likely to be traveled than a path that requires only two choices. You can solve the problem this way, but only if you apply a weight to each path where the weight indicates the probability of each path taken.

Incidently, a coworker of mine who has a degree in philosophy with a minor in math did the problem independently and also came up with the 7/32 answer that I did, so I feel more confident in that solution. Since there are differring answers, perhaps one of the programming people in the group can write a simulation for it as an external verification.

Davy

--- Steve Brazzellwrote: > Cliff, > > I'd say 1/4 of the time you will win (minus the time > you get stuck if you go West just before winning, > assuming you cannot backtrack).

I have to retract my 7/32 answer. I agree with Steve on this. I hastily assumed in my solution that the path A B D E F X that it mattered what you did at F. Really, you don't get a choice since the only way is south.

Davy

-----Original Message----- > From: Steve Brazzell [mailto:steve@brazzell.com] > Sent: Thursday, January 24, 2002 2:48 PM > To: CliffordPickover@yahoogroups.com > Subject: RE: [CliffordPickover] Tunnels of Callicrates > > > I put a very quick-and-dirty one up here (windows-32 executable). > In it, I > place a bunch of turnstyles to show where you have travelled. > > http://www.innovatum.com/carrotstop/maze.exe > > [Cliff says, > Steve, before we execute the program, > can you explain to us what this does > and what did you learned as a result of running it?]Sure. Basically, I placed gates at key points throughout the maze (note, not at junctions, but just before). Some of them send you to a junction while others just send you on through (no choices left, leaving the maze). I made each gate keep a counter of every time it was passed through and each gate calls a random function that returns 0 or 1 to determine which way the person will go at the next intersection. The gate then passes control to the appopriate next gate.

You release a new traveler through the maze by pressing the GO button and may specify how many travelers to let though at a time. Some sample runs are as follows:

12600 travelers. Gates (left to right): 12.33%, 39.25%, 23.67%, 24.78% 32767 travelers. Gates (left to right): 12.70%, 38.9%, 23.23%, 25.17% 1M travelers. Gates (left to right): 12.42%, 39.07, 23.52%, 24.99%

For those who care about the programming (in 2 words: Quick and Dirty), the program has one recursive procedure "Go" which is passed which gate is being visited along with the direction being travelled. Within the procedure, a case statement selects the gate and each gate calls the Go routine again, passing one of its two child gates. The "leaving the maze" gates don't call Go again, thus the procedure can exit.

> Steve, you mentioned bugs in your program. What's the latest > statistics for each tunnel of the puzzle? The bug was that the tunnel leading from leg 3 West to leg 2 was not being considered in one of the paths. With this fixed, a run of 150,000 marbles yielded: Tunnel 1: 12.61% Tunnel 2: 45.17% Tunnel 3: 17.17% Tunnel 4: 25.06%

Steve

[Cliff says, Steve are you assuming what I assumed:

(For this problem I want you to pretend you are a "mindless marble" always falling down or right or left (but not up). The marble keeps going until it reaches one of the four exits.)]

Yes. In fact, I had a strong urge to call the button "Drop" and the traveler "marble" even before you said that. No "north" or "180 degree turns" are allowed. The only thing unlike a marble is that during a straight drop downward the marble can turn left or right...well, that and also that the marble can roll straight over a drop without falling...geez)

(Incidentally, the 2nd and 3rd legs should be more like 45/17, not 39/24--there was a bug in version 1.0a)

From: "Steve Brazzell"for a layperson?....ok. Iss like Paw always say, heed say "boy, listen up! you see a fawk in the road, you take it." Well haff dem marbles you drop gone go right and about a half dozen othern gone go leff. Why'f att aint enuff, of the ones gone leff, half'o'dem gone drop on down dat secon shoot and the ones what don't, heck, they still got about half a chance a'fallin back t'ole number 2 'fo they get out. So number two gone get da line's share right off, even bleedin off some a'da ones what go right and drop down shoot 3. Now lookin at shoot 3 and 4, just eyeballin it says dey gone get filled up bout da same, so right after you done dropped 100 marbles you oughta see 50 of em goin right (50's bout half a 100). Half a dem gone end up in dat winnin shoot and d'utta half in shoot 3, 'cep like I done said 3's gettin bled off bout half the time towards ole 2.[Cliff says, Amazing. So if you were a betting man, you'd pick tunnel 2. Can you explain exactly why we see these probabilities in a way that a layperson can understand? Do you see something interesting or counterintuitive in this little experiment?]

qurqirishd@aol.comHAHA (The Qurqirish Dragon)

I am going to assume that you never backtrack (i.e. if you are going East when you hit an intersection, you will NOT choose West.) I will abbreviate all compass directions from now on.

At the first point, you can go E or W. If you go W, you cannot get to the goal. After two choices, (ignoring determined results) we have either EE or ES. Neither one is a guaranteed win or loss, so we continue. After 3 choices (again, ignoring determined results), we have EEE, EES, ESE, and ESS. EEE guarantees a win. ESS guarantees a loss. After 4 choices, we finish determination of all routes. We have EESS, which is a win; EESW, which is a loss; ESEE, which is a win; and ESES, which is a loss. Listing the results, we have:

W=loss, 50% chance EEE=win, 12.5% chances ESS=loss, 12.5% chance EESS=win, 6.25% chance EESW=loss, 6.25% chance ESEE=win, 6.25% chance ESES=loss, 6.25% chance.Adding these up, there is a 75% chance of losing, and a 25% chance of winning

Consider the parts of the diagram that look like this:

* * A*B * * * * * C*D * * Suppose you are travelling west, and reach C. Which of the following is the case? a) You are allowed to go north to B. The prohibition on going north only applies when you have a choice of direction. b) You can't go north to B, so you must turn around and go east to D. c) You can't go anywhere, you don't emerge anywhere, and you lose.Is there a 100% likelihood that you start out moving south?

Is there a 0% likelihood that you ever voluntarily come to a permanent stop before emerging?

When you reach an intersection, and you have two or three choices of direction, may you choose to reverse your previous direction (i.e. switch from east to west, or vice versa)?

Is there a 0% likelihood that you ever voluntarily change direction at a non-intersection before emerging?

I constructed a state machine in Excel and got the following results: with back-tracking allowed

1840 wins in 10000 trials or 18.40% with back-tracking not allowed 2282 wins in 10000 trials or 22.82%

Seems pretty simple to me. The only ambiguity lies in the framing of the puzzle. You don't say what happens if the mouse reaches a point at which he cannot move (i.e. the only option is north - which is against the rules). Does he turn around? I will assume he must.

Define all directions as seen by the mouse.

The sum of all possible correct routes is your answer.

1) Left, straight, straight (all options after this win) = 1/8 2) Left, straight, right, straight (ditto) = 1/16 3) Left, right, left, straight (ditto) = 1/16 All other directions fail. The answer is 1/4 or 25%The only other difference is if you intended "turn around" as a valid tunnel to follow at an intersection. I assumed not.

Problem 1

Problem 2

According to some people's false logic you, there is equal probability that the marble will get to the rightmost or leftmost exits. Doesn't it look funy to you? Yes, there are 6 different pasages, but don't forget that they don't have the same probability of occuring. The answer should be: on each junction the marble will go in the right direction with 50% probability. So, if we have four junctions, the answer is p=0.5*0.5*0.5*0.5=0.0625=6.25% and not 1/6 as you said. Thanks for your enjoyable site. It's delitious for Math-lovers. Jack Kuperman.

Steve Brazzell: If I'm looking at this right, in your second maze 1/2 of the marbles go left at the first junction. Of these, 1/2 fall out the first tunnel. So for the first tunnel, the odds are 1/2 of 1/2, or 1/4. Half of the marbles go right at the first junction. Of these, 1/2 go left at the next junction (leaving 1/4 of the marbles still going right). Of these, 1/2 fall out the next junction (leaving 1/8th). Of this 1/8th, 1/2 fall out at the last junction, leaving 1/16 winning.

Hi Steve (Brazzell), I think I'm latching on to your (valid) method. One way to solve this class of tunnel problems is to first draw all paths that get you to a "sure win" tunnel. At each intersection, you place a "1/2" representing a binary decision. For each path, you multiply the 1/2s along the route. And then, in the end, you sum the products for each path. Thanks for all the feedback, Cliff

Steve Brazzell says: Exactly. I think the general algorithm would be something like this, which would account for junctions (gates) with more than 2 choices:

ProbTotal = 0 For each UniquePathToWin ProbGate = 1 For each gate in path ProbGate = ProbGate * 1/(NumChoicesAtGate) endFor ProbTotal = ProbTotal + ProbGate endFor

Kris Musial

For Problem 1 p(4)=p(win)=3/16 and all probabilities are as follows: p (1)=1/16 p(2)=7/32 p(3)=13/32 p(4)=3/16 p(5)=1/8. For Problem 2 p(3)=p(win)=1/16 and all the probabilities are: p(1) =1/4 p(2)=11/16 and p(3)=1/16It easy to verify that the sum of all the probabilities for each case is 1, which means that the marble must fall somewhere. Kris

Mark Ganson: Hello Cliff,

I came across your website and thought I would tackle the Tunnel of Callicrates puzzle. I calculate the odds of winning to be 3:7 for a probability of roughly 42.86%. This assumes that the marble can never turn back on itself and go the other direction. For example, coming down from the Start position, we can go East or West. If we go West, we always lose unless when we get to the next tunnel where we can go West or South, we turn around and go back East. If you can turn back on yourself and do a 180 degree change of direction, you might get stuck in an infinite loop, bouncing East to West to East to West... and never win or lose. If i = infinity, I'd say the chances of getting caught in the infinite loop would be 1/(i-1) or thereabouts.

Here is an enumeration of the potential decisions that result in wins or losses (beginning from Start point) assuming you cannot make any 180 degree changes of direction:

S:W (lose) S:E:E:E (win) S:E:E:S:S (win) S:E:E:S:W (lose) S:E:S:S (lose) S:E:S:E:S (lose) S:E:S:E:E (win)Mark Ganson: I stand corrected. The probability of winning is 25%. Thanks for helping me understand this tricky puzzle.

I wrote a bit of Java code to verify that the odds of winning are 25%. I ran the simulation out to 100000 iterations, making 50% random direction decisions at each choice point. Each time it came back very close to 25% wins and 75% losses. If anybody want to see my java source, send me an e-mail with Tunnels.java in the subject line.

I think the problem with my original (incorrect) answer was that I was equating the different paths with having equal likelihoods of happening.

To recap:

C1:W -> (lose) C1:E -> C2:E ->C3:E (win) C1:E -> C2:E ->C3:S ->C6:S (win) C1:E -> C2:E ->C3:S ->C6:W (lose) C1:E -> C2:S ->C4:S (lose) C1:E -> C2:S ->C4:E -> C5:S (lose) C1:E -> C2:S ->C4:E -> C5:E (win)(wrong) odds of winning: 3:7 (42.86%) (wrong)

The problem with the above is that I was considering a loss at C1 to be the equivalent of a loss at C6. This was a mistake since getting a loss at C1 is much more likely to happen than getting a loss at C6 for the simple reason that every trip through the tunnel passes through C1 while only a fraction of them will go through C6. We can win or lose at the following choice points:

C1, C3, C6, C4, and C5.

We can refer to C1 as a 1st step choice point, C3 and C4 as 3rd step choice points, and C5 and C6 as 4th step choice points. The 4 4th step choice points present 2 ways to win and 2 ways to lose, so they cancel each other out. The 2 3rd step choice points present 1 way to lose and 1 way to win, so they cancel each other out. The lone 1st step choice point (C3) presents only 1 way to lose and no way to win directly, so it should have been given a greater weighting in my original analysis. As I had correctly stated in my original post, once you make the decision to go east from C1, you have a 50% probability of winning. So, we can simply by looking at the puzzle as having 1 key choice point, C1, where choosing west gets a loss and choosing east gets a 50/50 chance of winning.

C1:W -> lose C1:E -> 50/50 chance to win or lose

So, 50% of the time you will be in position to win 50% of the time. 50% of 50% = 25%. 50% of the time plus 50% of 50% of the time, you will lose. 50% + 50% of 50% = 75%. In the end, the probability of winning is 25%.

I repeat: the probability of winning is 1/4. Consider the following ASCII TEXT drawing:

Start | | | | ________________________| |___________________________ | _______ ____________C1___C2 _________C3 _______ | | | | | | | | | | | | | | | | | | | | | | |____ | | | |_________| | | | | __ | | | |C4___C5 ___C6| | | | | | |__| | _________| | | | | | | | | | |____ | | _______ | | | | | | | | | | | | | | | | | | |_______| | | | | |_______| | | | | | | _______ | | | | _________| | | | | | | | | | | | | | |___| | | | | | | | | | | _____| | |_______| | | | | | | | |_________ | | | | | | | | | lose lose lose win You have 6 choice points, which I have labeled C1, C2, C3, C4, C5, and C6. Here are the ways to win: 1) C1:East ->C2:East -> C3:East 1/2*1/2*1/2 == 1/8 2) C1:East ->C2:East -> C3:South ->C6:South 1/2*1/2*1/2*1/2 == 1/16 3) C1:East ->C2:South ->C4:East ->C5:East 1/2*1/2*1/2*1/2 == 1/16Thus, 1/8 + 1/16 + 1/16 == 1/4 probability of winning.

From: "kmusial_2000"

Mark provided an interesting twist on the Tunnels of Callicrates puzzle: "the marble can never turn back on itself and go the other direction". This version simplifies the original one because it eliminates loops from the marble movement.

The possible paths listed (inserted below) are valid, however, I think that the suggested probability of winning equal to 3/7 is wrong. It would be right if each of the paths is equally probable (3 successes out of 7 possibilities). But the paths are not equally probable and the longer the path the less likely it is: to be precise the probability is equal to 1/2 to the power of the number of decision points. For example the probability of the first path is 1/2, second - 1/8, third - 1/16, etc. If we count this way the probability of winning is 1/4. I confirmed this with the simulation, in which the marble was thrown 1 million times and it landed at each of the four gates with the ratios: 0.13, 0.45, 0.17 and 0.25 (win).

By the way: the simulation which allows the marble to move back (original puzzle) gave the following ratios: 0.13, 0.48, 0.19, 0.19.

Regards, Kris

mwganson@hotmail.com (Mark Ganson):

Tunnels of Callicrates

(The following assumes we cannot double back on ourselves. For example, if we are headed East and come to a point where we may choose to continue East or to go South, we may not choose to turn around and go West.)

Though there are many choices to be made in navigating the maze, we really only need to be concerned with 6 choices, which I have labeled C1 - C6.

Coming down the maze South from the Start, we find our first choice, which we will label C1. C1 offers 2 options, East or West. If we go West, we always lose. If we go East, we will win some and we will lose some. Going East from C1, we come to C2, where our options are South or East. Whether we go South or East, we can still win or lose. Continuing East from C2, we come to C3, where we can go East or South. East at C3 guarantees a win. If we go South at C3, we might win or we might lose. Going South from C3, we come to C6 where we can go South, which always wins, or West, which always loses. Let's go back to C2 and head South from there. The next choice we come to is C4 where we can go South or East. South always loses, East wins some and loses some. Let's go East where we find C5. At C5 South always loses and East always wins.

(Thanks to Bob Morris for his ASCI drawing, which I have modified here.)

Start | | | | ________________________| |___________________________ | _______ ____________C1___C2 _________C3 _______ | | | | | | | | | | | | | | | | | | | | | | |____ | | | |_________| | | | | __ | | | |C4___C5 ___C6| | | | | | |__| | _________| | | | | | | | | | |____ | | _______ | | | | | | | | | | | | | | | | | | |_______| | | | | |_______| | | | | | | _______ | | | | _________| | | | | | | | | | | | | | |___| | | | | | | | | | | _____| | |_______| | | | | | | | |_________ | | | | | | | | | lose lose lose winOnce you commit to going East from C1, you have a 50% chance of winning. Once you commit to going East from C2, you have a 66.67% chance of winning. If you go South from C2, bringing you to C4, you then have a 33.3% chance of winning. If you get to C6 from C3, your chances are 50% of winning. If you get to C6 from C5, you are assured of winning. So, the most critical choices, resulting in winning or losing are at C1, C3, C4, C5, and C6. Some of the choices are only decisive if you make the right/wrong decision, such as C1, C3, and C4. These three choices only produce one endgame result each (win or loss result). Two of the choices are the most critical of all. These are C5 and C6, which can result in both wins and losses. So, with these 2 choices, you have 4 potential decisive results. To recap: C1, C3 and C4 produce one decisive result each. C5 and C6 produce two decisive results each. This gives us a total of 7 decisive results, which means they produce a total of 7 win/loss results. Of these 7 decisive results, 3 are wins and 4 are losses. The odds of winning are therefore 3:7 (42.86%) Here is a list of the potential paths that result in wins or losses:

C1:W -> (lose) C1:E -> C2:E ->C3:E (win) C1:E -> C2:E ->C3:S ->C6:S (win) C1:E -> C2:E ->C3:S ->C6:W (lose) C1:E -> C2:S ->C4:S (lose) C1:E -> C2:S ->C4:E -> C5:S (lose) C1:E -> C2:S ->C4:E -> C5:E (win)odds of winning: 3:7 (42.86%)

mensanator@aol.com (mensanator)responds to the previoius analysis: That's not correct. The seven decisive results are NOT equally likely.

> > > Here is a list of the potential paths that result in wins or losses: > > C1:W -> (lose) The probability of this happening is 50%. > C1:E -> C2:E ->C3:E (win)So the sum of all C1:E paths must be 50% also. Since C2 will only be reached 50% of the time, the two choices are 25% W and 25% S. With C3 being reached only 25% of the time, the two choices are E 12.5% and S 12.5%. Thus, this winning path happens only 12.5% of the time.

> C1:E -> C2:E ->C3:S ->C6:S (win)

Each decision point cuts the probability in half. Since we arrive at C6 only 12.5% of the time, the critical choice (S) happens only 6.25% of the time.

> C1:E -> C2:E ->C3:S ->C6:W (lose) > C1:E -> C2:S ->C4:S (lose) > C1:E -> C2:S ->C4:E -> C5:S (lose) > C1:E -> C2:S ->C4:E -> C5:E (win)

C2:S is 25%. C4:E is half of that or 12.5%. And C5:E is half again or 6.25%

> > odds of winning: 3:7 (42.86%)

Summing the winning paths 12.5% + 6.25% + 6.25% = 25% >