Unrelated to Poker - Please Help

[ykcirT]

Totally unrelated to poker, but I know there are some people who are very good at probability here, so I
thought I would take a chance. Can anyone figure this out:

Assume you are in a room with 4 doors, each heading in a different direction. You have to leave the room, so you randomly decide on a door and go through it. Every subsequent room also has 4 doors, which also lead in different directions (assume each door is exactly N,S,E,W). In each room, you make a totally random choice as to which door to go through. What is the probability that you will return to the original room if you have to keep leaving the room you are in until you either 1) return to the original room, or 2) Get yourself far enough away that you cannot return in 2 consecutive room changes. This drawing might help:

x x x 3 x x x
x x 3 2 3 x x
x 3 2 1 2 3 x
3 2 1 S 1 2 3
x 3 2 1 2 3 x
x x 3 2 3 x x
x x x 3 x x x

In other words: start at S, and start moving around. If you enter a room #3, you will never get back.
What is the probability of returning to the start?

Thanks. Sorry this is so off topic,
Rich

[irchans]

After 1 step you are always at a distance of 1.

If you start at a distance of 1 and take two steps, I get that you will be at

S 1/4 of the time,
1 3/8 of the time, and
3 3/8 of the time.


So, by the 3rd step, there is a 1/4 chance of terminating at S, a 3/8 chance of terminating at distance 3, and 3/8 chance of being at distance 1.

Take two more steps. Now the chance that you have terminated at S is 1/4 + 3/8*1/4. The chance of terminating at distance 3 is 3/8 + 3/8*3/8. And the chance that you are at distance 1 is 3/8*3/8.

Take two more steps. Now the chance that you have terminated at S is 1/4 + 3/8 * 1/4 + 3/8 * 3/8 * 1/4. The chance of terminating at distance 3 is 3/8 + 3/8*3/8 + 3/8 * 3/8 * 3/8. And the chance that you are at distance 1 is 3/8*3/8*3/8.

If you keep repeating till infinity, you get that the chance of terminating at distance 3 is

3/8 + (3/8)^2 + (3/8)^3 + (3/8)^4 + ...

= 3/8 / (1 - 3/8) = 3/5.

So your chance of terminating at S is 1-3/5 = 2/5.

Hope that helps.

[lorinda]

/forums/images/icons/shocked.gif

I believe irchans is wrong....which would be amazing, even more amazing if I've got it right...here is my solution

Im starting in one room south, (symmetrical so irrelevant)

1/4 of the time I move north and "win"
1/4 of the time I move south to a "bad 2", with only one way back to a "1"
1/2 of the time I move east or west to a "good 2" with two ways back to a "1"

So...
1/4 of the time I win instantly
1/4 of the time (1/2 * 1/2) I die to a "Good 2"
1/4 of the time I go to a "Good 2" but return to a "1"
3/16 (1/4 * 3/4) of the time I go to a "Bad 2" and then die
1/16 of the time I go to a "Bad 2" to a "1"

Giving 1/4 win 7/16 dead 5/16 back in a "1"

= 4/16 win 7/16 dead 5/16 repeat

as the repeats follow same rules, i will win in a 4:7 ratio, or 36.36%

[irchans]

I agree with Lorinda. My incorrect solution did not distinguish between 'bad' 2 and 'good' 2.

[Ed Miller]

Ok... I got a slightly different answer from the other folks.

So, you are looking for the chance that you will return to the original square from a square that is labeled 1. I will call this P(1).

You will move to a 2 square with only one way back 1/4 of the time, to a 2 square with two ways back 1/2 of the time, and you will reach the honey pot 1/4 of the time. Thus,

P(1) = 1/4 + 1/4 * P(2_1) + 1/2 * P(2_2)

where P(2_1) is the probability of returning to S from the 2 square with only one way back, and P(2_2) is the probability of returning to S from the 2 squares with two ways back.

We also know that the chances of returning from the 2 squares are, respectively,

P(2_1) = 1/4 * P(1)
P(2_2) = 1/2 * P(1)

Thus,

P(1) = 1/4 + 1/4 * (1/4 * P(1)) + 1/2 * (1/2 * P(1))
P(1) = 1/4 + 1/16 * P(1) + 1/4 * P(1)
P(1) = 1/4 + 5/16 * P(1)
11/16 * P(1) = 1/4
P(1) = 4/11

Since you move initially to a 1 square from S every time, the probability you are looking for is just P(1). Thus, my answer is 4/11.

[Ed Miller]