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Alexsandro

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- Thread starter Alexsandro
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- #1

Alexsandro

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- #2

TenaliRaman

- 644

- 1

1. whenever the snake moves right, mark it as -1

2. whenever the snake moves straight, mark it as 0

3. whenever the snake moves left, mark it as 1

Thus the movements of snake can be encoded as a stream of -1's,0's and 1's.

This sequence terminates whenever we see 4 consecutive 1's or -1's.

The required probability then becomes,

P(length of sequence >=10)

= 1 - P(length of sequence<10)

I think this should be computable, albeit tedious (which is why i am avoiding doing it right now anyways).

-- AI

- #3

balakrishnan_v

- 50

- 0

-- and all mirror reflections state 1 whre the last 2 metres can be oriented in any way

_| " state 2 where the last 2 metres should not be oriented parallel to the first one

=| state 3

We can formulate the following:

p1(n+1)=1/3.p1(n)+1/3.p2(n)+1/3.p3(n)

p2(n+1)=2/3.p1(n)+1/3.p2(n)+1/3.p3(n)

p3(n+1)=0.p1(n)+1/3.p2(n)+0.p3(n)

which is of the form

P(n+1)=T.P(n)

So we have

P(9)=T^9.P(0)..as we need 9 movs to be alive so that it definitely moves 1 more metre

where P(0)=[1,0,0]'

SO we get

P(9)=[0.2317 0.3136 0.1108]

or pr(living)=0.6560

_| " state 2 where the last 2 metres should not be oriented parallel to the first one

=| state 3

We can formulate the following:

p1(n+1)=1/3.p1(n)+1/3.p2(n)+1/3.p3(n)

p2(n+1)=2/3.p1(n)+1/3.p2(n)+1/3.p3(n)

p3(n+1)=0.p1(n)+1/3.p2(n)+0.p3(n)

which is of the form

P(n+1)=T.P(n)

So we have

P(9)=T^9.P(0)..as we need 9 movs to be alive so that it definitely moves 1 more metre

where P(0)=[1,0,0]'

SO we get

P(9)=[0.2317 0.3136 0.1108]

or pr(living)=0.6560

Last edited:

- #4

balakrishnan_v

- 50

- 0

To be more precise it is

12913/19683

12913/19683

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