Insanely tough math question

[saintbert]

Suppose we are playing a board game with the following rules:

1. The board has 5 spaces on it.

2. Your playing piece starts out on space number 1.

3. Each move consists of moving the piece to a randomly chosen space other than the one which you currently occupy (for example, if you are on space number 2, you randomly move to either space 1, 3, 4, or 5, with each space having equal chance of being selected, but you cannot stay on space number 2).

4. You win if, after seven moves, you are back on space 1.



The probability of winning this game is p. Find 4096p.

 

[LavidDichterman]

solve it recursively

say you actually have x moves and at the x-1th move you are on a tile that is not labeled '1', if that is the case, the probability of landing on the '1' tile on the xth turn is 1/4. now you need to figure out the probablity of landing on a non '1' tile on the x-1th turn. this is = (1-p(x-1)) where p(x) is the probability of winning on the xth turn. so for your question:

p(1)=.25
p(2)=.25(1-.25)=.1875
p(3)=.25(1-.1875)=.203125
p(4)=.25(1-.203125)=.19921875
p(5)=.25(1-.19921875)=.2001953125
p(6)=.25(1-.2001953125)=.1999511719
p(7)=.25(1-.1999511719)=.200012207=p

4096p=819.25

 

[jagec]

Originally posted by: LavidDichterman
solve it recursively

say you actually have x moves and at the x-1th move you are on a tile that is not labeled '1', if that is the case, the probability of landing on the '1' tile on the xth turn is 1/4. now you need to figure out the probablity of landing on a non '1' tile on the x-1th turn. this is = (1-p(x-1)) where p(x) is the probability of winning on the xth turn. so for your question:

p(1)=.25
p(2)=.25(1-.25)=.1875
p(3)=.25(1-.1875)=.203125
p(4)=.25(1-.203125)=.19921875
p(5)=.25(1-.19921875)=.2001953125
p(6)=.25(1-.2001953125)=.1999511719
p(7)=.25(1-.1999511719)=.200012207=p

4096p=819.25

close, but remember, you START on square 1.

so,
p(1)=0 (your first move will NOT finish on square 1)
p(2)=.25*(1-0) (your second move has a .25 chance of ending on square 1)
p(3)=.25*(1-.25) (your third move's chances of landing on 1 are .25*the second move's chances of NOT landing on square 1)

...and so on, until p(7)=.25(1-.200195)=.1999511719, 4096p=819.

 

[LavidDichterman]

bah, i had that written down but didn't bother double checking... thanks for that... so what's the big deal about .25 :-D

 

[eilute]

Originally posted by: saintbertThe probability of winning this game is p. Find 4096p.

Are you asking for 4096*p?

 

[imported_Phil]

No-one else has the word "homework" popping into their heads?

 

[jman19]

This doesn't seem "insanely tough" to me. I agree with Phil, sounds like someone needs his probability hw done for them 😛