2003 AIME II problem 13 Probability of moving to a vertex
2003 AIME II problem 13
- A bug starts at a vertex of an equilateral traingle. On each move, it randomly selects one of the two
- vertices where it is not currently located and crawls along a side of the triangle to that vertex.
- Given that the probability that the bug moves to its starting vertex on its tenth move is m/n
- where m and n are relatively prime positive integers,
- find m + n
- So this lesson right here where m and n are relatively prime positive integers
- essentially says we have the positive version of this fraction in its simplified form
- So let's just think about the problem we have an equilateral triangle here
- with three vertices
- A B and C
- and our bug is going to start at lets say vertex A
- So this is our bug right here
- and on each move it will randomly select one fo the two vertices
- So on its first move it will either go to vertex B or then vertex C
- and then depending on whether it went there
- If it went to vertex C then on it's next move it wil either go to vertex B or vertex A
- if it went to vertex B on its next move it will either go to vertex A or vertex C
- So let's just think about it.
- We want to figure out the probability of it moving to vertex a on its tenth move
- So let's just think about what happened in each move.
- The probability of moving toward on of the vertices in a given move
- So let us say that we have move 1 over here in this column
- so we have vertex A,B and C
- so let's think about move one
- On move 1 what's the probablilty of moving to vertex A
- Well you are already on vertex A and it says that the insect in going to go to one of the other 2 vertices
- So the probability on move one of moving to vertex A is 0
- What's the probablilty of moving to vertex B?
- Well it is going to be 1/2
- what's the probablilty of moving to vertex C?
- It's going to be 1/2 as well
- half a chance of going there and half a chance fo going there
- Well obviously all of the probabilities have to add up to 1 because it's going to do something on that
- move. Now let us think about move 2
- So what is the probability of now moving to vertex A?
- Well there is a 1/2 chance that we are already at B. If we were already at B there will now be a 1/2
- chance of moving back to A
- So this is 1/2 times 1/2
- that's only the situation if we were already at B. If we were already at C
- then there would also be a 1/2 probability of moving to A
- So this is going be the 1/2 chance that we were already at C times the probability that we move to A
- So 1/2 times 1/2 = 1/4 + 1/2 times 1/2 = 1/2
- So what is the probability of moving to B in the second turn? Now the only place other than B
- that the bug can move to is C in the second turn.
- This is because we know that at the end of the first turn the bug will not be at A
- It will either be at B or C
- If the bug is already at B there is no way that it will go back to B
- So the only situation is if the bug is already at C there is a 1/2 chance of that happening
- so that is this number right here.
- So if the bug is already at C and that is the 1/2 probablility then there is going to be a further 1/2
- probability that it moves to B
- So it is going to be 1/2 times 1/2 = 1/4
- Now what is the probability of moving to C? Well it is the same logic.
- The bug is not going to be on vertex A after the first move there is a 1/2 probability that it is going
- to be at B; and then if it is at B then there is a 1/2 probability that it will go to C
- So this is going to be 1/4 as well
- In general let's say that in move N the probablility of being at vertex A is
- the probability of A
- The probability of being at vertex B is probability of landing on B at move N
- and the probability of C is Probability of moving ot vertex C on move N
- Now let us think of the probability of being at each of these nodes on N+1
- So in order to get to A on move n+1, you either have to be at B or C. You cannot be at A and stay at
- So the probability of moving to A in your n+1th move is going to be the probability of being at B times
- 1/2 (right because if you are at B there is a one half chance that you are going to go to A) + the probability
- of being at C at the end of your last move times 1/2
- There is a half chance that if you are at C you will move to A. There is a half chance that if you are
- at C you will move to A.
- Now what's the probability of going to B?
- Well there are two ways that will get ot you B.
- You can start at A so you are going to start at A
- So the probablilty of being at A times 1/2
- or you could start at C. So it's plus the probability at C times 1/2
- If you were doing this problem in a situation that had time constraints you would not neccesarily have
- to go through this, but I just want ot make it all clear.
- And finally what is the probability of landing in C at you n+1th term
- It's going to be the probability of being at A times 1/2
- + the probability of being at B times 1/2
- So hopefully this makes the pattern of what is happening little bit clearer
- So no matter what A is going to be the average.
- A in the next move is essentially going to be the average the probabilities of being at B and C
- And you can see that. The average of 1/2 and 1/2 is 1/2
- The probability of being at B in the n+1th move is going to be the average of the probabilities of being
- at A and C in the last move the average of 0 and 1/2 is 1/4
- and the same logic for C. The average for 0 and 1/2 is 1/4
- Now that will help me simplify things a little bit more for our brains
- Let's keep going this way and let me continue the columns
- Rows A,B and C
- and now we are on move 3. I am just continuing it down here
- So what is the probability of the move being A in move 3
- Well we just said it's going to be the average of B and C and B and C is 1/4
- So the average of 1/4 and 1/4 is going to be 1/4
- What is the probability of being at B?
- Well the average of A and C in move two is 1/2 and 1/4
- that's the same thing as the average of 4/8ths and 2/8ths
- that's 3/8ths
- and then we are going to get the exact same value for C. I think 3/8ths
- Another pattern here is that B and C are always going to be the same thing
- and since B and C are always going to be the same thing the average is going to be that value
- and it's going to be the probability of being at A in the next turn
- So let us just keep doing this
- and we can go all the way to ten but maybe we will see some type of pattern here
- So on out fourth move what is the probability of being at A? Which is the average of B and C
- So it is going to be 3/8ths
- So we can just take this value here and then what is the probability of being at B?
- Well it is going to be the average of A and C
- and so let's see this is 4/16 + 6/16 = 10/16
- and then we have to divide it by two which is 5/16
- Did I do that right? Let me just do this on the corner. I don't want to make a careless mistake here.
- 1/4 is 4/16 + 6/16 and then we want to divide that by two.
- We are taking the average. So that's 10/16 divided by 2 which is equal to 5/16
- Alright so let me clear theis out so that I do not waste valuable real estate
- Alright. So then this is also going to be 5/16
- because it's going to be the average of 3/8 and 1/4 as well
- so it's going to be 5/16
- Now we are on move 5
- And we could keep going like this all the way to move ten but hopefully we willl see a pattern here
- So move five the probability of moving to A is going to be 5/16 the average of these two
- The probability of moving to B (We can already see a pattern) it seems like we went from having
- 2,4,8 in the denominator to 16 in the denominator
- I'm guessing that we are going to have to go to 32 in the denominator
- and so we are going to take the average of 10/32 and 12/32 and that is 11/32
- And this is also going to be 11/32
- Let's go to move 6 for A it's going to be 11/32 (it seems like the pattern is holding) this is going
- ot be something over 64 and so this is the average if you multiply this times 4 you get 20
- and then this is 20/64 and this is 22/64 the average is going to be 21
- So let us keep extrapolating this and see if there is a way
- and we could actually just keep doing the math for 7,8,9,10
- Feel free to do that on your own if you are interested and hopefully you will get the right answer.
- But let's see is you can spot a pattern here
- This is also going to be 21/64 move 7 for A is easy. That's just going to be the average of these two
- guys, 21/64
- But let's see if there is a pattern forming
- So it looks like the (and remember the question is we just want to look at the probability of it moving
- to A on the tenth move.)
- and so if you look at A you start off with a denominator
- So this is 2 to the move minus 1. 1/2 to the move minus 1
- It looks like it's always cuz its 1/2 to the 0. This is 1/2 to the 1
- and you have 1/2 squared that's 1 less than three
- 1/2 to the 3rd
- So it looks like if you fast forward to the nth move or even better let us fast forward to the tenth
- move. So 7, 8, 9 and 10
- This is not actually a lot of math to fill in. But let us just figure it out
- So 7,8,9, 10, this is going to be 16, 32, 64
- This is going to be 128 if we go with the pattern
- This is going to be 256 and then this row over here is going to be over 512. So we actually already figured
- out the n part. And lets see is we can figure out a pattern for the numerators here.
- We went from 1 to 3 to 5
- So let us see 5 is 3+2 not 1
- Then we go, let's see you have 5 and it looks like you are taking 2times this and adding it to that to
- get that. 2 times 1 plus 3 is five. 2 times 3 plus 5 is 11. 2 times 5 plus 11 is 21.
- So let us just go with that
- So 2 times 11 is 22 plus 21 is 43 and then 2 times 21 is 42 plus 43 is 85
- and then this is going to be 86 plus 85 is let's see 85 times two is 170 so its going to be 171
- So if we believe what we just did we get out answer
- m + n = 171 +512
- Which is let's see we are going to have two plus 1 is 3
- 1 plus so this becomes ten and then we get 703. Wait I made a bone headed mistake on the addition
- Let me do that again
- So 1 plus 7 is 8 not 10. I don't know why my brain was thinking that and then 5 plus 1 is 6. So we get
- 683 as our answer
- So that is m + n and the probability of moving to vertex A on the tenth turn is 171/512
- and you can verify this for yourself
- Or we could even prove to ourselves is we are interested that each successive term here is really equal
- to the previous term times two times the term before that
- Actually for fun let us just prove it to ourselves
- So let us just say that this is the nth term
- Our probability of being at A is A/2^n
- and what ends up happening is that the probability of being at b and C let us call that B/
- This power is always 1 higher power of 2. 2 ^n plus 1 and this right over here is also going to be the
- same probability 2/n+1
- Then on the nth + 1 term the probability of being at A is just going to be the average of these two guys
- which is going to be B times 2 to the n plus 1
- and the probability of being at, this is B
- this is C, this is A
- The probability of being at B is going ot be the average of 2A/ 2^(n+1) (I just multiplied the numerator
- and the denominator by 2) plus B/2^(n+1)and then all of that over 2
- So we are going to divide the whole thing by 2
- and that's going to be the same value for C. And actaully if we do a little math here this is going
- to be [2A + B]/[2^(n+2)]
- And so on the nth+2 move our value for A is going to be this thing right here
- The average of these two guys which is the exact same thing so it's going to be
- [2A+B] / [2^(n+2)]
- So this verifies the pattern that we just thought about
- Our powers of two are increasing and so if we go two terms forward
- This term is = to 2 times 2 terms before plus the terms before. 2A+B
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