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ultimate probability question (1 Viewer)
- Thread starter hainsay
- Start date
Macquarie's actuarial program is generally regarded as better because it's been operating for much longer. UNSW has only had it for roughly 5 years if I'm not mistaken.
Yeah, 2nd year Actuarial/Maths. I say our UNSW Maths faculty rules supreme though.
I could be wrong, I didn't think too hard about it. Our answers could be the same anyway. One is decimal, one is fraction, I can't be bothered to check.
Basically my method:
There are 10 states.
My notation is as follows:
STATE : COWBOYS ALIVE (COWBOY CURRENTLY SHOOTING)
0 : 1, 2, 3 (1)
1 : 1, 2 ,3 (2)
2 : 1, 2 ,3 (3)
3 : 1, 2 (1)
4 : 1, 2 (2)
5 : 1, 3 (1)
6 : 1, 3 (3)
7 : 1
8 : 2
9 : 3
It is not possible to have a state where cowboy 1 is dead, and cowboys 2 and 3 are alive.
Basically, certain states feed into other states.
From state 0 we can go to state 1 (p = 2/3) or state 4 (p = 1/3), and so on with various probabilities for various states.
Eventually we end up in state 7, 8, or 9, and are stuck there. The state is called recurrent. We can't get back out.
So I created the Markov Matrix, denoted by M, and then calculated M^1000 in MATLAB. (basically M^n as n approaches infinity).
Next, I looked at the probabilities of being in state 7, 8, or 9 after "1000 turns" (ie. 1000 bullets are shot) given you started in state 0 as the question specified.
That's how I got those decimal answers.
1000 turns will be more than accurate enough for an answer to 4 decimal places.
Alternatively, I could have calculated fraction answers by hand if I wished to find the limiting probabilities of states, but by that time I was bored and decided MATLAB would do the job.
Of course, this will largely make absolutely no sense to year 12 NSW students since matrixes aren't in the syllabus (yet!).
Disclaimer: I could quite easily have made a mistake. I didn't double check any figures I put into my matrix, etc.
Yeah, 2nd year Actuarial/Maths. I say our UNSW Maths faculty rules supreme though.
I could be wrong, I didn't think too hard about it. Our answers could be the same anyway. One is decimal, one is fraction, I can't be bothered to check.
Basically my method:
There are 10 states.
My notation is as follows:
STATE : COWBOYS ALIVE (COWBOY CURRENTLY SHOOTING)
0 : 1, 2, 3 (1)
1 : 1, 2 ,3 (2)
2 : 1, 2 ,3 (3)
3 : 1, 2 (1)
4 : 1, 2 (2)
5 : 1, 3 (1)
6 : 1, 3 (3)
7 : 1
8 : 2
9 : 3
It is not possible to have a state where cowboy 1 is dead, and cowboys 2 and 3 are alive.
Basically, certain states feed into other states.
From state 0 we can go to state 1 (p = 2/3) or state 4 (p = 1/3), and so on with various probabilities for various states.
Eventually we end up in state 7, 8, or 9, and are stuck there. The state is called recurrent. We can't get back out.
So I created the Markov Matrix, denoted by M, and then calculated M^1000 in MATLAB. (basically M^n as n approaches infinity).
Next, I looked at the probabilities of being in state 7, 8, or 9 after "1000 turns" (ie. 1000 bullets are shot) given you started in state 0 as the question specified.
That's how I got those decimal answers.
1000 turns will be more than accurate enough for an answer to 4 decimal places.
Alternatively, I could have calculated fraction answers by hand if I wished to find the limiting probabilities of states, but by that time I was bored and decided MATLAB would do the job.
Of course, this will largely make absolutely no sense to year 12 NSW students since matrixes aren't in the syllabus (yet!).
Disclaimer: I could quite easily have made a mistake.
I didn't double check any figures I put into my matrix, etc.
Constip8edSkunk
Joga Bonito
ok heres what i did:
after 3 (or atmost 3) shots theres 5 possibilities:
a - cowboy 1 remains - prob. = 1/3*1/2*1/3 + 1/2*2/3*1/3 = 1/6
b - cowboy 2 remains - prob = 1/3*1/2 = 1/6
c - both cb1 + cb2 remains - prob = 1/3*1/2*2/3 + 1/2*2/3*2/3 = 1/3
d - both cb1 + cb3 remains - prob =2/3*1/2*2/3 = 2/9
e - all 3 remains - prob = 1/3 *1/2 *2/3 = 1/9
now from c, there are 2 possibilities either cb1 or cb2 can remain. the prob. of cb2 remaining is the infinite sum 1/2(1 + 1/3 + 1/3<sup>2</sup> ...) = 3/4
similarly for 1 to remain, the prob. = (1/3*1/2)(1 + 1/3 + 1/3<sup>2</sup> ...) = 1/4
from d, either cb3 or cb1 can remain alive, for cb1 to remain alive, the prob. is the infinite sum 1/3(1 + 1/3*2/3 + (1/3*2/3)<sup>2</sup> + ...) = 3/7
prob. of cb3 remaining is (2/3*2/3)(1 + 1/3*2/3 + (1/3*2/3)<sup>2</sup> + ...) = 4/7
from e, one can repeat the 3 shots once more, so for whatever the prob. of each cowboys being alive, you multiply that by the infinite sum of GP with r = 1/9 , which is 9/8
so prob . of cb1 of being alive at the end is:
9/8(1/6 + 1/3*1/4 + 2/9*3/7) = 87/224
prob. of cb 2 of being alive at the end is:
9/8(1/6 + 1/3*3/4) = 15/32
and prob of cb 3 being alive at the end is:
9/8(2/9*4/7) = 1/7
...which checks up with mills' answers. ND, i think u missed the cases where after only cowboys 1 and 2 are left and each misses again (and again and again and so on... ie the attached GP)
blah markov would have been so much simpler than this massive case bash i think
after 3 (or atmost 3) shots theres 5 possibilities:
a - cowboy 1 remains - prob. = 1/3*1/2*1/3 + 1/2*2/3*1/3 = 1/6
b - cowboy 2 remains - prob = 1/3*1/2 = 1/6
c - both cb1 + cb2 remains - prob = 1/3*1/2*2/3 + 1/2*2/3*2/3 = 1/3
d - both cb1 + cb3 remains - prob =2/3*1/2*2/3 = 2/9
e - all 3 remains - prob = 1/3 *1/2 *2/3 = 1/9
now from c, there are 2 possibilities either cb1 or cb2 can remain. the prob. of cb2 remaining is the infinite sum 1/2(1 + 1/3 + 1/3<sup>2</sup> ...) = 3/4
similarly for 1 to remain, the prob. = (1/3*1/2)(1 + 1/3 + 1/3<sup>2</sup> ...) = 1/4
from d, either cb3 or cb1 can remain alive, for cb1 to remain alive, the prob. is the infinite sum 1/3(1 + 1/3*2/3 + (1/3*2/3)<sup>2</sup> + ...) = 3/7
prob. of cb3 remaining is (2/3*2/3)(1 + 1/3*2/3 + (1/3*2/3)<sup>2</sup> + ...) = 4/7
from e, one can repeat the 3 shots once more, so for whatever the prob. of each cowboys being alive, you multiply that by the infinite sum of GP with r = 1/9 , which is 9/8
so prob . of cb1 of being alive at the end is:
9/8(1/6 + 1/3*1/4 + 2/9*3/7) = 87/224
prob. of cb 2 of being alive at the end is:
9/8(1/6 + 1/3*3/4) = 15/32
and prob of cb 3 being alive at the end is:
9/8(2/9*4/7) = 1/7
...which checks up with mills' answers. ND, i think u missed the cases where after only cowboys 1 and 2 are left and each misses again (and again and again and so on... ie the attached GP)
blah markov would have been so much simpler than this massive case bash i think