hah hah

Dear ALvin,

Once again your solution agrees with mine. I think I will present it as a tree diagram since that will be easiest for readers to follow.

Also, the result for the A-B duel can be gotten recursively, without using geometric series. If x is the probability that A survives, then x = P(A kills B) + P(A misses B)P(B misses A)(x), so: x = 1/3 +(2/3)(1/3)x, and thus x = 3/7.

Best,

MB

-----"Loo Chee Wee" wrote: -----

To:
From: "Loo Chee Wee"
Date: 12/28/2008 11:32PM
Subject: RE: Triple shootout: Charlie goof

Hi sir

yar sure you are free to post up my solution
Here's my try-out for the infinite shootout

everyone's intention is to be last man standing or survive as long as possible
they move in sequence A,B,C
shootings are independent of each other
B and C are still trying to kill each other as top priority

scenario 1: A shoots B
A hits: C then shoots A and C survives
A misses and B shoots C, if B hits then A and B engages in a 1-1 shootout with A making first move, probability of A surviving is 4/21 from a geometric progression...if B misses C, c kills B with his shoot and A must hit his next shoot at C for any chance of survival, probability of surviving is 2/27

summing everything: under this case A has 50/189 chances of surviving

Scenario 2: A shoots C
A hits, B and A then engages in 1-1 combat with B making first move, probability of A surviving is 1/21, from a geometric progression
A misses, B shoots C, if B hits, B and A engages in 1-1 combat with A making first move, probability of A surviving is 4/21 using geometric progression
if B misses, then C kills off B with his shot and A must then hit C to survive, probability of A surviving is 2/27

summing everything: A has a 59/189 chance of surviving

Therefore: A should start with shooting C in the infinite case

I'm not so confident of this solution though but it's certainly more fun
Hope you can advise on this

Regards
ALvin

hah

This is correct -- at least, it agrees with what I calculated for a 1-shot triple shootout.

The case of an "infinite" shootout (participants continue shooting until someone wins) is even more interesting. Why not give it a "shot"?

With your permission, I will post your solution for a the 1-shot case.

Thanks for writing.

M. Bridger

-----"Loo Chee Wee" wrote: -----

To:
From: "Loo Chee Wee"
Date: 12/28/2008 06:13AM
Subject: Triple shootout: Charlie goofs

Hi sir
Here's my try-out for the question you posted
Hope you can advise

P(A misses)=2/3
P(B misses)=1/3
P(C misses)=0

suppose they shoot in the sequence of A,B,C and each only has 1 bullet
Assume that the shootings are independent, that is C's probability of hitting target doesn't change whether A,B hits or not

scenario 1: Suppose A aims B

(1) :P(A misses)=2/3 and A will survive as B will aim for C and vice versa....(A)
(2): P(A hits)=1/3 and A dies and only C remains and he/she never miss
summing: P(A survives)= 2/3 = 6/9

scenario 2: Suppsoe A aims C

(1): P(A misses)=2/3 and A will survive with argument from (A)
(2): P(A hits)=1/3 and B will shoot A with 2/3 chance of killing him/her
summing: P(A surviving)= 2/3 + (1/3)(1/3) = 7/9

therefore A should aim for C

Regards
ALvin

Never knew i could contribute to a American TV show like that haha

More shows down

Starship troopers 3
the storm riders animated
Forgetting sarah marshall
step brothers

my 4D/TOTO strategy

twin primes: pairs of primes such that both P and P+2 are primes
(5,7) (11,13) (17,19),(29,31),(41,43)

Cousin primes: pairs of primes such that P and P+4 are primes
(3,7) (7,11) (19,23) (37,41)

Prime numbers:
2,3,5,7,11,13,17,19,23,29,31,37,41,43

perfect numbers: numbers that is the sum of all its divisors
6=1+2+3
28=1+2+4+7+14

Fibonacci numbers:
0,1,2,3,5,8,13,21,34

obviously i only paid attention to those below 45

Gauss: "Mathematics is the queen of sciences and number theory is the queen of mathematics"

Sia lah

Management science: A
Managerial economics: B..wtf
Algorithmic graph theory: A-...lucky never lose face
AODE: B+
Complex analysis 2: B-...little surprise considering that TPC is Chan heng huat's descendant

Before s/u: 4
After: 4.25

p.s: what is there left to fight for?
p.s.s: remind me to tell Darren and pong that the strategy of taking extra electives in order to push up CAP should be done with some SUs on hand as an insurance

a bit late...reviews and thoughts

1: Ok so Mr Bush was (nearly)hit by a shoe in Iraq, seems that he finally found the "weapon of mass destruction"...well shoes causes blisters to many people, so i'm right right?

2: Lehman brother bank sage: ok so people are wondering if there's going to be payout
three options:
a: no compensation
b: full compensation
c: compromised compensation
But anyway since town council is involved in the decision, if its an honest one, shall be clear
If not send the town council to cow council then

3: Ex-NKF chairman's wife sentenced only to 22 months jail? Come on lah lets dig out all the past records of people "washing black money"
For all these people let's divide the amount of money by their sentence and record the "dirty ratio" and we will then see that Js in SG don't seem to pass PSLE math

4: right when i was i year1 i proposed (for fun) setting up bubble tea shop and hairdresser in NUS and both are implemented now. One more to ensure NUS earn till siao
SINGAPORE POOLS

5: When i'm free one day i'm going to spend a weekend at a singapore pools outlet and record the number of people betting that day. Each person i will just multiply by $2. And then i will multiply by the number of outles across the island and then by 2 and 52.
Sui ah

progress(3)

1: try to run the program done by ah fu on MATLAB, prof has done n=4 on MAPLE and that is his limit, must try to out-do him
Lucky got "divine help"on this one

2: define an association scheme for my graph...should be an analog to the Johnson graph(scheme)

3: go find out any results i can use on symmetric groups and then project it down onto my graph

P.S: proof on eigenvector for smallest eigenvalue of graph has been done