Monday, December 29, 2008

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

Sunday, December 28, 2008

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

Wednesday, December 24, 2008

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"

Tuesday, December 23, 2008

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

Monday, December 22, 2008

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

Saturday, December 20, 2008

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

Thursday, December 18, 2008

more

forgot to mention that

Bolt is also down
together with
ZEITGEIST, The Movie

more shows

Another 3 movies down,


The savages
The miracle
Fred Claus

I'm not going to watch hulk and iron man, those Hollywood crap
If you believe, anything can happen

Tuesday, December 16, 2008

2 theorems

In combinatorics, the Erdős–Ko–Rado theorem of Paul Erdős, Chao Ko, and Richard Rado is a theorem on hypergraphs, specifically, on uniform hypergraphs of rank r.

The theorem is as follows. If n\geq2r, and A is a family of distinct subsets of {1,2,...,n}, such that each subset is of size r (thus making A a uniform hypergraph of rank r), and each pair of subsets intersects, then the maximum number of sets that can be in A is given by the binomial coefficient

{n-1} \choose {r-1}.
Baranyai's Theorem

If k|n, then the complete k-uniform hypergraph on n vertices decomposes into 1-factors, where a 1-factor is a set of n/k pairwise disjoint k-sets. Brouwer and Schrijver (1979) give a beautiful proof using the maximum flow, minimum cut theorem of network flows.

Monday, December 15, 2008

progress(2)

things to do

--> understand the proof for baranyai's theorem(ongoing)
--> prove the eigenvector for the smallest eigenvalue at hoffman equality (halfway done)
-->read up basic association schemes(1 page)
-->study association scheme for my graph(havent start)

Tuesday, December 9, 2008

simpson paradox

Simpson's paradox (or the Yule-Simpson effect) is a statistical paradox wherein the successes of groups seem reversed when the groups are combined. This result is often encountered in social and medical science statistics,[1] and occurs when frequency data are hastily given causal interpretation;[2] the paradox disappears when causal relations are derived systematically, through formal analysis.

Batting averages

A common example of the paradox involves batting averages in baseball: it is possible for one player to hit for a higher batting average than another player during a given year, and to do so again during the next year, but to have a lower batting average when the two years are combined. This phenomenon, which occurs when there are large differences in the number of at-bats between years, is well-known among sports sabermetricians such as Bill James.

A real-life example is provided by Ken Ross[12] and involves the batting average of baseball players Derek Jeter and David Justice during the years 1995 and 1996:[13]


1995 1996 Combined
Derek Jeter 12/48 .250 183/582 .314 195/630 .310
David Justice 104/411 .253 45/140 .321 149/551 .270

In both 1995 and 1996, Justice had a higher batting average (in bold) than Jeter; however, when the two years are combined, Jeter shows a higher batting average than Justice. According to Ross, this phenomenon would be observed about once per year among the interesting baseball players. In this particular case, the paradox can still be observed if the year 1997 is also taken into account:


1995 1996 1997 Combined
Derek Jeter 12/48 .250 183/582 .314 190/654 .291 385/1284 .300
David Justice 104/411 .253 45/140 .321 163/495 .329 312/1046 .298

[edit] Kidney stone treatment

This is a real-life example from a medical study[14] comparing the success rates of two treatments for kidney stones.[15]

The first table shows the overall success rates and numbers of treatments for both treatments (where Treatment A includes all open procedures and Treatment B is percutaneous nephrolithotomy):

Treatment A Treatment B
78% (273/350) 83% (289/350)

This seems to show treatment B is more effective. If we include data about kidney stone size, however, the same set of treatments reveals a different answer:


Treatment A Treatment B
Small Stones Group 1
93% (81/87)
Group 2
87% (234/270)
Large Stones Group 3
73% (192/263)
Group 4
69% (55/80)
Both 78% (273/350) 83% (289/350)

The information about stone size has reversed our conclusion about the effectiveness of each treatment. Now treatment A is seen to be more effective in both cases. In this example the lurking variable (or confounding variable) of stone size was not previously known to be important until its effects were included.

Which treatment is considered better is determined by an inequality between two ratios (successes/total). The reversal of the inequality between the ratios, which creates Simpson's paradox, happens because two effects occur together:

  1. The sizes of the groups, which are combined when the lurking variable is ignored, are very different. Doctors tend to give the severe cases (large stones) the better treatment (A), and the milder cases (small stones) the inferior treatment (B). Therefore, the totals are dominated by groups 3 and 2, and not by the two much smaller groups 1 and 4.
  2. The lurking variable has a large effect on the ratios, i.e. the success rate is more strongly influenced by the severity of the case than by the choice of treatment. Therefore, the group of patients with large stones using treatment A (group 3) does worse than the group with small stones, even if the latter used the inferior treatment B (group 2).

[edit] Berkeley sex bias case

One of the best known real life examples of Simpson's paradox occurred when the University of California, Berkeley was sued for bias against women applying to graduate school. The admission figures for fall 1973 showed that men applying were more likely than women to be admitted, and the difference was so large that it was unlikely to be due to chance.[16][3]


Applicants % admitted
Men 8442 44%
Women 4321 35%

However when examining the individual departments, it was found that no department was significantly biased against women; in fact, most departments had a small bias against men.

Major Men Women

Applicants % admitted Applicants % admitted
A 825 62% 108 82%
B 560 63% 25 68%
C 325 37% 593 34%
D 417 33% 375 35%
E 191 28% 393 24%
F 272 6% 341 7%

The explanation turned out to be that women tended to apply to competitive departments with low rates of admission even among qualified applicants (such as English), while men tended to apply to less-competitive departments with high rates of admission among qualified applicants (such as engineering). The conditions under which department-specific frequency data constitute a proper defense against charges of discrimination are formulated in Pearl (2000).

[edit] 2006 US school study

In July 2006, the United States Department of Education released a study[17] documenting student performances in reading and math in different school settings.[18] It reported that while the math and reading levels for students at grades 4 and 8 were uniformly higher in private/parochial schools than in public schools, repeating the comparisons on demographic subgroups showed much smaller differences, which were nearly equally divided in direction.

[edit] Low birth weight paradox

The low birth weight paradox is an apparently paradoxical observation relating to the birth weights and mortality of children born to tobacco smoking mothers. Traditionally, babies weighing less than a certain amount (which varies between countries) have been classified as having low birth weight. In a given population, low birth weight babies have a significantly higher mortality rate than others. However, it has been observed that low birth weight children born to smoking mothers have a lower mortality rate than the low birth weight children of non-smokers.[19]


Vector interpretation

Simpson's paradox can also be illustrated using the 2-dimensional vector space.[21] A success rate of p / q can be represented by a vector \overrightarrow{A}=(q,p), with a slope of p / q. If two rates p1 / q1 and p2 / q2 are combined, as in the examples given above, the result can be represented by the sum of the vectors (q1,p1) and (q2,p2), which according to the parallelogram rule is the vector (q1 + q2,p1 + p2), with slope \frac{p_1+p_2}{q_1+q_2}.

Simpson's paradox says that even if a vector \overrightarrow{b_1} (in blue in the figure) has a smaller slope than another vector \overrightarrow{r_1} (in red), and \overrightarrow{b_2} has a smaller slope than \overrightarrow{r_2}, the sum of the two vectors \overrightarrow{b_1} + \overrightarrow{b_2} (indicated by "+" in the figure) can still have a larger slope than the sum of the two vectors \overrightarrow{r_1} + \overrightarrow{r_2}, as shown in the example.

decoy effect

In marketing, the decoy effect (or asymmetric dominance effect) is the phenomenon whereby consumers will tend to have a specific change in preference between two options when also presented with a third option that is asymmetrically dominated. An option is asymmetrically dominated when it is inferior in all respects to one option; but, in comparison to the other option, it is inferior in some respects and superior in others. In other words, in terms of specific attributes determining preferability, it is completely dominated by (i.e., inferior to) one option and only partially dominated by the other. When the asymmetrically dominated option is present, a higher percentage of consumers will prefer the dominating option than when the asymmetrically dominated option is absent. The asymmetrically dominated option is therefore a decoy serving to increase preference for the dominating option. The decoy effect is also an example of the violation of the independence of irrelevant alternatives axiom of decision theory.

Example

For example, if there is a consideration set involving MP3 players, consumers will generally see higher storage capacity (number of GB) and lower price as positive attributes; while some consumers may want a player that can store more songs, other consumers will want a player that costs less. In Consideration Set 1, two devices are available:

Consideration Set 1

A B
price $400 $300
storage 30GB 15GB

In this case, some consumers will prefer A for its greater storage capacity, while others will prefer B for its lower price.

Now suppose that a new player, C, is added to the market; it is more expensive than both A and B and has more storage than B but less than A:

Consideration Set 2

A B C
price $400 $300 $450
storage 30GB 15GB 25GB

The addition of C—which consumers would presumably avoid, given that a lower price can be paid for a model with more storage—causes A, the non-dominated option, to be chosen more often than if only the two choices in Consideration Set 1 existed; C affects consumer preferences by acting as a basis of comparison for A and B. Because A is better than C in both respects, while B is only partially better than C, more consumers will prefer A now than did before. C is therefore a decoy whose sole purpose is to increase sales of A.

Conversely, suppose that instead of C, a player D is introduced that has less storage than both A and B, and that is more expensive than B but not as expensive as A:

Consideration Set 3

A B D
price $400 $300 $350
storage 30GB 15GB 10GB

The result here is similar: consumers will not prefer D, because it is not as good as B in any respect. However, whereas C increased preference for A, D has the opposite effect, increasing preference for B.

Monday, December 1, 2008

things to do

cut hair...done
Find way to sell shades to make space for new one...on the way
Plan Vietnam trip...on the way
massage...done
paddle...on going
Gym...on going
run...on going
Do HT...resumed
Meet Prof before he goes for Christmas break...soon
Refurnish room...decide to postpone
do balance sheet...soon
watch movies...M2 and 4xmas done...