c**i
发帖数: 6973 | 1 Nathan Hodge, Navy’s Drone Death Ray Takes Out Targets at Sea. Wired, May
28, 2010.
http://www.wired.com/dangerroom/2010/05/navys-drone-death-ray-takes-out-targets-at-sea/
Note:
(1) Click "LaWS" and you will reach an undated page from US Navy titled
Laser Weapon System (LaWS), whose first paragraph states,
"What is it? LaWS is an ORDALT to the Mk 15 CIWS that adds a laser weapon
to provide multi-mission, 'zero time of flight,' low per shot cost, and deep
magazine capabilities to counter asymmet |
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b********n
发帖数: 38600 | 2 The USD has been the defacto one-world currency for 70 years. The problem is
that the USA has abused and weaponized that currency as part of asymmetic
warfare against regimes who do not support their agenda. The current
pushback by the BRICS is long overdue. The predictable efforts by the USA to
defend this illegitimate source of world power is also to be expected.
Does anyone really believe the the recent problems in Brazil and Venezuela
were due to internal economic and political issues only? |
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p******i
发帖数: 1358 | 3 两种情况：
1.toss a UNFAIR coin say probability 0.6 head/0.4 tail, if it comes out a
head, you get $1,otherwise you lose $1, what is the prob. you have win $a
before you lose $b
2.toss a FAIR coin, if head, you get $2,other wise you lose 1, same question |
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p*****k
发帖数: 318 | 6 suppose a and b are positive integers.
for the 1st problem, it's easier to construct the martingale: (3/2)^(-xn), where xn is the dollar amount you win after n rounds, and apply optimal stopping theorem. in general the martingale is [p(H)/p(T)]^(-xn). the desired prob satisfies the eq.:
p*(3/2)^(-a)+(1-p)*(3/2)^(b)=1,
thus p=[(3/2)^(a+b)-(3/2)^a]/[(3/2)^(a+b)-1].
for the 2nd problem, i will interpret the question as to find the probability of winning at least $a before losing $b (note the forw |
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p******i
发帖数: 1358 | 7 nice solution,thx
where xn is the dollar amount you win after n rounds, and apply optimal
stopping theorem. in general the martingale is [p(H)/p(T)]^(-xn). the
desired prob satisfies the eq.:
probability of winning at least $a before losing $b (note the forward jump
is 2 - the random walk could pass "a" without hitting it). the optimal
stopping theorem does not work too well here, as one could stop at either a
or a+1. instead denote p(n) as the desired prob if one starts with $n. the
recur |
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