p******i
发帖数: 1358 | 1 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 | J*****n
发帖数: 4859 | 2
question
Markov Chain
【在 p******i 的大作中提到】
: 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
| p******i
发帖数: 1358 | 3 详细讲讲？
【在 J*****n 的大作中提到】
:
: question
: Markov Chain
| p*****k
发帖数: 318 | 4 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 | p******i
发帖数: 1358 | 5 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
【在 p*****k 的大作中提到】
: 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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