t*****e
发帖数: 38 | 1 Just had an interview with Jane Street, a nice guy not like the average
interviewer (I gauge him against the threads on this board). Hear are the
questions:
1. a clock splashes into 3 pieces with each containing numbers sum up
equally, describe the configure.
2. Elaborate my research.
3. We toss a fair coin, you toss 5 times and I toss 4. I win if I got at
least same number of heads compared with yours. What is the chance that I
win?
4. Two games: in game one, you toss a fair coin 4 times, and w | n****e
发帖数: 2401 | 2 1.
11, 12, 1, 2
10, 9, 3, 4
8, 7, 6, 5 | K*V
发帖数: 192 | 3 4 Game two
怎么我算的概率大于1啊。。疑惑。。。
average
are the
up
got at
that I
if you
【在 t*****e 的大作中提到】
: Just had an interview with Jane Street, a nice guy not like the average
: interviewer (I gauge him against the threads on this board). Hear are the
: questions:
: 1. a clock splashes into 3 pieces with each containing numbers sum up
: equally, describe the configure.
: 2. Elaborate my research.
: 3. We toss a fair coin, you toss 5 times and I toss 4. I win if I got at
: least same number of heads compared with yours. What is the chance that I
: win?
: 4. Two games: in game one, you toss a fair coin 4 times, and w
| o*p
发帖数: 77 | 4 2. 3/5
by symmetry; first, suppose both persons run a 4 toss times' games,
the toss results are
A>B p
A=B 5/5^2
B>A p
1=2p+5/5^2 => p=2/5
p(Awin)=p+5/5^2=3/5
then add the 5th time toss to one person(A), you will find the 5th toss not
change the result. so the result is 3/5
3. pwin=1-(5/6)^4 for the first game,
pwin=1-(5/6)^24 for the second game,
prefer the second game
you
【在 t*****e 的大作中提到】
: Just had an interview with Jane Street, a nice guy not like the average
: interviewer (I gauge him against the threads on this board). Hear are the
: questions:
: 1. a clock splashes into 3 pieces with each containing numbers sum up
: equally, describe the configure.
: 2. Elaborate my research.
: 3. We toss a fair coin, you toss 5 times and I toss 4. I win if I got at
: least same number of heads compared with yours. What is the chance that I
: win?
: 4. Two games: in game one, you toss a fair coin 4 times, and w
| d**********9
发帖数: 5215 | 5 game 1: 1-(5/6)^4
game 2: 1-(2/3)^24
not
【在 o*p 的大作中提到】
: 2. 3/5
: by symmetry; first, suppose both persons run a 4 toss times' games,
: the toss results are
: A>B p
: A=B 5/5^2
: B>A p
: 1=2p+5/5^2 => p=2/5
: p(Awin)=p+5/5^2=3/5
: then add the 5th time toss to one person(A), you will find the 5th toss not
: change the result. so the result is 3/5
| o*p
发帖数: 77 | 6 a little correction to problem 2,when the 5th time toss is added to one
person(A), you will find the 5th toss will have 1/2 probability to change
the result.if A is just one head less than B, when A gets head at the fifth
toss, A will win, so there has 4/5^2 probability to happen
then A wins will increase to 3/5+4/5^2=19/25.
so the total prob A wins will be 1/2*5/3+19/25*1/2=17/25.
not
【在 o*p 的大作中提到】
: 2. 3/5
: by symmetry; first, suppose both persons run a 4 toss times' games,
: the toss results are
: A>B p
: A=B 5/5^2
: B>A p
: 1=2p+5/5^2 => p=2/5
: p(Awin)=p+5/5^2=3/5
: then add the 5th time toss to one person(A), you will find the 5th toss not
: change the result. so the result is 3/5
| h*******n
发帖数: 614 | 7 For question 3: the answer is 1/2 | h*******n
发帖数: 614 | 8 both toss 4 times B toss 1 more time Rsult
A
A=B B doesn't change B win
A=B B++ A win
A>B whatever A win
Therefore the chance is 1/2 | h*******n
发帖数: 614 | 9 both toss 4 times B toss 1 more time Rsult
A
A=B B doesn't change B win
A=B B++ A win
A>B whatever A win
Therefore the chance is 1/2 | h*******n
发帖数: 614 | 10 both toss 4 times B toss 1 more time Rsult
A
A=B B doesn't change B win
A=B B++ A win
A>B whatever A win
Therefore the chance is 1/2 | m*****n
发帖数: 2152 | 11 1 (1 2 11 12) (3 4 9 10) (5 6 7 8)
3 对于任意n+1和n都是一样的.
先比n vs n,三种情况A>B, A=B, AB) =p(A
so 2x+y = 1
n+1 vs n, p(A>B) + 1/2p(A=B) = x + y/2 = 0.5
4 题目错了吧，应该是两个都是4次才make sense。
Game 1: 1 - (5/6)^4
Game 2: 1 - (5/6)^4
两个应该是一样。
怎么这么简单，我上次电面了一个小公司，上来就是n dimension sphere的表面积和体积怎么求？
然后就是对 random variables X1, X2 in U[0,1], 求correlation of max(x1,x2)
min(x1, x2)，给出积分和非积分的两种方法。md，一堆公式在电话里都没法给他解释。 | f***a
发帖数: 329 | 12 A=B 的写反了吧
【在 h*******n 的大作中提到】
: both toss 4 times B toss 1 more time Rsult
: A: A=B B doesn't change B win
: A=B B++ A win
: A>B whatever A win
: Therefore the chance is 1/2
| p*****k
发帖数: 318 | 13 Q3, there is a nice solution exploiting the H/T symmetry.
see, e.g., attm's post in the thread below:
http://www.mitbbs.com/article_t/Quant/31218659.html |
|
