f*****s
发帖数: 141 | 1 In a game, 10 balls, labeled 0,1,...,9, are put in a bag. Randomly pick out
5 balls and add the number. How do you make the market for this game? | S*****H
发帖数: 90 | 2 Mean total 22.5. bid:ask = 22.5+/d.
What profit margin do you want? | f*****s
发帖数: 141 | 3 5 * (0+1+...+9)/10 = 22.5
But this is just for picking 5 balls with replacement. Here, the question is
without replacement, so I think the answer shall be > 22.5, right? | S*****H
发帖数: 90 | 4 Do yo mean the maximum of the sum of the first 5, and the sum of the remaining 5?
You need to make your question clear.
is
【在 f*****s 的大作中提到】
: 5 * (0+1+...+9)/10 = 22.5
: But this is just for picking 5 balls with replacement. Here, the question is
: without replacement, so I think the answer shall be > 22.5, right?
| p********2
发帖数: 9939 | | f*****s
发帖数: 141 | 6 In the game, you pick out one ball, then another, until pick out 5 balls.
Now, add the number of all the selected 5 balls.
remaining 5?
【在 S*****H 的大作中提到】
: Do yo mean the maximum of the sum of the first 5, and the sum of the remaining 5?
: You need to make your question clear.
:
: is
| t***s
发帖数: 4666 | 7 then things are more complicated. will you see the number on the ball
after it's got picked? will you be allowed to add bet after picking
a ball?
【在 f*****s 的大作中提到】
: In the game, you pick out one ball, then another, until pick out 5 balls.
: Now, add the number of all the selected 5 balls.
:
: remaining 5?
| s*******u
发帖数: 35 | 8 This should be correct. Let X_1 be the number on the first ball, ..., X_5 be
the number on the 5th ball. Then we
want to find E(X_1+...+X_5); By linearity of expectation, this is equal to E
(X_1)+...+E(X_5)=5*E(X_1).
is
【在 f*****s 的大作中提到】
: 5 * (0+1+...+9)/10 = 22.5
: But this is just for picking 5 balls with replacement. Here, the question is
: without replacement, so I think the answer shall be > 22.5, right?
| f*****s
发帖数: 141 | 9 I doubt it, since EX_2 is conditional on X_1, EX_3 is conditional on X_1 and
X_2,.......
be
E
【在 s*******u 的大作中提到】
: This should be correct. Let X_1 be the number on the first ball, ..., X_5 be
: the number on the 5th ball. Then we
: want to find E(X_1+...+X_5); By linearity of expectation, this is equal to E
: (X_1)+...+E(X_5)=5*E(X_1).
:
: is
| s*******u
发帖数: 35 | 10 Actually by symmetry, a number appears on the second place with the same
probability as the first. In other
words, X_2 and X_1 have the same distribution. So the expectation should be
the same.
and
【在 f*****s 的大作中提到】
: I doubt it, since EX_2 is conditional on X_1, EX_3 is conditional on X_1 and
: X_2,.......
:
: be
: E
| s*****y
发帖数: 33 | 11 I think you can use the symmetric property of the sums. The sum ranges from 10 to 35 and the Pi=Prob(sum=10+i)=Prob(Sum=35-i). i=0,1,...12.
So E(sum) =Sigma Pi*(10+i)+Pi*(35-i)=45 * Sigma Pi =22.5
out
【在 f*****s 的大作中提到】
: In a game, 10 balls, labeled 0,1,...,9, are put in a bag. Randomly pick out
: 5 balls and add the number. How do you make the market for this game?
| c**********s
发帖数: 295 | 12 the five balls remained in the bag has the same expect value as the five
picked. so 22.5 is correct. | B*****9
发帖数: 48 | 13 这个不对吧
E(X_1)和E(X_2)不会一样吧
取第一个球是 可以是 0,1,2....9 十个中的一个
第二个球只剩9个中一个了
be
E
【在 s*******u 的大作中提到】
: This should be correct. Let X_1 be the number on the first ball, ..., X_5 be
: the number on the 5th ball. Then we
: want to find E(X_1+...+X_5); By linearity of expectation, this is equal to E
: (X_1)+...+E(X_5)=5*E(X_1).
:
: is
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