b***k
发帖数: 2673 | 1 ☆─────────────────────────────────────☆
trenchant (N/A) 于 (Tue Aug 28 00:21:25 2007) 提到:
谢谢！
1 You have a fair coin. Calculate the expected number N of throws
That you needed to find two HH at the end of sequence?
2 A deck of cards with numbers on: 1,2,...,N. Draw a card randomly from the
deck, keep it at hand, then draw another one; if the new one is larger than
the largest number in the hand, keep it, otherwise, discard it. Stop when
you have the card N. Question: What is the expectatio | p***y
发帖数: 59 | 2 The solution for the second question S(N+1)=S(N)+1/(N+1) can be easy to
understand. S(N) is the average cards kept in hand if we have N cards to
play. Now what is S(N+1)? The last card on the table could be any one of 1,2
,....,N+1, but only if the last one is N+1 can we keep it in hand.
Probability is 1/(N+1).
Notice that S(N) remains in the above equation even when we have N+1 cards.
For example, if the last card on the table is 5, then we could simply
consider the cards 6,7,8,...,N (already d | x**********n
发帖数: 13 | 3 does everyone agree on the solution of 2nd question?
first, I agree S(2)=3/2
if you try N=3, E(3)=1/3*(1+S(2))+1/3*(1+X)+1/3*1 {for first draw of 1,2,3}
where X is the expected number of draws under the rule of this game to get 3
if you have 2 in first draw and card 1, 3 left
X=1/2*1+1/2*(1+X)=> X=2
Then E(3)=1/3+1+5/6=13/6
However, according to your solution, E(3)=1+1/2+1/3=11/6
Please let me know if I did anything wrong | o****g
发帖数: 314 | 4 如果第一张是 2，那么剩下的1和3，1肯定被抛弃，3肯定留下来，所以
X = 1
3
【在 x**********n 的大作中提到】
: does everyone agree on the solution of 2nd question?
: first, I agree S(2)=3/2
: if you try N=3, E(3)=1/3*(1+S(2))+1/3*(1+X)+1/3*1 {for first draw of 1,2,3}
: where X is the expected number of draws under the rule of this game to get 3
: if you have 2 in first draw and card 1, 3 left
: X=1/2*1+1/2*(1+X)=> X=2
: Then E(3)=1/3+1+5/6=13/6
: However, according to your solution, E(3)=1+1/2+1/3=11/6
: Please let me know if I did anything wrong
| x**********n
发帖数: 13 | 5 哦，我好像理解错了。但如果第一张是2，剩下的你要从1，3里抽，抽到1后扔掉而不是
放回，这样你还是要抽一遍3，所以X=1/2*1+1/2*2=3/2。就像只有一张牌E(1)=1而不是
0。
所以 E(N) = E(N - 1) + 1/N 的结果还是不对。 |
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