Probability

bearwing

小玲真係偉大[[bao_29]]
感動

[S]【YU】

just name them as A and B

possible events
Case 1. A hv more $ than B
Case 2. B hv more $ than A
since both of them hv no evidence, the probability should always be 1/2

Case 1.
If A don't change, then value = 0
1/2 x 1/2: 0
If A change, then value= -original amount /2
1/2 x 1/2: -x/2

Case 2
If A don't change, then profit = 0
1/2 x 1/2: 0
If A change, then value = original amount = x
1/2 x 1/2: x

Expected profit = (x-x/2)/4 = x/8

lose half to gamble against win double, we can conclude that the game is not fair.
Expected outcome =/= true outcome.
Profit and loss depends on original amount hold (x)
i.e., u hold more, u have the chance to earn more, but also lose more

High risk, High expected return

This is investment...

Whether they will choose to participate in the game depends on whether he is risk adverse or risk taking.

I don't think it is a probability question....
the question has not stated what probability it is asking for.

[ 本帖最後由 [S]【YU】 於 26-11-2008 00:24 編輯 ]

bearwing

Sorry but i cannot get the meaning of "1/2 x 1/2: -x/2" >_<
i guess you are calculating the expected profit in case I (or II) :
i.e. 0 + (-x/2) = -x/2

However, i think it can be calculated by probability....

My Method:

Let x be the smaller value of the two.
i.e. One money bag have $x, another $2x.

Probability being the one with more money (2x): 1/2
Probability being the one with less money (x): 1/2

If you don't change, obviously you will not gain or loss.
E(profit) = 0

If you change, then your expected value of money you will get:
(1/2)(2x) [this is the case that you have less money]
+ (1/2)(x) [this is the case that you have more money]
= x + 1/2(x) = 1.5x

Your original expected money:
(1/2)(2x) + (1/2)(x) = 1.5x

Therefore, your expected value of profit:
1.5x - 1.5x = 0

This is a fair game.

(ps: I don't have the answer of this question as i only copy it from other places
so far the answer has not been given
so pls point out my mistake if you think this is wrong
This answer may not be correct.)

[S]【YU】

If you don't change, obviously you will not gain or loss.
E(profit) = 0

expected value = weighted average of value from "all" events on probability of occurance of the event.
This is not the correct usage of expected value. You can only say it's the outcome in one particular event(the value u get if u don't change), not reflecting the whole game.

(1/2)(2x) [this is the case that you have less money]
+ (1/2)(x) [this is the case that you have more money]
= x + 1/2(x) = 1.5x

both outcome are postive, where is the loss?
u have not account for the cost of participation.

if change:
if u have less money, u will receive 2x, but your actual gain is only x
if u have more money, u will receive x/2, so your actual loss is -x/2

Your original expected money:
(1/2)(2x) + (1/2)(x) = 1.5x

i also dunno wt u are doing here.

probability that u have more money or less money are equal = 1/2
the probability that u change or not change is also 1/2
1/2 x 1/2 repesent the probability in one path in the 2X2 tree diagram (total 4 outcomes)

my calculation is very straight forward.
Expected value = P(more $ & no change) x 0 + P(less $ & no change) x 0 + P(more $ & change) x (-x/2)+ P(more $ & no change) x (x)
= (x - x/2) x 1/2 x 1/2 = x/8
where all P = 1/2 x 1/2

The game is not fair as i said.
U can win double amount, but only lose half the amount, provided that all probability = 1/2.
If u don't agree, let me quote an example.
買大細
買中賠雙,買唔中渣都無. (1/2 x 1 + 1/2 x -1 = 0) fair game.
家下你買中賠雙,買唔中都仲有一半係手
你話fair唔fair?

bearwing

我以純邏輯的角度推論答案:
由於小明和小丙的情況完全一樣
兩者所得與所失 總和必定為0
否則 如果兩人的expected value 都 >0
那誰是輸家?

original expected money:
(1/2)(2x) + (1/2)(x) = 1.5x

因為你有1半機率 擁有$2x, 1半機率 擁有$x
你expect自己有的$ = 1.5x

(因為你根本不知自己有的$ 是x or 2x)

[ 本帖最後由 bearwing 於 26-11-2008 18:35 編輯 ]

[S]【YU】

原帖由 bearwing 於 26-11-2008 18:31 發表
我以純邏輯的角度推論答案:
由於小明和小丙的情況完全一樣
兩者所得與所失 總和必定為0
否則 如果兩人的expected value 都 >0
那誰是輸家?

Your original expected money:
(1/2)(2x) + (1/2)(x) = 1.5x

因 ...

未玩已先expect自己會多左(if 少),or少左錢(if 多),it's gg~
然後再用計出黎既expected - 呢個expected
跟本就係自己減自己,緊係0啦.


the one with more $(2x), can have no gain, or lose half. (-x)
the one with less $(x), can have no gain or gain double. (x)

So the one with more $ have -ve expected return & the one with less $ have +ve expected return, when we consider the 2 participants together.
The expected value of the game = 1/2(x) + (1/2)(-x) = 0.
i think this tell the real situation.
rich guy lose x, poor guy gain x, and vice versa.  and both of them have equal probability to be the rich guy or the poor guy.

but why is it different from my previous calculation?

In my previous calculation, the situation is
if i am the one holding x, the other one may be holding 2x, or x/2
such that if i gain, i can gain x , but if i lose, i only lose x/2.
here i only consider the case of one participant only.
To him only, it appears to be gain more than lose.

The same situation to another guy. Both of them can enjoy gain more than lose. So, they will play!
x is a fixed amount. but it appear to the rich one as only his half, but appear to the poor one as his whole.
So they have different perspective to play the game even though the actual transfer of amount is fixed at x.

[ 本帖最後由 [S]【YU】 於 26-11-2008 19:07 編輯 ]

bearwing

好似明 又好似唔明[[tk_33]]
躲在牆角思考下先[[tk_23]]

呀怒

原帖由 piyopiyo 於 24-11-2008 06:11 發表

wht if I think in this way
I hv $y
1/2 prob to gain $y
1/2 prob to lose $y/2
expected gain
= (1/2)y = y/2
or
= (1/2)(-y/2) =-y/4
-----------------
呢個遊戲贏ge回報大過輸[[tk_22]] 係我就會玩 ...

no no no...錢包悖論
http://zh.wikipedia.org/w/index.php?title=%E9%92%B1%E5%8C%85%E6%82%96%E8%AE%BA&variant=zh-tw