赌博问题

QL · 发表于 2005-11-8 06:36:47

去年出的一题，好象大家没太注意，连我自己都忘了。不难，但有点意思。

甲乙两人参加一个电视节目，住持人告诉他们，每人前面的钱匣内各有一定数量但不相同的钱，他们或者现在

就把自己前面的钱拿走，或者同意作如下赌博：开匣数钱，钱少的把两份钱都拿走，钱多的分文不得。

甲这样想：如果赌博，有一半的可能性我比乙钱多，那我最后收入为0，另一半的可能性我比乙钱少，那我得到乙的钱，因此最后总收入超过现在的两倍。平均而言，参加赌博的收入超过我现有的钱。因此赌博对我有利。

乙也同样想。

问题是，这是一个0和游戏，怎可能对两人都有利呢？问题出在哪？

sean9991 · 发表于 2005-11-8 11:21:47

Assume the amount of money in the boxes are X and Y, respectively. X > Y.

No gambling: the expectation is 1/2 * X + 1/2 * Y

Gamble: the expectation is 0 + 1/2 (X + Y)

What's the difference?

QL · 发表于 2005-11-8 16:26:41

You're right, of cause.

I guess the question is not about 'what is the right way to think about the question', but rather, 'what is wrong with the original logic'.

Let me repeat the question in a slightly different (hopefully more confusing) way: A player, although not knowing exactly the amount of money in his box, he can assume that it is A dollars. If he decides to play the game, with 1/2 probability(when more money in his box than that in his counterparty's), he will get 0; with 1/2 probability(when less money in his box), he will get something more than 2A. So on average, he will get more than A dollars. Hence he should play the game. Anything wrong with this logic?

大头羊 · 发表于 2005-11-8 22:51:58

既然“甲乙二人”跟主持人赌，主持人出钱，那“零和”二字从何而来？一定是“甲乙二人”赢主持人嘛，这是“常和博弈”。只要他俩参加，这个二人集团就会收益。

constant · 发表于 2005-11-8 23:00:24

好象不是特别简单。在已知甲钱多少的条件下，乙比甲多不一定是1/2。可能和乱弹吵架的题有关系。（Sorry 乱弹，忘了那题的 detail 了。）

ob · 发表于 2005-11-9 01:38:34

平均而言，参加赌博的收入超过我现有的钱。This statement is not true. IF you think in different way, then you will get different conclusion. Suppose A has more money than B, then A should think that if he has more money, then he lose everything. Even the judge set a new rule that the guy whose has more money win, averge wise, he get less than twice of his money. So, if both A and B think this way, they will draw a different conclusion.

The reality is that a gambler always optimistic...

QL · 发表于 2005-11-9 05:17:53

That the host will contibute money is a given fact, once this is given, what ever A gains (losses) is exactly what B loses (gains), so it is 0 sum in this sense.

QL · 发表于 2005-11-9 05:29:39

I'd like to hear others' opinion

大头羊 · 发表于 2005-11-9 07:00:24

A赢，A得的不是B的钱，是第三者的钱

B赢，B得的不是A的钱，是第三者的钱

所以，跟什么“零和”完全不沾边。

零和，必须是“A赢B输或B赢A输”。A赢B不输，B赢A不输，明明是“常和”

sean9991 · 发表于 2005-11-9 07:44:51

To answer this question, we need to understand the value of money, i.e., the utility of money. People want money not for the money itself, but its utility. A normal person will assume a utility function which is concave, so called risk-adverse. The marginal increment in utility is less than 1, so for large number of A, utility(2A) < 2 * utility(A). We can consider this example, if you think one envelop has 10 million, the other is 10 million plus 1 dollar. Do you want to gamble? 10 million is already very good, 20million plus 1 dollar is not much better. If gamble, you have 1/2 probability end up empty-handed. Why not just take that 10 million? A bird in hand is better than .... On the other hand, if you think the amount are just in the range of 10 dollars, you might incline to gamble.

The calculation in the question assume a linear utility function, which is not true for most people. Therefore, the conclusion drawn will be not reasonable for most people. But if you were really a risk-seeker, you would gamble.

QL · 发表于 2005-11-9 16:29:18

But gamblers are risk-lovers.

Let's not make this problem too complicated, here just assume that people are risk neutral, that they will take a chance as long as the expected outcome is favorable.

QL · 发表于 2005-11-9 16:45:43

I guess this is just a game of words, nothing essential.

Let's assume that, when you were young, your father give both you and your brother 10 yuans, and you two decided to gamble with the money, winner took all. Suppose your brother won, would you say that you lost 10 yuans or that you didn't lost anything, but your father lost 10 yuans?

sean9991 · 发表于 2005-11-9 18:09:30

For a risk lover, the conclusion is obvious: gamble. There is no problem at all when both decide to take the risk.

The reason why you sense there is a problem in their decisions is because you still keep your normal risk-adverse utility function in your mind, but do math using risk neutral formula. It's this inconsistency that causes the problem.

QL · 发表于 2005-11-9 19:48:35

As you pointed out earlier (which I agree with), that under risk neutral assumption, the players should be indifferent to the choices, gamble or not.

On the other hand, also under the risk neutral assumption, the original logic claims that the player should favor entering into the gamble.

I don't think my attitude toward risk (which is risk-averse, incidently ) have anything to do with the situation, the question is: why two seemingly `reasonable' logics, both based on risk neutral attitude lead to different decisions?

Can you point out, where in the original logic, a non-risk-neutal assumption is used or implied?

乱弹 · 发表于 2005-11-9 23:19:26

你不能假定所有事件等概率出现。

QL · 发表于 2005-11-12 17:27:35

I agreed with Constant, that given the amount of money A in the box, the probability of winning the gamble may not be 0.5 anymore, that is where the orginal logic breaks.

For illustration purpose, assume everyone knows that A can not be larger than 10, then the player should esitmate his expectation as:

P(A=1)*P1+P(A=2)*P2+.... P(A=10)*P10, where Pi is the probability of win when A = i.

Obviously 1=P(A=1)>P(A=10)=0

If we do not assume there is any limit in the amount of money, then we have to assume there is a distribution on all the positive integers (or more accurately on all pairs of positive intergers), it can not be uniform, as LuanTan pointed out, and a direct proof using the above method can be tricky, because we may end up with infinite expectations. Sean9991's proof is a better approach, he acctually showed that given the amount of money in both boxes x,y (A = x, B=y or A = y, B=x), the expectation is the same, gamble or not.

好好活着 · 发表于 2005-11-13 14:00:03

让我们假设一个钱包有8元,另一个钱包有10元,赌博的结果是:

甲或者拿走0元,或者拿走18元, 则他的平均收入是9元,也就是总钱数的一半.

乙也是同样的.

所以赌博对于甲和乙完全是机会均等的.

错觉的原因在于,甲在推测另一种结果时,假定自己拿到8元的钱包,继而把8元误认为是自己应得的收入,而实际甲如果不赌博,应得的收入可能是10元或者8元,平均来说还是9元.

同理, 甲也可以这样想: 我要是拿到10元的钱包,我不赌博的话,就是我自己的了; 如果赌博我就得不到钱了, 而我就算赢了,最多也不过拿到18元, 假设有一半的可能性会赌输,那还是不赌博来得划算.

QL · 发表于 2005-11-13 15:55:26

你前半段证明是对的，和SEAN9991类似。

我认为原逻辑主要问题在于当假设他钱包里的钱为某值时，输赢概率仍各为1/2。

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我的e文不好,给个我自己笨想的中文答案,大家评评看[:P]

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