酒和茶

constant · 发表于 2006-4-11 21:11:56

难度：++.5

一间酒吧卖两种饮料：酒5毛钱一杯，茶1毛钱一杯。后来老板改了规矩：每个顾客交4毛钱，然后和老板玩游戏。老板和顾客每人可以出一个或两个手指头，如果两人都出一个手指头，顾客得一杯酒；如果两人都出两个手指头，顾客得两杯酒；如果两人手指头不一样多，顾客得一杯茶。

老板说，4种情况，平均4毛2分5，我只要4毛钱，比以前赚钱少了。

老板说的对不对？

LOTUSEATER · 发表于 2006-4-11 21:29:31

1.6/3=0.5333333333333333333333333333333333333333333333333333...........................

????

本贴由[LOTUSEATER]最后编辑于：2006-4-11 13:36:43

ob · 发表于 2006-4-11 22:06:42

If the game is totally random, there are four cases: (O-Owner, C-Customer, M-Owner make, CG-Customer get)
O C M CM
1 1 0.4 0.5
1 2 0.4 0.1
2 1 0.4 0.1
2 2 0.4 1.0

Case 1.
If both parties do the game randomly, it is correct.

Case 2.
if the owner always 出一个手指头 and the customer do it randomly, the owner can make 20 cents more than he usually make after two times.

Case 3.
If the customer is greedy and put out two fingers every time, then the owner can make 0.30 more each time.

Case 4.
If the owner put out one finger three out of four time, after 8 times, he can make .30 more than before if the customer do it randomly. If the customer put out one finger every time, it is the same as before.

Use case 4 method, the owner will make more money and not be blamed cheating.

大头羊 · 发表于 2006-4-11 22:11:54

于是老板每次都出一个手指头

，众人大喊上当。

于是众人过了几天后，每次都出一个手指头

老板一看不好，又过了几天后，改成两个手指头

。于是众人过了几天后，每次都出两个手指头

老板一看不好，又改回了一个手指头....于是周而复始，趋近于随机试验。但是，这只是理想情况。所以，从长远讲，如果众人的智商都足够的话，老板说的是对的。

但如果是黑箱操作，那他赢大了。

sean9991 · 发表于 2006-4-13 21:24:18

The optimal solution for both the boss and customer is:

1 finger - with 0.6 probability

2 fingers - with 0.4 probabilty

Therefore, the expected payout is $0.388 per game.

sean9991 · 发表于 2006-4-13 21:49:16

The probability should have been 9/13, 4/13.

The payout is then $0.376923.

Hopefully i don't mess up the calculation again. :-)

constant · 发表于 2006-4-14 18:25:29

9/13和4/13是正解

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酒和茶

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9/13和4/13是正解