Monty Hall Revenge - My 2 cents on solution

[vishals]

The real question is to find out a stategy of maximising the probability of finally able to select the bag with most money. This boils down to finding a strategy to know when to stop and say this is "the bag with most money"

Please note - down in explanation, I have used - "dream bag " for "bag with most money"

So first get some clear assumptions laid down -
If out of total 100 bags, I have already checked 'n' bags and now opened 'n+1' bag.
The attribute the money in the bag must have is - It should be higher then the amount of money in any of the first n bags.

So if 'n' is 1 an the the second bag is having higher money then first then should we stop and say this is the dream bag. Absolutely not!!
On other hand if 'n' is 98 and n+1 bag comes with bag with most money in all 99 bags. Should we say this is out dream bag - Absolutely yes

So what are we getting at -
The bottem line is we will stop and say this is the dream bag only if 2 conditions are fulfilled -

1) The current bag should have more money then any of the previous bags.
2) The probability of current bag benig the dream bag is highest.
Lets say n bags are already checked and n+1 bag has most money then any of previous bags. Then probability of this bag being the dream bag is = 1/(100-n) ------ (1)
The probality looks odd, but it is actually not. See it this way,all the numbers are independant.So the current bag is having money greater then first n bags. So the competition is only among the last 100 - n bags. So its probabilty of being the dream bag is just 1/(100- n)
And the probabilty of any bag after 'n+1' bag being the dream bag = (100-(n+1))/100
well the reason is all the bags have equal probability of being the dream bag,so probability of any random bag being dream bag = 1/100. Continuing the same logic, probabiltity of any bag out of (100 - (n+1)) bags being the dream bag
= (100 - (n+1))* (1/100)
=(100-(n+1))/100 ------------------ (2)

Now we have two equations (1) and (2)
Lets equate the two and find value of n
1/(100-n) =(100-(n+1))/100

Solving,approx - value of n is around 90.

So if after opening 89 bags. Anybag comes with most money then claim it May be you are driving out with the big prize !!!


Vishal

[solvah]

You are wrong. refer to montys revenge perfect solution....

[MacPhearson]

Or how about just open the bags until you find one equaling the value of the new station wagon? And even if it's not the most money you could have gotten, you'd still have enough money for a down payment on a new station wagon from your good 'ol Honest Abe car salesman.