
There are five boxes remaining
box 1,2,3,4,5
The odds of finding any of the boxes is
5+4+3+2+1=151 is this true?
After next pick
4+3+2+1=101 ??
last pick
3+2+1=61??
I'm afraid I don't understand the question (I don't watch the show).
Can you explain in more detail?
slashnburn99 wrote:  There are five boxes remaining
box 1,2,3,4,5
The odds of finding any of the boxes is
5+4+3+2+1=151 is this true?

Er, wouldn't it simply be 1 out of 5, as 1 box is the winner, and the rest are not, and you can only end up with one?
I think that show amounts to just a guessing game really, aside from the possibility of selling your box to the bank, you might as well pick a box, and just accept whatever is inside the first one you picked, regardless of the rest of the show and all their fake suspense and strategy.
(Though with the addition of the banker, the best strategy is to decide how much you want, and play until the banker offers you that, then sell out.)
No. As ocalhoun said, the probability of having the winning box, when you have 5 left, is simply 1/5. It's as simple as that. Now if you had to find them in a particular order, then your logic would be correct. Actually, it wouldn't. I just looked at it in detail and saw that you were adding, not multiplying. It's incorrect no matter what. To find out what the probability of finding them in a particular order is you'd have to do 5*4*3*2*1, not 5!
Here's a related problem that's rather interesting. It's relatively old, and it's based off of a game show titled "Let's Make a Deal." I never saw the show, but my friend showed me this interesting problem a few years back. The problem was originally written by Marilyn vos Savant in Parade magazine. This problem was also featured in the semirecent movie, 21. Anyways... it's titled the Monty Hall Problem.
Quote:  Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? 
Afaceinthematrix wrote: 
Quote:  Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? 

I did finally find an easy way to explain the solution to that... you have to make a diagram.
ocalhoun wrote:  Afaceinthematrix wrote: 
Quote:  Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? 

I did finally find an easy way to explain the solution to that... you have to make a diagram. 
There are hundreds of ways to explain it. I usually just use a diagram... I used to try explaining it with more doors but that just confuses people even more. So I usually just draw out the diagram so that the person realizes that every time their original guess is incorrect (about 2/3 of the time), switching will be beneficial.
