[Guest]

this might have been done before but i couldnt find it. Its a simple problem that some of you have probably seen before(so dont answer it if you have), but I found it made me understand probability more when I figured it out so...

There are 5 closed chests. two of them have 2 dollars in them: 3 are empty. The game is simple pay 1 dollar and pick a chest. I show you one of the empty chests (i always show a different chest then you picked). Then you can either stay with your chest or switch to another chest. You then open your box and the games over. What do you do? y?

what is the expected gain or loss for the better option? the worse?

[Guest]

See my answer:

You should always switch. When you pick the first chest(out of 5) the odds that you are going to pick one of the 2 that have a dollar is 40% (2 in 5 chance). After you are shown an empty chest, the other 4 chests not shown have 2 dollars in them (including the one I already picked). Therefore, switching at this point gives me a 50% chance (2 in 4) of getting a dollar.

this might have been done before but i couldnt find it. Its a simple problem that some of you have probably seen before(so dont answer it if you have), but I found it made me understand probability more when I figured it out so...

There are 5 closed chests. two of them have 2 dollars in them: 3 are empty. The game is simple pay 1 dollar and pick a chest. I show you one of the empty chests (i always show a different chest then you picked). Then you can either stay with your chest or switch to another chest. You then open your box and the games over. What do you do? y?

what is the expected gain or loss for the better option? the worse?

[Guest]

I don't think this is the typical Monty Hall problem where it's always beneficial to switch. In this case, before the game starts, you have a 2 in 5 chance of getting $2.

Once you pay your dollar and are shown a losing chest, you are now sitting with a 2 out of 4 chance of winning $2. If you pay ANOTHER dollar to get a different chest, you still have the same 2 out of 4 chance to win $2. But it cost you a dollar, so there's no point.

I guess the expected value of the best option is 2($2) + 2($0) / 4 chests = $1. And it cost you a dollar, so the total expected value of the game is $0. If you do another pick, the expected value is even less, $-1. Curious what others think.

[Guest]

you don't have to pay a second dollar to change. I actually was going to put that in there to add a trip for people that have seen the problem but even on top of that

nothing right so far

[Guest]

I would not switch being shown an empty box does not increase my chances if i have already chosen a box that contains $2 all it does is make me second guess myself

[Guest]

this might have been done before but i couldnt find it. Its a simple problem that some of you have probably seen before(so dont answer it if you have), but I found it made me understand probability more when I figured it out so...

There are 5 closed chests. two of them have 2 dollars in them: 3 are empty. The game is simple pay 1 dollar and pick a chest. I show you one of the empty chests (i always show a different chest then you picked). Then you can either stay with your chest or switch to another chest. You then open your box and the games over. What do you do? y?

what is the expected gain or loss for the better option? the worse?

As this forum has taught me (and others), the answer is... it depends on your criteria for picking that other chest. Do you always show an empty chest 100% of the time? You stated that you always show a different chest, and that, this time in particular, you show one of the empty chests. Can we assume that you always show an empty chest as part of the rules of the game?

[Guest]

this is a problem i am not trying to trick u with my words. you can take anything at face value. I show you an empty chest always and I always show a different chest then you picked. Also i want to say i chose this problem (modified a little obviously) because alot of people know the answer and can state reasons y. But they dont really understand y. That and its a cool little problem that once you understand helps alot in thinking about probability problems.

[bushindo]

Assume that you always show an empty chess, but not the one picked by the player. Then

Always switch. If you switch, the chance of you winning is

(2/5)*(1/3) + (3/5)*(2/3) = 8/15

So your expected winnings per game, if you switch, is

-1 + 2*(8/15) = 1/15 = .07 dollar.

[Guest]

nice bushindo now lets see if i can solve yours

[Guest]

this is a problem i am not trying to trick u with my words. you can take anything at face value. I show you an empty chest always and I always show a different chest then you picked. Also i want to say i chose this problem (modified a little obviously) because alot of people know the answer and can state reasons y. But they dont really understand y. That and its a cool little problem that once you understand helps alot in thinking about probability problems.

As I see it, your response always reduces to one effective outcome to match my five available choices.

My first choices are:

A: 2/5 chance of a full chest

B: 3/5 chance of an empty chest

Given either choice A, you have 1/1 chance of choosing an empty chest, leaving 1 non-empty chest and 2 empty chests should I choose to switch.

Given either choice B, you have 1/1 chance of choosing an empty chest, leaving 2 non-empty chests and 1 empty chest should I choose to switch.

If I don't choose to switch, I have 2/5 = 6/15 chances of having selected a non-empty chest.

If I choose to switch, I have (2/5) * (1/3) + (3/5) * (2/3) = 2/15 + 6/15 = 8/15 chances of having selected a non-empty chest

Edit: Corrected my math-- I had six chests in choice B...

Edited May 22, 2009 by Phatfingers