Fun with Cones

[Dave H.]

Three cones, labeled A, B, C are on a table. A gold coin is hidden under one of the cones. I give you the opportunity to guess which cone hides the gold coin and your prize will be that coin if you guess correctly.

You guess is the cone labeled B. I now lift cone A, under which there is no gold coin.

I give you the opportunity to stay with your original guess or change your guess to cone C.

Should you change your guess? Why or why not?

[dansalmo]

This has been covered many times here and in many books.

You should always switch because when you chose you had a 1/3 probability of choosing correctly. 2/3 of the time the coin will be under on of the other two cones. When one of the two remaining cones is lifted and shown to be empty, the other cone must be 2/3 likely to have the coin.

Anyone who doubts this should imagine the same game played with 100 cones. You choose 1, I then turn over 98 others and show there is no coin. Do you still want to keep your cone?

By the way, I would have chosen A, not B, since that where I think you really put the coin. [img]/images/graemlins/smirk.gif[/img]

[Dave H.]

It was really a piece of chocolate!

[Kellon]

I sure would like a reference or two to some of the other posts or books that cover this, b/c I don't understand the explanation. It seems to me that initially, you're a 2:1 dog to pick the right cone (each cone is 1/3 likely to have the coin). If one cone is turned over and is empty, you're now even money on either of the other two. I don't understand how turning over a second empty cone means that the cone with the coin was somehow transformed into 2/3 likely to have the coin, as each was 1/3 likely at the start. Similarly, if you've got a hundred and turn over 98 blanks, you now have 1:1 odds that your original choice is correct.

Not saying it isn't so, just want to see a more detailed analysis. Sorry if I'm being dense, but I just don't get it.

[dansalmo]

I will play a 10 cone game with and give you 2:1 odds on your money when you win. How does $1000 a game sound?

This is classic because it shows how counter intuitive probabilty problems can really be. Most people do not get it immediatley and some will never believe it, but it is true. Try it for yourself with 10 cups and coin for enough times unitl you are convinced. This is also known as the monty hall problem.

[Dave H.]

You're not being dense in the least! Dansalmo's first post explained it exactly, but let me just rephrase it. Sometimes someone else's words hit differently.

When you started out, you had a 2/3 probability of being INCORRECT in your choice, which means that the coin was twice as likely to be under one of the other two cones.

Does that do it for you?

[pudley4]

[ QUOTE ]
I sure would like a reference or two to some of the other posts or books that cover this, b/c I don't understand the explanation. It seems to me that initially, you're a 2:1 dog to pick the right cone (each cone is 1/3 likely to have the coin). If one cone is turned over and is empty, you're now even money on either of the other two. I don't understand how turning over a second empty cone means that the cone with the coin was somehow transformed into 2/3 likely to have the coin, as each was 1/3 likely at the start. Similarly, if you've got a hundred and turn over 98 blanks, you now have 1:1 odds that your original choice is correct.

Not saying it isn't so, just want to see a more detailed analysis. Sorry if I'm being dense, but I just don't get it.

[/ QUOTE ]

This is correct only if the cones are turned over randomly.

[Dave H.]

Why would randomness have anything to do with this? I don't understand. When you made your initial choice, you had a 2/3 probability of being incorrect, meaning that it was twice as likely that the coin was under a different cone than the one you selected.

[dansalmo]

Jeff A is correct. Turning over one of the remaining two randomly is the equivalent of the following:

I put a coin under one of three cones and let you choose two of them. The remaining cone has a 1/3 chance of containing a coin. I then allow you to look under one of the two you chose. When there is a coin under one of the two you chose, 1/2 the time you will see a coin under the first one you uncover, since there is a 1/3 chance it is under each. When you do not see a coin, there is still 1/3 chance that it is under each of the remaining cones, so you then have a 50/50 chance of having the coin under the cone you are left with.

[Kellon]

Nope. After you turned over one of the cones, and it was a blank, you asked me to choose again. (I know that you actually asked if I wanted to change my mind, but that's technically the same as asking me to choose again.) This second choice is an independent act, based on new circumstances, with new odds. At the beginning, each cone had a 1/3 chance of winning and was a 2/3 loser. After turning up one blank, to stay with the original odds, each remaining cone is still a 1/3 winner or 2/3 loser, when viewed from the perspective of the original situation.

Dave, it seems to me that this is a bit of the reverse of some of the difficulties that were had with the earlier odds questions that you raised, in which we were struggling with changing circumstances and changing, or reassessing, odds.

I don't know if this really is analagous, but let me ask something: Take all of the spades out of a deck except the A of spades. Shuffle. Fan the remaining 40 cards. I point to one card as the potential As.

What are the odds against this card being the As? 39:1, with a probability of 1/40 or .025, right?

38 cards are turned over with no As, leaving two cards. Are you all saying that, at that point in time, the odds of the card I chose being the As is still 39:1 against (or .025), and that the probability of the other card being the As is .975?

If so, then dansalmo is probably right. I may be one of those poor souls who will never get it. But I will probably believe that y'all don't get it either, so there! [img]/images/graemlins/laugh.gif[/img]