Posted: 4/20/2006 6:05:39 PM EDT
|
I love the Monte Hall problem. Wars have been fought over this problem. Religions rise and fall in the wake of this problem. So I pose it here, for you. The mathematical explanations all make perfectly reasonable sense, the simulations all support the mathematical explanations, and yet it is maddeningly difficult to analyse the problem. There are three doors. Behind one door is a prize. Behind each of the other two doors is something of no value. The participant choooses one door. That door remains closed. --At this stage, the door the participant has chosen has a 1 in 3 chance of hiding the prize, but a 2 in 3 chance of hiding something of no value. --At this stage, the other two doors hide two things in one of three scenarios: Prize and item of no value, item of no value and prize, or two items of no value. The host now opens one of the two unchosen doors. The host always opens a door hiding an item of no value. --At this stage, there are now only two doors closed, behind one of which is a prize, and behind the other is an item of no value. Having chosen one door and seeing that one of the other doors reveals an item of no value, the participant is now given the opportunity to switch his/her choice to the other door. The question is: Does the participant have a better chance staying with the original choice, a better chance switching, or does it make no difference? Mathematicians will explain it one way which seems reasonable, and I'll post it here if you want. Another explanation that arrives at the same conclusion as the mathematicians is this: If no doors were opened, and you picked one door (having a 1/3 chance of winning the prize), would you give up your door if the host would let you have BOTH other doors? If you keep your door, you have a 1/3 chance of winning. If you switch to BOTH other doors, you have a 1/3 + 1/3 = 2/3 chance of winning. You double your chances of winning if you switch to choose both doors. In the original statement of the problem, the host reveals one of the two non-chosen doors to be a non-winner. Switching your choice after that non-winner is revealed gives you a 2/3 chance of winning. And experiments support this conclusion. Switching your choice gives you a 2/3 chance of winning, no matter what you choose first. But, wait...there are a finite number of possible outcomes, which I list below. They are arranged in this manner: Choice/Prize/Reveal, where the Reveal is which door is opened showing an item of no value. Here are the possible outcomes: Choice/Prize/Reveal 1/1/2 * 1/1/3 * 1/2/3 $ 1/3/2 $ 2/1/3 $ 2/2/1 * 2/2/3 * 2/3/1 $ 3/1/2 $ 3/2/1 $ 3/3/1 * 3/3/2 * In these cases, I have placed a * after each case where you will win if you stay with your original choice. I have placed an $ after each case where you will win if you switch from your original choice. There are six *'s and six $'s, so out of twelve possible cases, half have you winning if you switch, and half have you winning if you stay. Another situation is this: Two people are on the same team. One enters the arena, chooses a door, the host opens one of the remaining doors and shows the item of no value, and then the original participant, host, and the door with the item of no value are removed from the arena. The second participant enters the arena and is asked: 1. To choose a door, or 2. To choose whether to keep the door chosen by the original participant, or to switch to the door not chosen by the original participant. We know that behind one door is a prize, and behind the other door is an item of no value. If the second participant simply chooses between the two, there is a 1 in 2 or 50% chance of finding the prize. Does this chance if the participant, without knowing what door was chosen by the original participant, simply chooses to switch to the door not originally chosen? Empires have risen and fallen over this problem, so I present it here for you to discuss. Choose your door. |
Not true. After the host reveals the item of no value and gives you the option to switch, you are essentially choosing between two doors: one with a prize and one without. It's a 50% chance of getting it right. Whether you stay with your door or not is statistically irrelevant. |
Ok, let me think about that for a second. |
|
I also enjoy this little problem. Sometimes a diagram can help and wikipedia's got a great little treatment: Wild, Wild, West...wiki wiki |
And yet, running the simulations, this is not the case. Simulations (or simply playing the game) show that the players choosing to switch have a 2 in 3 chance of winning, while the players choosing to keep the original choice have a 1 in 3 chance of winning. Annoying, isn't it? |
I get it now. mcgredo's illustration did just that. |