I had this terrible dream in which Monty had five doors and let you pick two. Then he showed you that two of the three unchosen doors had goats and not the car. Now he offered to trade your two doors for the one unchosen and unopened door. Awful dream.
And this is what you came up with?
I can’t understand what you’re trying to say, and I already understand the Monty Hall problem. How do you expect to explain it to someone who doesn’t get the concept when you can’t express it clearly to someone who does understand?
A simple demonstration that makes the math more clear is to use a standard deck of cards. Assign one card as the money card, or car. Say Ace of Diamonds. Then shuffle the deck, assign one person to be the contestant and another to be the host. Have the contestant pick a card, then have the host look at all the other cards, and turn up all but one of them to show they are not the Ace of Diamonds.
Now how confident are you, as the contestant, you guessed right the first time? Do you want to keep your initial guess, or take the card that the Host looked at and did not turn up? Is that guess still 50:50? (Hint: no.)
LOL…you should still trade though, as I’m sure you know. This is sort of the opposite of the deck of cards or 99 doors in that it would be even more difficult for less mathematically inclined people to see why you would do something as crazy as trading two doors for one. 
Right. So in my dream I did switch. The door opened…and behind it was my mother-in-law!
So I ran and ran until I found myself in another game. This time there were seven doors. And worse I found myself with another poor sod who was in the same situation.
It was explained to me that when there are two players and seven doors each of us players gets to pick two doors leaving Monty with three. Monty then opens two of his doors showing goats. He then picks one of the players and opens one of his doors showing a goat. He goes to the other player and does the same. Monty then selects one of the two players, opens his remaining door and shows goat. That player never escapes the nightmare. The player still alive in the game is then always given the opportunity to switch. If he gets the car he also escapes the dream.
Well in my dream I was the remaining player. After Monty opened his two doors he opened one of mine and one of the other guys. All goats . Then he opened the other guy’s remaining door. Goat. I can still hear that guy screaming. So Monty turned to me and offered a switch, my remaing door for his remaining unopened door. Should I switch?
What if you really, really like goats?
I had someone pull that on me when I was trying to explain the classic Monty Hall program. Mexicans cook up goats, and probably a lot of other people do as well. So I’ve gone to dropping the goats whenever I try to explain it.
I still have problems with this, but this table from Wikipedia draws it out for me nicely.
Your initial pick has a 33% chance of winning. At that point the other 2 doors have a 66% chance of winning. Remember that. A 66% chance of winning would be better but you can only pick one door. Then when Monty opens a non-winning door he makes it possible for you to switch to the other door and inherit that 66% chance of winning that set of two doors presented.
Hey! They mention Cecil in that article. I followed the link on his name and I love this opening line:
Also, that’s good wording. It’s tough to explain this to people. You use the word “inherit”, someone else uses “transfer”, I keep looking for a good word to describe what happens. The two doors have a 66% of winning combined. Once you know one of them doesn’t have the prize the remaining door has the 66% chance. I suppose that it is simply deduction.
This is admittedly *extremely *nitpicky, but I’m wondering why everyone seems to be truncating to 66% rather than rounding to 67% (which also allows the percentages to sum to 100)?
Saves time not moving one finger a half inch to the right when typing. Or maybe it’s because 66% better conveys that that the chances are doubled compared to 33%. I suppose 1/3 and 2/3 could do the same thing without losing precision.
I will point or one flaw in increasing the number of choices to exemplify the problem, namely the more choices are eliminated, the further your initial odds drop, so the more tempting a 50:50 choice appears.
If I know I only had a 1/52 chance to begin with, then even if I think the odds are now 50:50, I still get improved odds on the rethink. At 1/3 in the original, people are more likely to trust their gut independent of “minor” odds improvement. But 1/52 to 1/2 is a huge improvement.
Plus, there may be something psychological to watching the host view the cards and select which not to show over just knowing he’s doing it. It’s the act of looking that makes it seem more deliberate.
Good point!
In the last Monty Hall thread I mentioned that it’s closely related to a situation that arises in card-play. I’ll provide more detail. (It arises in Contract Bridge, but non-players may be able to understand the question.)
Contracters are missing 4 Hearts, the Q,J,3,2 (Queen, Jack, Trey and Deuce). These four cards are split between West and East hands, but you have no information to go on about how they’re split. On the first round of hearts, West plays the Deuce and East the Jack. On the second round, West plays the Trey and you must guess the original holdings before East plays. (You’re guessing whther to take the finesse or play for the drop.) If you guess right you win; wrong you lose.
There are only two possibilities. The original holdings could have been
West: 32 … East: QJ
or
West: Q32 … East: J
The former appears to be slightly more likely (12/23) a priori than the the latter (11/23). But which do you play for?
“Appears to be”? If you start with a billion shuffles, the former split will outnumber the latter almost exactly 12-to-11.
Beginners are taught “Eight ever, nine never.” That is, when contracters have nine cards in the suit as here, never try to finesse against the Queen; instead play to drop her.
But does the rule apply here?
In the early pages of this thread, I really struggled to get to grips with this, but here’s how I managed to figure it out.
Three doors:
[A] ** [C]
One has the prize. Each door is 1:3 chance of guessing right. I pick door C:
[A] ** [+]
1/3 1/3 1/3
Open one door that definitely does not have the prize, door A:
** [+]
0/3 2/3 1/3
At this stage I am still 1:3 in my guess. Door A is still there, but now is 0/3, so that leaves 2/3 for door B. Swapping improves my chances.
Does that seem right? The key is that the eliminated door is guaranteed to not have the prize behind it, removing a failure point from your guess, and technically still part of the decision-making equation.
Monty Hall threads never fail to amuse (and bemuse …)
An analogous, though not identical, problem , which may have been mentioned at some point in this thread (I haven’t read all the posts) is the one which Martin Gardner featured in his Scientific American column years ago.
You have three cards, one is double-sided black, one is double-sided white, and one is black on one side and white on the other.
You place the cards in a bag, shake them about, then lay the bag on a table and slide a card out so that only the top side is visible.
This top side is black. What is the probability that the other side is also black ?
It is quite amazing how many people argue (erroneously) that the probability is 50%.
I guess I’ll have to confess to my ignorance here. There are two cards with black sides. If they’re both facing up, either of them has a 50% chance of being black on the other side as well (and a 50% chance of being white). What’s the catch?
There are three equal possibilities if you’re seeing a black side:
- The card is B/W and you’re seeing side 1
- The card is B/B and you’re seeing side 1
- The card is B/B and you’re seeing side 2
What are the odds that the other side of the card you are seeing is black?
There are six sides (3 white sides, 3 black sides). Two of the black sides belong to the same card. We know that we have a black side. The probability that it is from the card with two black sides is 2/3.
Well explained.
Another way to look at it is to say that before you even draw the card, the probability of drawing the B/W card is 1/3, so the probability of drawing either of the double-sided cards is 2/3. The top side must be either black or white, so the probability of it being a double sided card remains at 2/3, irrespective of which color is showing.