# The BEST and FINAL answer on the MONTY HALL PARADOX! **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [June 18, 2001, 1:50pm UTC](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218 "2001-06-18T13:50:13Z")\ **Posts on this page:** 11\ **Page:** 2
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [June 27, 2001, 12:24pm UTC](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218/21 "2001-06-27T12:24:13Z")
**december,** it sounds like you’re saying something like the following: _You are engaged in a card game with two opponents. At the beginning of the game, you determine that both the Jack and Queen of spades have been dealt, and are not in your hand (thus, they must be your opponents’ hands). During the course of the game, you find that Opponent A holds the Jack of Spades. What is the probability that he (or she) also holds the Queen of Spades?_ In this case, it seems to me that, at the beginning of the game, there are four potential situations, each equally possible: 1. Jack held by A, Queen held by A. 2. Jack held by A, Queen held by B. 3. Jack held by B, Queen held by A. 4. Jack held by B, Queen held by B. During the course of the game, you eliminate possibilities 3 and 4, leaving 1 and 2. As far as I can see, the probability of 1 or 2 is still approximately 50/50 (not exactly, because the total number of cards is still finite). In this case, since the Jack and the Queen are differentiable, the bit of information you obtain (Opponent A holds the Jack of Spades) serves to eliminate two of the four possibilities. This is in contrast to the “intended” two-children problem, where the children are not differentiable, and the bit of information you obtain (one of the children is a girl) serves to eliminate only one of the four possibilites. However, by your earlier description, you seem to be angling for a 1/3-2/3 solution. Am I misinterpreting? Comments? ---
**Author:** ![december](https://avatars.discourse-cdn.com/v4/letter/d/838e76/32.png) [@december](https://boards.straightdope.com/u/december)\ **Post date:** [June 27, 2001, 1:11pm UTC](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218/22 "2001-06-27T13:11:26Z")
> [@](#): > > \*Originally posted by zut \* > \*\ ***december,** it sounds like you’re saying something like the following: > > _You are engaged in a card game with two opponents. At the beginning of the game, you determine that both the Jack and Queen of spades have been dealt, and are not in your hand (thus, they must be your opponents’ hands). During the course of the game, you find that Opponent A holds the Jack of Spades. What is the probability that he (or she) also holds the Queen of Spades?_\*\* Yes, that’s a good summary of the problem, with one additional fact. The way you found out that an opponent held the Spade Jack is that he played that card. At that point in the hand, he would have played a lower card, if he had held one, so you know that his Spade holding was only the Jack or only the Queen and Jack. If he had both the Queen and the Jack, but no lower card, he might have played either the Queen or the Jack. If he had the Jack alone, he would have played it. Hope this helps… ---
**Author:** ![december](https://avatars.discourse-cdn.com/v4/letter/d/838e76/32.png) [@december](https://boards.straightdope.com/u/december)\ **Post date:** [June 27, 2001, 2:25pm UTC](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218/23 "2001-06-27T14:25:18Z")
An urn holds 3 coins, one with two heads, one with two tails, and one with a head and a tail. Without looking, you select one coin and place it on the table. You see a Head side up. What is the probability that the other side of this coin is a Head? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [June 27, 2001, 4:23pm UTC](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218/24 "2001-06-27T16:23:26Z")
There’s three heads in the urn, and two of those three heads have a head on the other side. 67% that the side down is a head. I’m still not convinced on the bridge problem. Suppose we have a “special” deck of cards, where the jack and queen are the only spades. That shouldn’t change the problem, should it? But then, we already know that his spade holding is either the JS or the JS and QS. What’s the difference with what **december** ’s saying? ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [June 27, 2001, 4:44pm UTC](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218/25 "2001-06-27T16:44:35Z")
> [@](#): > > \*Originally posted by december \* > If he had both the Queen and the Jack, but no lower card, he might have played either the Queen or the Jack. If he had the Jack alone, he would have played it. > > Hope this helps… I think so. Judging from your quoted comment, there’s a key point I missed earlier, along the lines of: _At one point during the game, Opponent A is required to play **either** the Jack of Spades **or** the Queen of Spades (but not both), if he holds them._ Given that, before play begins, there are clearly five different possibilies: 1a) Opponent A has both Q and J and plays Q. (1/8 probability) 1b) Opponent A has both Q and J and plays J. (1/8 probability) 2) Opponent A has J only and plays J. (1/4 probability) 3) Opponent A has Q only and plays Q. (1/4 probability) 4) Opponent A has neither Q nor J (and plays neither). (1/4 probability) Since A played a J, possibility _1a,_ 3, and 4 are eliminated, and thus the probability that A also has the Q is only 1/3. Ha! I get it. ---
**Author:** ![december](https://avatars.discourse-cdn.com/v4/letter/d/838e76/32.png) [@december](https://boards.straightdope.com/u/december)\ **Post date:** [June 27, 2001, 6:33pm UTC](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218/26 "2001-06-27T18:33:32Z")
> [@](#): > > \*Originally posted by zut \* > \*\*Ha! I get it. \*\* Yep. ---
**Author:** ![spacecat](https://avatars.discourse-cdn.com/v4/letter/s/ccd318/32.png) [@spacecat](https://boards.straightdope.com/u/spacecat)\ **Post date:** [June 29, 2001, 4:56am UTC](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218/27 "2001-06-29T04:56:42Z")
Cecil sez: There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.) I disagree. This is analogous to the following question: If your friend had two coins, coin A and coin B, and he agreed to flip both coins and tell you what coin A landed on, but make you guess about coin B, you would have a 1/2 chance of guessing correctly. I would agree with the 2/3 answer if Cecil’s question was worded as follows: There is a family with two children. You have been told this family has a daughter. One day, you visit them and one of the children is away. The child who is at their house when you visit is a girl. What are the chances that the other child is a girl too? ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [June 29, 2001, 12:48pm UTC](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218/28 "2001-06-29T12:48:34Z")
**spacecat** : I disagree. Let’s look at Cecil’s problem. Before knowing anything else, you know that the family has two children. From this, there are four potential combinations, all equally probable:[ul]1) Child A is a girl, Child B is a girl. 2) Child A is a girl, Child B is a boy. 3) Child A is a boy, Child B is a girl. 4) Child A is a boy, Child B is a boy.[/ul]The additional fact that this family has a daughter eliminates option (4). In two of the three options left, the other child is a boy, so the answer is 2/3. [/quote] In your problem, you add the twist of also seeing on of the children. So, before knowing anything else, you know there are eight potential combinations, all equally probable:[ul]1) Child A is a girl, Child B is a girl, you see Child A. 2) Child A is a girl, Child B is a boy, you see Child A. 3) Child A is a boy, Child B is a girl, you see Child A. 4) Child A is a boy, Child B is a boy, you see Child A. 5) Child A is a girl, Child B is a girl, you see Child B. 6) Child A is a girl, Child B is a boy, you see Child B. 7) Child A is a boy, Child B is a girl, you see Child B. 8) Child A is a boy, Child B is a boy, you see Child B.[/ul]The additional fact that the family has a daughter eliminates options 4 and 8. The further fact that you see a girl eliminates options 3 and 6 (and, actually, would also eliminate options 4 and 8, so the previous statement that “You have been told this family has a daughter” is redundant). The options left are 1, 2, 5, and 7. In two of these four, the other child is a girl, so the answer is 1/2. ---
**Author:** ![spacecat](https://avatars.discourse-cdn.com/v4/letter/s/ccd318/32.png) [@spacecat](https://boards.straightdope.com/u/spacecat)\ **Post date:** [June 29, 2001, 5:26pm UTC](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218/29 "2001-06-29T17:26:22Z")
You’re right! Thanks for making it more easy to visualize. Aaron Ackerson ---
**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\ **Post date:** [June 29, 2001, 5:35pm UTC](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218/30 "2001-06-29T17:35:32Z")
Wumpus wrote: > [@](#): > > The original Monty Hall problem predates Maralyn’s column on it. I believe that Tom and Ray Magliozzi should get the credit for bringing this into the mainstream. I read Marilyn’s column when it was new, and I read Cecil’s column when it was new, and before each I was already familiar with the full analysis of the problem because it was a “weekly puzzler” on their car talk show before either Cecil or Marilyn addressed it. ---
**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\ **Post date:** [June 29, 2001, 8:06pm UTC](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218/31 "2001-06-29T20:06:15Z")
KidCharlemagne wrote: > [@](#): > > Foolsguinea, I was just curious if you liked the name because it’s generally melodious or if you knew it was the name of a Steely Dan song. (with the greatest guitar solo possibly ever recorded) Please credit Larry Carlton for his excellent work. One thing I love about Steely Dan is how they got the best musicians and allowed them freedom to create their own parts. Larry Carlton was also responsible for the excellent guitar work in _Don’t Take Me Alive_. [Previous page](https://boards.straightdope.com/t/the-best-and-final-answer-on-the-monty-hall-paradox/69218.md?page=1)