A Puzzling conundrum

[Tomcat]

A game show host shows you three face down playing cards and states that there are two Jokers and a queen.
If you pick the Queen you win $100,000.

You then decide on the middle card B.

The host the turns over the first card A and shows that it is a Joker . He then offers you the chance to switch your choice B, with the other face down card C

A                      B                     C
Joker            your pick?            ?

The conundrum is that by switching card B for card C you double your chances of getting the Queen. You would now have a two thirds chance of getting the Queen if you switch to card C.

people have argued about this and guys have been bludgeoned to death in wine bars but im afraid that it is true

[seanbeag7]

In Ireland that is called the 3 card trick, but with a difference.

The difference is that you try to find the king of spades and the other 2 cards are the oueen and jack of spades but the way the man does it is that when he is spinning the cards around the table he trys to distract you and then he slips the king of the table so that u can never win.

The tinkers used to do this at horse fairs, and there were always to spotters in the crowd to keep an eye out for the police.

I never saw anybody win while playing this but i did see the guys do a runner when one spotted a policeman coming there way.

[PigDogg]

In American statistical circles this is known as the Monty Hall problem.   Here's more than one needs to know about it.    

Monty Hall Problem

[bigmick22]

Sean - I think Tomcat's point relates to statistical probability and not a scam scenario

[jb8]

Sounds like a good illustration of the difference between statistics and reality. With 2 cards left you are either
right or wrong, if you switch picks you still will be either right or wrong. If you switch picks it's true you will have had 2 chances and a 66% shot at getting it right but the reality is if I walk up when there are 2 cards remaining, and offer to take the one you don't choose we both have a 50-50 shot.

I'm not a math wiz but thats the way I see it.

[PigDogg]

(bigmick22 @ Mar. 06 2009,16:09) a scam scenario

In New York, the scam scenario is know as Three Card Monte,

Three Card Monte

[PigDogg]

(jb8 @ Mar. 06 2009,17:50) but the reality is if I walk up when there are 2 cards remaining, and offer to take the one you don't choose we both have a 50-50 shot.

This is true only if I'm choosing randomly. Then we both have a 50-50 shot.

But if I'm choosing based on prior knowledge such that I have a 2/3 chance of winning then the leftover choice has a 1/3 chance.

Or to put it another way, let's say we're about to watch a tape of the Super Bowl. If I hadn't heard who won I'd have a 50-50 chance of picking the winner (assuming two evenly matched teams in a game where the spread was pick em) and your leftover team would have a 50-50 shot.

But if I picked the Steelers cause I knew they had already won then you'd have a 0% chance.

Stat guys refer to this as conditional probabilty. If you bet the ponies it's known as past posting.

[jb8]

Pigdogg, I do bet the ponies and I'll take the odds if you want to lay them. There's 2 cards left as long as you haven't seen if your original choice is right or wrong, you're welcome to switch picks if you like . I'll take whichever card you don't want. You give me $52 if I'm right and I'll give you $50 if you are right, it won't come close to evening up the supposed 2/3 to 1/3 statistical advantage but if we do it 100 times I know i'll end up on the winning side.

[PigDogg]

JB8, Tomcat's original post is a bit too vague.  Let's do 100 simulations (with u getting $52 for each win and me getting $50 for each win) and I know I'll come out ahead if you accept the following rewording:

Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats [that is, booby prizes].

The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows:

After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly.

After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.

Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?"

Is it to your advantage to change your choice?  

Btw guys, this is not a straightforward problem such as with cats and suitcases.  This is complicated and has generated numerous academic papers.  Just like with JB8 and I, both sides are usually absolutely sure that they're correct with both sides being very smart, sharp people both bookwise and streetwise.      

[manarak]

It is true.

I can show this with an easy example.
Imagine there are 100 cards on the table, one winning card.

After you chose, the guy turns 98 cards.
So, what are the chances that you will win by switching?

Chances that you chose the right card at the beginning: 1/100.
Chances that you didn't chose the right card (and then it is the other remaining card that wins): 99/100

[jb8]

I'll take any odds you want to give me over even money and stick with my card every time, because it doesn't matter how many we start with. There are two cards left, one is right and one is wrong. In Manaraks situation if I had 1 card and Pigdogg had the other should PD now switch too?

It doesn't matter how many choices, each card of the original 100 had a 1/100 chance of being the right one.
That's my story and I'm stickin to it

Pd I like the way you think(the part about us both being smart , sharp people) Too bad I'm not smart enough to figure out how to pull a partial quote from a post

[jb8]

I would honestly be interested to hear the opinion of a top notch card player like Doyle Brunson, someone with card and gambling and statistic smarts.

Anyways thanks Tomcat for posting something to make us think a little bit!

[Ozzie]

(jb8 @ Mar. 08 2009,10:35) I would honestly be interested to hear the opinion of a top notch card player like Doyle Brunson, someone with card and gambling and statistic smarts.

Anyways thanks Tomcat for posting something to make us think a little bit!

This is all about conditional probabililty.

It is the probability of some event A, given the occurrence of some other event B. The second sentence is the key here. Also assuming non independent events.

There is a good example in Conditional probability

Consider the simple scenario of rolling two fair six-sided dice, labelled die 1 and die 2. Define the following three events:

A: Die 1 lands on 3.
B: Die 2 lands on 1.
C: The dice sum to 8.
The prior probability of each event describes how likely the outcome is before the dice are rolled, without any knowledge of the roll's outcome. For example, die 1 is equally likely to fall on each of its 6 sides, so P(A) = 1/6. Similarly P(B) = 1/6. Likewise, of the 6 × 6 = 36 possible ways that a pair of dice can land, just 5 result in a sum of 8 (namely 2 and 6, 3 and 5, 4 and 4, 5 and 3, and 6 and 2), so P© = 5/36.

Some of these events can both occur at the same time; for example events A and C can happen at the same time, in the case where die 1 lands on 3 and die 2 lands on 5. This is the only one of the 36 outcomes where both A and C occur, so its probability is 1/36. The probability of both A and C occurring is called the joint probability of A and C and is written , so . On the other hand, if die 2 lands on 1, the dice cannot sum to 8, so .

Now suppose we roll the dice and cover up die 2, so we can only see die 1, and observe that die 1 landed on 3. Given this partial information, the probability that the dice sum to 8 is no longer 5/36; instead it is 1/6, since die 2 must land on 5 to achieve this result. This is called the conditional probability, because it is the probability of C under the condition that A is observed, and is written P(C | A), which is read "the probability of C given A." Similarly, P(C | B) = 0, since if we observe die 2 landed on 1, we already know the dice can't sum to 8, regardless of what the other die landed on.

On the other hand, if we roll the dice and cover up die 2, and observe die 1, this has no impact on the probability of event B, which only depends on die 2. We say events A and B are statistically independent or just independent and in this case


In other words, the probability of B occurring after observing that die 1 landed on 3 is the same as before we observed die 1.

Intersection events and conditional events are related by the formula:


In this example, we have:


As noted above, , so by this formula:


On multiplying across by P(A),


In other words, if two events are independent, their joint probability is the product of the prior probabilities of each event occurring by itself.

[bigmick22]

(jb8 @ Mar. 08 2009,04:21) Too bad I'm not smart enough to figure out how to pull a partial quote from a post

just quote the whole post  ... then scroll down to the text box containing "Original post to quote" and delete the bits of original post you don't want quoted! .. then go back to first text box and type your response!!!

[kahuna]

To hell with your doors...Here's the real conundrum...

I'm in the bar and there are 3 ladyboys...

Ladyboy A is young, small and cute but a bit shy...

ladyboy B is older, experienced and very beautiful...

Ladyboy C is a slut and has her cock out before I can even say hello...

Who do I pick?