Math Is Fun Forum

Simon

Member

Registered: 2006-06-06

Posts: 41

Another Monty Hall question - the host's state of knowledge

In previous discussions, we concluded that the host's state of knowledge was crucial.

But is this always true?

With a single game we said it was

Translate it to three cards face down - two Jacks and a Queen. The Queen represents the prize. You pick a card. You estimate correctly that there is 2/3 chance you chose a Jack. The host now reveals one of the other two cards. If he shows a Queen, the game is voided. As it happens, he reveals a Jack. You then ask him "Did you know that the card you turned over was going to be a Jack?"

If he says "Yes", your original 2/3 chance of having chosen a Jack remains intact and there is a therefore a 2/3 chance the other card is the Queen. If he says "No", the odds are reduced to 50/50.

The explanation for the 50/50 odds is clear when you consider the possible outcomes with a host who doesn't know where the Queen is.

Here are the six equally possible games using three cards - Jack1, Jack2 and the Queen. The card the player selects is X. The one the ignorant host turns over is Y. The remaining card is Z.

     X    Y   Z

  -------------

1.  J1  J2   C

2.  J2  J1   C

3.  J1  Q   J2 voided

4.  J2  Q   J1 voided

5.  Q   J1  J2

6.  Q   J2  J1

In those six games you will most likely originally have chosen 4 Jacks and 2 Queens. However, because the host doesn't know where the Queen is, he is most likely to nullify two out of the six games. This cancels your advantage. Nevertheless, it remains true that you originally picked 4 Jacks and 2 Queens. It is simply that the ignorant host cancelled two of the games where you chose a Jack. So whether you stick or swap, you end up with 2 Jacks and 2 Queens.

BUT

What if six games were played but none were voided?

Imagine them as six rows of cards face down, each with two Jacks and a Queen.
You selected one card randomly from each row.
You estimated that you picked 4 jacks and 2 Queens.
The host revealed a card from each row. They were all Jacks.

No games were voided.

You asked the host the same question: "Did you know that the cards you turned over would be Jacks?"

Does his answer affect your estimate of having originally picked 4 Jacks and 2 Queens.

What if his answer was no?

Should you retrospectively revise the probability to 50/50 and beleive it is more likely you picked 3 Jacks and 3 Queens?

Or does the original probability of your selection stand?

Simon

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Another Monty Hall question - the host's state of knowledge

I think the answer is the same as in the original Monty Hall problem.

This is pretty much the same thing, but just phrased a bit differently. Instead of knowingly choosing a jack, he just happens to always choose one because he's lucky. But the fact remains that the way the problem is posed means that it's impossible for him to pick a queen, and so the same logic from the original problem will follow.

---

Why did the vector cross the road?
It wanted to be normal.

Simon

Member

Registered: 2006-06-06

Posts: 41

Re: Another Monty Hall question - the host's state of knowledge

the way the problem is posed means that it's impossible for him to pick a queen, and so the same logic from the original problem will follow.

Therein lies the point. In those six rows, it was not impossible for the host to have revealed a Queen. It simply turned out that he didn't do so. The way the problem is posed is to consider what happens in that event. This does not rule out other events.

Let's clarify the rules. Whenever the host reveals a Queen, that particular row is null and void, but you consider the rows that haven't been voided. The most probable scenario is that the ignorant host will randomly reveal 2 Queens. In that event, there is no reason not to stand by your initial guess that you selected 2 Queens and 4 Jacks. You know that two rows where you chose a Jack have been cancelled. This means that in the four uncancelled rows, you probably selected 2 Queens and 2 Jacks. For those rows, this does indeed confirm the odds as 50/50 whether you stick or swap.

So what happens when, inevitably, you get a game in which by chance the host revealed six Jacks?

Is the intial guess for your selection - 4 Jacks and 2 Queens - still the right one? Do you in fact have a 2/3 advantage in swapping?

As far as it affects the probability of your selection, what you're saying is:

The host revealing six Jacks by chance = The host revealing six Jacks on purpose.

If so, might it not follow that in a single game:

The host revealing one Jack by chance = The host revealing one Jack on purpose.

But it has been well established in Monty Hall that what the host reveals must be based on certain knowledge. It is only then that it is impossible for the host to reveal a Queen. This is what is said to give you the 2/3 advantage.

I appreciate your position. On average, your initial selection is always likely to be 4 Jacks and 2 Queens. You're saying, therefore, that in each game, the probability of that initial selection must remain the same, regardless of what the host randomly or knowingly reveals afterwards.

If, as expected, the ignorant host cancels two rows by showing a couple of Queens, the 50/50 odds do apply, but only to the four remaining rows, from which you probably selected 2 Jacks and 2 Queens. This would still translate to your original selection having been 4 Jacks and 2 Queens.

If the host happens not to cancel any of the six rows, nothing is removed from your initial selection. A reasonable assumption is that this selection, made before the host revealed anything, probably contains exactly what was originally likely - 4 Jacks and 2 Queens.

But doesn't this contradicts the importance of the host's state of knowledge, so crucial to the original Monty Hall solution.

If, in a single game, the host was ignorant but didn't reveal the prize - there's no end of websites that explain why this reduces the odds to 50/50.

But if true, shouldn't those odds apply to each of the six rows in a game where the host was ignorant but didn't reveal a Queen? Surely each individualcard you selected should now have a revised 50/50 chance of being a Jack or Queen - because the host had no knowledge when he revealed a Jack from each row? Wouldn't your entire six card selection, therefore, now most likely contain 3 Jacks and 3 Queens,

This would mean that the 4-Jack-2-Queen probability of your initial selection can be retrospectively changed by what a host happens to do randomly, after the fact.

A strange but true conclusion?

Simon

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Another Monty Hall question - the host's state of knowledge

After thinking about it a bit more, I've changed my mind.
The host needs to select the queen to void a game, so he'll void around half of the games where you chose a jack, and none of the ones where you chose a queen.

Before, we had 1/3 of the games with the player choosing a queen, and 2/3 of them with the player choosing a jack.

Now, we have 1/3 queens, 1/3 jacks, and 1/3 voided.
Of the unvoided games, there are equal amounts where the player chooses a queen and where the player chooses a jack, and so swapping shouldn't give an advantage.

(My original thinking was based on the wrong assumption that the voided games were distributed evenly.)

---

Why did the vector cross the road?
It wanted to be normal.

Simon

Member

Registered: 2006-06-06

Posts: 41

Re: Another Monty Hall question - the host's state of knowledge

Before, we had 1/3 of the games with the player choosing a queen, and 2/3 of them with the player choosing a jack.

Now, we have 1/3 queens, 1/3 jacks, and 1/3 voided.

Agreed. More precisely:

1/3 of the time I choose the Queen - and the game proceeds
1/3 of the time I choose a Jack - and the game proceeds
1/3 of the time I choose a Jack - and the game is cancelled

This describes what happens on average, which includes 1/3 of the games actually being voided by the host. Note that 2/3 times you still choose a Jack. It is just that half of those games are cancelled. The probability of your initial selection remains unchanged.

But Mathys, you have yet to explain the situation where you have a sample of successive games that happen not to have been voided. Hence, the six row scenario where you reasonably guess you picked 4 Jacks and 2 Queens. Subsequently the host reveals 6 Jacks. He could have voided a couple of games, but didn't.

Has the probability of your selection, made before the host's actions, now changed to 3 Jacks and 3 Queens, because of the host's lack of knowledge?

Simon

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Another Monty Hall question - the host's state of knowledge

I would say yes. If we're given that a game has proceeded, then there's an equal chance for jack or queen.

---

Why did the vector cross the road?
It wanted to be normal.

Simon

Member

Registered: 2006-06-06

Posts: 41

Re: Another Monty Hall question - the host's state of knowledge

I would say yes. If we're given that a game has proceeded, then there's an equal chance for jack or queen.

But this presents an apparent contradiction. The probability of your selection was originally 4 Jacks and 2 Queens. There are subsequently six pairs of cards remaining. You already know there is at least one Jack in each pair. As it happens, the host reveals a Jack from each pair. Apart from not cancelling any row,  the host has simply confirmed what you already knew.

What information have you got that could change the 4-Jack-2-Queen probability of your selection, which was made before the host entered?

Last edited by Simon (2007-08-06 15:46:13)