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#1 2007-01-27 00:28:27

Simon
Member
Registered: 2006-06-06
Posts: 41

The Six Card Trick - an advanced variation on Monty Hall

The following should be solvable by everyone, but will be of particular interest to those familiar with Monty Hall.

There are six cards face down - three are Kings, three are Queens. There is a host present who knows what each card is. Naturally, you don’t.

1. You choose one of the cards.

2. From the five remaining cards, four cards must now be revealed and eliminated. The rule is that three of a kind must not be shown. No game can ever be voided.

3. You have three choices as to how this is done.

a) You can instruct the host to reveal all four cards. He will use his knowledge to reveal 2 Kings and 2 Queens in any order you wish.
b) You can turn over the first two cards yourself. and the host will turn over the opposite.
c) You can turn over the first and third card yourself and the host will turn over the opposite.

Whichever method is used, exactly two Kings and two Queens are always removed. You are left with a King and a Queen. Your chosen card is one of them.

Which is your card more likely to be? Are the odds 50/50?
Does it make a difference who revealed what or is the probabilty constant?

Consider the following three games that were played using each method.

In game a), you instructed the host to remove two Kings and then two Queens.
In game b) you turned over two Kings and the host showed two Queens
In game c)  you alternated with the host in a sequence of King, Queen, King, Queen.

In each game there is a King and Queen left. Your chosen card could be either. Let's say you want the Queen.

Should you stick or swap? Does it make no difference? Do the odds vary, depending on the method of elimination?

Have fun! smile

Simon

Last edited by Simon (2007-01-27 00:29:22)

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#2 2007-01-27 02:23:54

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: The Six Card Trick - an advanced variation on Monty Hall


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

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#3 2007-01-27 02:50:04

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

Re: The Six Card Trick - an advanced variation on Monty Hall

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#4 2007-01-27 20:18:53

Devantè
Real Member
Registered: 2006-07-14
Posts: 6,400

Re: The Six Card Trick - an advanced variation on Monty Hall

I hate card tricks. Very predictable at times.

I'll read this.

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#5 2007-01-27 20:40:35

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: The Six Card Trick - an advanced variation on Monty Hall

Last edited by luca-deltodesco (2007-01-27 20:40:48)


The Beginning Of All Things To End.
The End Of All Things To Come.

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#6 2007-01-28 00:22:55

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: The Six Card Trick - an advanced variation on Monty Hall

I've been thinking a bit more, and I've changed my mind.


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

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#7 2007-01-28 14:27:10

Simon
Member
Registered: 2006-06-06
Posts: 41

Re: The Six Card Trick - an advanced variation on Monty Hall

Apologies for throwing you all in the deep end!

I'm going to make it much simpler.

Forget six cards. Let's do it with four, and only have one card eliminated.
By the way, this is not a trick. Just a probability exercise!

There are four cards face down - two Kings and two Queens. There is a a host on hand who knows what the cards are.

1) You choose a card.

2) You decide who is going to eliminate one of the three remaining cards – you or the host.

a) If it's you, then simply turn over one of the cards.
b) If it's the host, you choose whether he reveals a King or Queen. He will use his knowledge to comply.

Based on what is revealed, what are the odds that your chosen card is a King or Queen?
Do the other two cards have the same odds?
Does it make a difference who performed the elimination?

As an example, imagine two games were played and the same card was shown each time.

In game 1, a Queen was revealed. You turned over the card youself.
In game 2, a Queen was revealed. You told the host to reveal a Queen.

In both cases you were left with two Kings and one Queen.

In each game:

Where is the Queen most likely to be?
Are the odds the same for each card?
Does it matter who performed the elimination?

Hope that's more straightforward!

Simon

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#8 2007-01-29 01:03:08

Simon
Member
Registered: 2006-06-06
Posts: 41

Re: The Six Card Trick - an advanced variation on Monty Hall

So far, Mathsyperson has come pretty close in his solution to the six card version, so I'll start with that.

It does indeed matter who revealed what. Yes, the odds are 50/50 if the host revealed all four cards. Also if you participated and revealed one of each kind, the odds are still 50/50.

However, if you revealed two of a kind - say Kings - and the host revealed two Queens, then no matter what order they were removed, your card now has a 3/4 probabilty of being a Queen. (If you revealed just one King, and the host removed another King and two Queens, the odds would be 3/5 your card is a Queen.)

With the four card version the answer very simple but quite unbeleivable.

Lets take those two possible games where the same card is revealed.

a) You choose one of four cards. From the remaining three cards a Queen is revealed. You turned it over yourself. There are definitely two Kings and one Queen left. Result? The odds are 1/3 that your card is a Queen. Identical odds apply to the other cards.

Most people agree with the above but dispute the following:

b) You choose one of four cards. From the remaining three cards a Queen is revealed. The only difference is you instructed the host to reveal it. There are two Kings and one Queen left. Result? The odds are 50/50 that your card is a Queen. For each of the other two, the odds are 1/4.

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