card problem 9

[BobRodes]

You have three cards in a hat. One is white on both sides, one is red on both sides, the third is white on one side and red on the other. You pull a card out of the hat and lay it down. The face is white. What are the chances that the other side of the card is also white?

 

[kwbMitel]

@ Strongm and CajunCenturion
[hide]
My Brute force attacks do not agree with your results. I have arranged my cards as follows[tt]
A B
Card 1 - W / W
Card 2 - W / R
Card 3 - R / R[/tt]

My formula works like this
Card Shows White and other side is White = True
Card Shows White and other side is Red = False
Card Shows Red = False

Card 1 Side A - Shows White Other Side White = True
Card 1 Side B - Shows White Other Side White = True
Card 2 Side A - Shows White Other Side Red = False
Card 2 Side B - Shows Red - False
Card 3 Side A - Shows Red - False
Card 3 Side B - Shows Red - False

2 out of 6 = 1/3
[/hide]

*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.

 

[CajunCenturion]

IPGuru - You're treating the calculation of the odds of the other side being white as beginning after the white side has been shown. In other words, you're viewing this as a two-card problem with one card having the other side white and the other card with the other side red. You're viewing this as a two-card single step problem.

It's not.

This is a three-card, two step problem. The odds of the other side being white must take into account the initial conditions of all three cards and the initial step of choosing a white side to begin with.

I ask you, given the initial conditions of three cards, what are the odds of pulling a card showing a white side?
How did you determine that?

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[kwbMitel]

Strongm and CC ignore ignore ignore previous.

As soon as I hit send I realised my mistake.

I 100% agree with your results.
[hide]
The card combinations that come up facing red do not apply as the puzzle states that the card face is white. That leaves only 3 possibilities (not 6 as in my example above) as 2 of those possibilities are true the answer is definitely 2/3.
[/hide]

*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.

 

[CajunCenturion]

@ kwbMitel -
[hide]Please read the problem again, and I think you'll see that your results confirm what strongm and I are claiming. You're providing the odds of the other side being white regardless of the original side shown. Only have of your brute force results pertain to a white side being shown, and in 2/3 of those, the other side is white.[/hide]


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[kwbMitel]

CC - I was retracting as you were writing. My second post explains my error.

*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.

 

[CajunCenturion]

Yeah, we were posting at the same time. No problem.

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Good Luck
To get the most from your Tek-Tips experience, please read
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[guitarzan]

IPGuru: Maybe this explanation will help?

Spoiler

You have three cards in a hat. One is white on both sides, one is red on both sides, the third is white on one side and red on the other
So let's label the cards as Card1, Card2, and Card3 as follows
Card1 = is white on both sides
Card2 = is red on both sides
Card3 = is white on one side, and red on the other.

Now, let's label the sides as SideA or SideB, as follows:
Card1SideA = White
Card1SideB = White

Card2SideA = Red
Card2SideB = Red

Card3SideA = White
Card3SideB = Red

You pull a card out of the hat and lay it down. The face is white.
This means that you are NOT looking at Card2, since it has no sides that are white. So we can only be looking at Card1 or Card3.
Of the four sides that exist on those cards, there are only three sides we could be looking at: Card1SideA, Card1SideB, or Card3SideA (Card3SideB is red, so we are not looking at that side).

So what is under those cards?
Card1SideA = underside is Card1SideB which is White
Card1SideB = underside is Card1SideA which is White
Card3SideA = underside is Card1SideB which is Red

What are the chances that the other side of the card is also white?
2 in 3, as strongm said.

 

[IPGuru]

you are all falling in to a simple mathamatical trap
ingnoring the real world

Spoiler

you think red red is iliminated because the initial set up says one side white this is correct

you rightly say that white white has side B & side B - Correct
however as A = B it has no affect on this puzzle

if the card is red/ white then answer false

if white/white then side A or side B has alredy been defined in the setup condition (although invisible to the naked eye) leaving only the 1 remaining side to test
the question should realy consist of cards A/B a/red red/red

if the card shows A what are the chances of the reverse showing B

bottom line you have 2 cards 1 & 2
the chances of it being card 2 are 50/50 - it des not mater which way up.

 

[kwbMitel]

@ IPguru
[hide]
It is ironic that your last post claims that someone else is ignoring the real world when in fact that is the exact flaw in your premise

there are 6 possible outcomes of drawing a card.
the 3 outcomes that result in red face and are eliminated

Of the 3 outcomes that result in a white face 2 have a white face on the opposite side

Answer = 2/3

You are only counting the White/White card once. There are 2 ways to draw that card and both MUST be counted.[/hide]

*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.

 

[strongm]

>you are all falling in to a simple mathamatical trap

Actually the person falling into a trap is you, I'm afraid.

And if, after all the examples of the card states that have been provided, you still can't see that then I don't see that I am going to be able to convince you. But let's have one more go[hide]
Let's assume that on one side of the white/white card is printed a big black A, and on the other side a big black B (note that doing so does not alter the conditions of the puzzle)

Are you prepared to grant that now, to have a white face showing after pulling a card from the hat we have three starting possibilties

W (from the R/W card)
WA
WB

?

And turning the card over results in:

W -> R
WA -> WB
WB -> WA

i.e in 2 of the 3 situations we have a W.

Putting the big A and B is just to help you visualise the situation by showing that each of the white sides is a different case. The mathematics, however, does not require the letters.[/hide]

 

[guitarzan]

Spoiler

you rightly say that white white has side B & side B - Correct

No one said that. No one understands that, either. Are you saying that a white/white card only has one side? Every card has two sides. The sides may share the same color, but they are still DIFFERENT SIDES and must be treated separately.

Reread my earlier post, and if it makes you happy, take three sheets of paper (which have a combined 6 sides) and label them as I explained above.

If you disagree with my setup from earlier, explain where you disagree. If you agree, go through EACH AND EVERY possibility (there are only six!) You will find that only three outcomes actually match the original premise (white side facing up), and of those three, only two give you a favorable outcome (white on the underside)

 

[CajunCenturion]

Putting the big A and B is just to help you visualise the situation by showing that each of the white sides is a different case. The mathematics, however, does not require the letters.

Great explanation - I'm going to borrow that.

--------------
Good Luck
To get the most from your Tek-Tips experience, please read
FAQ181-2886
As a circle of light increases so does the circumference of darkness around it. - Albert Einstein

 

[randy700]

Spoiler

Some of you continue to state there are 6 possible draws, which is true. However, the original problem poses a question of what could happen AFTER the draw. Therefore, there are no longer 6 possible draws. The only draw to be concerned with has ALL READY TAKEN PLACE. Based on that premise, if you are looking at a white face, there are only 2 possible colors on the other side - red (if you have the red/white card) or white (if you have the white/white card). It doesn't matter which side of the white/white card you see. The problem only asks about the possible colors on the reverse side. The red/red card does not exist at this stage of the problem.

Randy

 

[strongm]

Randy700, please see my explanation involving big As and Bs to help you too to visualise what is actually going on.

 

[guitarzan]

Spoiler

Some of you continue to state there are 6 possible draws, which is true. However, the original problem poses a question of what could happen AFTER the draw. Therefore, there are no longer 6 possible draws.

Agreed. There ARE six possible draws, but three of them can be eliminated because they show red on top, and the original question clearly states that the card selected shows white on top.

The only draw to be concerned with has ALL READY TAKEN PLACE.

Ummm, yes. That is the premise.

Based on that premise, if you are looking at a white face, there are only 2 possible colors on the other side - red (if you have the red/white card) or white (if you have the white/white card).

True. But what you are missing is that there are TWO DIFFERENT WAYS to get white on the other side. strongm's explanation illustrates that nicely.

It doesn't matter which side of the white/white card you see.

This is where your error is. It does matter. Again, strongm's big black letters on the cards should illustrate that for you.

The problem only asks about the possible colors on the reverse side.

No it doesn't. It asks more than that. Re-read the question.

The red/red card does not exist at this stage of the problem.

Yes. And we established that the red/red card couldn't have been drawn, as it would show a red on top, and the premise clearly states that white is showing on top.

 

[CajunCenturion]

randy700 - You're an Access programmer, so please try this

Open up an Access form and add a button (cmdCards) and two text boxes (txtRed and txtWhite). Then drop in this code in the click event for the button.

Private Sub cmdCards_Click()

    Dim cards(3, 3)     As String
    Dim idx             As Integer
    Dim cardpicked      As Integer
    Dim sidepicked      As Integer
    Dim otherside       As Integer
    
    cards(1, 1) = "Card 1"
    cards(1, 2) = "White"
    cards(1, 3) = "White"
    cards(2, 1) = "Card 2"
    cards(2, 2) = "White"
    cards(2, 3) = "Red"
    cards(3, 1) = "Card 3"
    cards(3, 2) = "Red"
    cards(3, 3) = "Red"
    
    Me.txtRed.Value = 0
    Me.txtWhite.Value = 0
    Randomize
    
    For idx = 1 To 10000
        cardpicked = Int((3 * Rnd) + 1)                     ' Pick one of the three cards
        sidepicked = Int((2 * Rnd) + 1) + 1                 ' Pick which side to show
        If cards(cardpicked, sidepicked) = "White" Then
            If sidepicked = 2 Then                          ' Figure out which is the other side
                otherside = 3
            Else
                otherside = 2
            End If
            If cards(cardpicked, otherside) = "White" Then  ' Increment the totals
                Me.txtWhite.Value = Me.txtWhite.Value + 1
            Else
                Me.txtRed.Value = Me.txtRed.Value + 1
            End If
        End If
    Next idx
    
End Sub

Please run it and explain the results.

--------------
Good Luck
To get the most from your Tek-Tips experience, please read
FAQ181-2886
As a circle of light increases so does the circumference of darkness around it. - Albert Einstein

 

[randy700]

CajunCenturion

Spoiler

The code you provide does, in fact, support your argument that the odds are 2 in 3. However, I contend that this is flawed because you no longer have 3 cards to choose from because the card has all ready been drawn and we know it has at least 1 white side. How about this modification to your code?

Private Sub cmdMyResponse_Enter()
Dim cards(2, 3) As String
Dim idx As Integer
Dim cardpicked As Integer
Dim sidepicked As Integer
Dim otherside As Integer

cards(1, 1) = "Card 1"
cards(1, 2) = "White"
cards(1, 3) = "White"
cards(2, 1) = "Card 2"
cards(2, 2) = "White"
cards(2, 3) = "Red"

Me.txtRed.Value = 0
Me.txtWhite.Value = 0
Randomize

For idx = 1 To 10000
cardpicked = Int((2 * Rnd) + 1) ' Pick one of the TWO cards
sidepicked = 2 ' MUST SHOW WHITE SIDE
If cards(cardpicked, sidepicked) = "White" Then
If sidepicked = 2 Then ' Figure out which is the other side
otherside = 3
Else
otherside = 2
End If
If cards(cardpicked, otherside) = "White" Then ' Increment the totals
Me.txtWhite.Value = Me.txtWhite.Value + 1
Else
Me.txtRed.Value = Me.txtRed.Value + 1
End If
End If
Next idx

End Sub

Randy

 

[CajunCenturion]

[hide]

Some of you continue to state there are 6 possible draws, which is true. However, the original problem poses a question of what could happen AFTER the draw. Therefore, there are no longer 6 possible draws.

No, there are no longer six possible draws, but you're still in one of three valid states.
[tt]Card 1: 1Ra - Invalid state - No Flip
1Rb - Invalid state - No Flip
Card 2: 2Ra - Invalid state - No Flip
2Wb - Valid state - Flip to Red
Card 3: 3Wa - Valid state - Flip to White
3Wb - Valid state - Flip to White[/tt]
Even though the flip takes place after the draw, you still can't ignore the states that the draw put you in. You're flipping from one of the three valid states created by the initial draw. You cannot combine states 3Wa and 2Wb into a single state 3Wb state before taking the next step and executing the flip. Doing so ignores the initial conditions of the problems and skews the results.[/hide]

--------------
Good Luck
To get the most from your Tek-Tips experience, please read
FAQ181-2886
As a circle of light increases so does the circumference of darkness around it. - Albert Einstein

 

[IPGuru]

I can see the reasoning behind the 2-3 answer but it makes an incorrect presumption

it is assuming that we have 3 possibilities for the unknow side

a b or c where a & b are +ve results giving 2:3

however a=b therefore the unknows are actualy a & c
giving 1 : 2

You are calculating using sides but failing to take into account there a 4 possible sides not 3

only 2 of these 4 sides results in a +ve result.

using only 3 sides would violate the laws of physics

 

[CajunCenturion]

@randy700
[hide]==> However, I contend that this is flawed because you no longer have 3 cards to choose from because the card has all ready been drawn and we know it has at least 1 white side.
No, the flaw is incorrectly combining the valid states created by the initial draw. It's not two one-step problems; it's one two-step problem.[/hide]

--------------
Good Luck
To get the most from your Tek-Tips experience, please read
FAQ181-2886
As a circle of light increases so does the circumference of darkness around it. - Albert Einstein