Homework - Late Crammer (1 Viewer)

[AnthonyGerrard]

The answer is indeed 2/3.

There's 3 possible red dots that could be face up on the table.

2/3 of them have a red dot on the other side.

(each face is equally likely to have been selected originally, but there are 2 possible ways of choosing a red opposite a red, only one of picking a red opposite a white., So 2/3 you have a red on the opposite face)

 

[AnthonyGerrard]

I never mentioned previous draws!

I took the question seriously in my first post, but the question was then changed so I became facetious as when there is a choice of two it is always 50 - 50 even in the threads I posted a link to but there Anthony talked a load of tosh to try to indicate that previous draws matter, strangely many agreed.

Read my post 11 before being critical.

Brian

The question never changed was just very generously clarified by myself - I presume you missed a smilie from the rest of it!

 

[Frothingslosh]

The answer is indeed 2/3.

There's 3 possible red dots that could be face up on the table.

2/3 of them have a red dot on the other side.

(each face is equally likely to have been selected originally, but there are 2 possible ways of choosing a red opposite a red, only one of picking a red opposite a white., So 2/3 you have a red on the opposite face)

That is incorrect.

You're assuming there is a set of three possibilities (the number of dots), when there are in fact only two possibilities (two cards with a dot on at least one side). Either you have the card with two dots, or you have the card with one dot and one blank. That's it - the two dot card is no more likely to have been drawn than the one dot card.

The entire puzzle, as given, can be summed up as "The card you drew from the set described above has at least one dot on it. What are the odds it is the card with two dots rather than the card with only one dot?".

The question never changed was just very generously clarified by myself - I presume you missed a smilie from the rest of it!

Actually, I'm fairly certain he was speaking to me.

 

[AnthonyGerrard]

That is incorrect.

You're assuming there is a set of three possibilities (the number of dots), when there are in fact only two possibilities (two cards with a dot on at least one side). Either you have the card with two dots, or you have the card with one dot and one blank. That's it - the two dot card is no more likely to have been drawn than the one dot card.

The entire puzzle, as given, can be summed up as "The card you drew from the set described above has at least one dot on it. What are the odds it is the card with two dots rather than the card with only one dot?".



Actually, I'm fairly certain he was speaking to me.

Its is correct. 2/3 is the correct answer.



Look at it as a dice - two dots opposite, two blanks opposite and a dot opposite a blank.

The chances of rolling a dot opposite a dot are twice that of rolling a dot opposite a blank.


2/3 of the dots are opposite dots.


If you are not convinced get a dice - some stickers - and roll away.


--
PS - its who Brian was talking about - I was more concerned with.

 

[scott-atkinson]

Its is correct. 2/3 is the correct answer.



Look at it as a dice - two dots opposite, two blanks opposite and a dot opposite a blank.

The chances of rolling a dot opposite a dot are twice that of rolling a dot opposite a blank.


2/3 of the dots are opposite dots.


If you are not convinced get a dice - some stickers - and roll away.


--
PS - its who Brian was talking about - I was more concerned with.

I disagree... I believe the answer is indeed 50%

The drawn card that sits on the table with the red dot facing up is the controlling factor in this, the other cards are irrelevant as they have not been drawn.

Knowing that the reverse of the card on the table will either be a Blank or a Red narrows the chance to 50%

The outcome would be the same with if any card was drawn, either Blank face up or Red face up, the chance of the reverse being the same as the front is always 50% as there are only two possible outcomes. The undrawn cards are not part of the equation.

 

[Frothingslosh]

As Scott said, the question is about the cards, not the dots. The card on the table has one dot showing. The question boils down to whether this is the card with a dot on each face or the card with a dot on one face and a blank on the other. One of these two cards has been dealt to you. You know there is a dot on the top - the question is whether the other side is blank (making it card 2) or has a dot (making it card 3). The odds of getting card three are not multiplied just because both sides have dots - you're still drawing from a pool of two cards.

Try rephrasing it:

You have two cards. One has a dot on the front, one does not. One card at random is dealt face-down. What are the odds that it has a dot on the face?

This is the exact same situation - all the dot on the face-up side does in the original question is eliminate the double-blank card from the set of possible cards. Once that has been done, the face-up side no longer has any bearing on the problem, and we're left determining the odds of the face-down side having a dot.

 

[SmallTomato]

I change my answer to 2/3 now that I had time to sleep on it.

Yes, your first outcome is already decided with 100% probability (red dot facing up). However, for 50% you're assuming the odds of it being the Red/Red or the Red/Blank are 50/50.

They are not. Since you are MORE likely that the original card is the R/R.

 

[SmallTomato]

Think of it this way, what are the odds the dot facing up belongs to the Red/Red card compared to the Red/Blank card.

 

[SmallTomato]

So in the end, you have a 2/3 chance of it being a 100% chance of a red dot on the other side and a 1/3 chance of it being a 0% chance of a red dot on the other side. It's 2/3rds.

 

[Frothingslosh]

Well, FML. I totally overlooked the fact that pulling a random SIDE as well as a random card changes things.

God DAMN I hate being wrong like that.

Sorry about that, folks.

For anyone still thinking it's 50% (*grumble*), here's a simulator I threw together that, once I corrected it to allow freely random draws and to randomly choose which side is visible, proves SmallTomato and the rest are correct. Feel free to run through my code to verify.

Now if you'll excuse me, I have some crow to start cooking up.

 

[scott-atkinson]

Well, FML. I totally overlooked the fact that pulling a random SIDE as well as a random card changes things.

God DAMN I hate being wrong like that.

Sorry about that, folks.

For anyone still thinking it's 50% (*grumble*), here's a simulator I threw together that, once I corrected it to allow freely random draws and to randomly choose which side is visible, proves SmallTomato and the rest are correct. Feel free to run through my code to verify.

Now if you'll excuse me, I have some crow to start cooking up.

I cannot get your attachment to work...

So I cannot verify your solution, so until then I stand by mine, 50:50 chance of the drawn card having a Dot on both side...

 

[Frothingslosh]

I cannot get your attachment to work...

So I cannot verify your solution, so until then I stand by mine, 50:50 chance of the drawn card having a Dot on both side...

Well, that's annoying - what error message are you getting, if any? I created it with Access 2003, for what it's worth.

There's a button in it that will draw any number of times. I've attached a photo of the results of a million draws.

* * *

Here's some more straightforward logic:

Each card can be drawn 1/3 of the time.
Card 1 will always show a blank face.
Card 3 will always show a dot.
Card 2 will show blank half the time, and a dot half the time.

That means that for every 6 draws, on average, you'll get card 1 showing blank 2 times, card 3 showing a dot 2 times, card 2 showing blank 1 time, and card 2 showing a dot 1 time.

Now we know that the selected card shows a dot. We've already seen that cards are distributed in such a way that it is card 3 twice as often as it is card 2. That means that there's only a 33% chance that this is card 2, meaning there's a 67% chance that the flipside has a dot.

The mistake you and I were making was that we'd left out the fact that card 2 was only half as likely to be dealt dot-side-up in the first place.

Edit: And here's the code behind the mass-drawing calculations. The controls should be obvious, and the table the recordset is based on simply has three columns - a card ID, a yes/no for Face A, and a yes/no for Face B. There are only three entries, laid out as explained in the photo I attached.

Private Sub cmdExecute_Click()
 
On Error GoTo cmdExecute_Err
 
Dim rs As DAO.Recordset
Dim Iterations As Long
Dim x As Long
Dim LengthOfRS As Long
Dim VisibleFace As Long
Dim LogFieldName As String
Dim RandID As Long
Dim CardID As Long
Dim ControlToUpdate As Control
Dim FaceA As Boolean
Dim FaceB As Boolean
 
    Iterations = Me.txtDrawXCards.Value
    Set rs = CurrentDb.OpenRecordset("Select * FROM tblCards;", dbOpenSnapshot)
 
    DoCmd.Hourglass True
 
    SysCmd acSysCmdInitMeter, "Drawing cards...", Iterations
 
    With rs
        .MoveLast
        LengthOfRS = .RecordCount
        For x = 1 To Iterations
            SysCmd acSysCmdUpdateMeter, x
            RandID = Int(LengthOfRS * Rnd)
            .MoveFirst
            .Move RandID
            CardID = .Fields("CardID").Value
            FaceA = .Fields("FaceADot").Value
            FaceB = .Fields("FaceBDot").Value
 
            'Flip a coin to determine which face is visible.
            Select Case Int(2 * Rnd + 1)
                Case 1
                    'Assign A as visible and B as hidden.
                    VisibleFace = FaceA
                Case 2
                    'Assign B as visible and A as hidden.
                    VisibleFace = FaceB
            End Select
 
            'Update the log.
            Select Case VisibleFace
                Case -1     'Dot
                    LogFieldName = "txtDot" & CardID & "Count"
                Case 0      'Blank
                    LogFieldName = "txtBlank" & CardID & "Count"
            End Select
 
            Set ControlToUpdate = Me.Controls(LogFieldName)
            ControlToUpdate.Value = ControlToUpdate.Value + 1
            Set ControlToUpdate = Nothing
 
        Next x
    End With
 
    SysCmd acSysCmdRemoveMeter
 
    DoCmd.Hourglass False
 
cmdExecute_Exit:
    If Not ControlToUpdate Is Nothing Then Set ControlToUpdate = Nothing
    If Not rs Is Nothing Then Set rs = Nothing
    Exit Sub
 
cmdExecute_Err:
    DoCmd.Hourglass False
    MsgBox "Error " & Err.Number & vbCrLf & vbCrLf & "Description: " & Err.Description
    Resume cmdExecute_Exit
End Sub

 

[scott-atkinson]

Makes more sense now...