Homework - Late Crammer

[AnthonyGerrard]

There is a set of 3 cards

Card 1 - Blank on both sides
Card 2 - Red Dot on both side
Card 3 - Red dot on one side and blank on the other.

The set is shuffled and randomly flipped (so the faces uppermost are randomised) totally randomly a card is dealt onto the table.

Seeing the card has a red dot facing up from the table - what are the chances the other side of the card (face down) has a red dot on it?

 

[Brianwarnock]

It depends what you mean by reversed, ie which way up they are, it is either evens or 100%

Brian

 

[pr2-eugin]

7 out of 8 Chances its has Red facing down !?!

Well after realizing that how stupid I sounded, checked again. There is a fair possibility everytime the card has a red dot, the bottom one could be a red too.

 

[AnthonyGerrard]

It depends what you mean by reversed, ie which way up they are, it is either evens or 100%

Brian

I changed it to randomly flipped - to be clearer. So which way up they are is randomised.

 

[pbaldy]

I'm going with 50%

 

[Brianwarnock]

I changed it to randomly flipped - to be clearer. So which way up they are is randomised.

Ok, so scrub my previous answer, i'll go for 2 out of 3

Brian

 

[scott-atkinson]

50% chance of there being a Red Dot on the other side of the card.

We can discount the Total Blank card, as by your statement the card in question face up has a Red Dot, as the blank card has no Dots it has to be removed from the Chance Calculation.

This then leaves the Dot/Blank and Dot/Dot cards, hence a 50% Chance that the card will have another Red Dot on the other side.

 

[Brianwarnock]

Ah, but we don't know which way up they are, there are 2 chances it may be the red-red and and one that it maybe the redblank.

Brian

 

[scott-atkinson]

Ah, but we don't know which way up they are, there are 2 chances it may be the red-red and and one that it maybe the redblank.

Brian

Brian,

The Control Statement in the original question was;

Seeing the card has a red dot facing up from the table - what are the chances the other side of the card (face down) has a red dot on it?


Therefore the card has already been dealt and the question is given the fact that a Red Dot is present so what are the chances of that card having a Red Dot on the otherside.... Answer = 50%

 

[scott-atkinson]

Ah, but we don't know which way up they are, there are 2 chances it may be the red-red and and one that it maybe the redblank.

Brian

It does not matter which way up the cards are, one card we know is Blank Blank, one is Red Blank, the other is Red Red..

If you perform the same test and a Blank is shown face up, you still have a 50% chance of the otherside being either a Blank or a Red.

Any card drawn will automatically exclude one of the other cards from the chance equation.

 

[Brianwarnock]

I agree that it should be evens, however I remember the thread as below and it's predecessor so presumably some mad professor somewhere has used a bunch of students to prove that what happened before is still relevant to the result. I thought I would show that I can still learn at my ripe old age.


http://www.access-programmers.co.uk/forums/showthread.php?t=254540

Brian

 

[AnthonyGerrard]

I agree that it should be evens, however I remember the thread as below and it's predecessor so presumably some mad professor somewhere has used a bunch of students to prove that what happened before is still relevant to the result. I thought I would show that I can still learn at my ripe old age.


http://www.access-programmers.co.uk/forums/showthread.php?t=254540

Brian

So which answer are you going for? The box example does not change the odds when the first box is opened?

 

[Frothingslosh]

There is a set of 3 cards

Card 1 - Blank on both sides
Card 2 - Red Dot on both side
Card 3 - Red dot on one side and blank on the other.

The set is shuffled and randomly flipped (so the faces uppermost are randomised) totally randomly a card is dealt onto the table.

Seeing the card has a red dot facing up from the table - what are the chances the other side of the card (face down) has a red dot on it?

This is basic statistics.

One card and only one card has been dealt. That means there are no previous cards to worry about, meaning we're dealing with the full set of three cards.

We know for a fact that the card that was dealt has a red dot facing upwards. Because of that, we know that it CANNOT be card number 1. Both card 2 and card 3 have a red dot on one side, but only one of them has a red dot on both. That means there is a 1 in 2 chance that the side facing down has a red dot, or 50%.

This is only a trick question insofar as that people who don't actually think about it a second will tend to give a knee-jerk reaction of 1:3. The previous draws Brianwarnock mentioned do not matter, because there ARE no previous draws. The quesiton here is not referring to the cards in the original set [Cards in the deck], but rather, the cards in the FINAL set [Cards from the deck with a red dot on one or more faces].

 

[Brianwarnock]

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

 

[Vassago]

The answer is 2 out of 3 chances, or 66.6%. As Brian said, you cannot remove known results from the equation. You knew that was a possibility at the beginning. Just because you know what it is now doesn't remove it as a possibility until you mix them up and take that one away without showing results.

 

[SmallTomato]

It is 50%.

 

[Frothingslosh]

It is not 2 in 3. The question, given in post 1, is:

Seeing the card has a red dot facing up from the table - what are the chances the other side of the card (face down) has a red dot on it?

So here is the data:

• You have three cards.
• Card 1 is blank on both sides.
• Card 2 is blank on one side and has a dot on the other.
• Card 3 has a dot on both sides.
• A single card was dealt.
• The card that was dealt has a dot on the side showing.

Now, look again at the question that was asked. It is NOT "What are the odds of picking the one card from the original set that has a dot on both side". The question was "You have one card from the set, and that card has a dot on the side you can see. What are the odds that the other side ALSO has a dot?".


So, logic time:

• Of the three cards in the initial set, only two have dots on at least one side.
• The dealt card MUST be one of those two cards.
• That means that since we have one card and two possibilities for which it is, the odds are 1 in 2 or 50%.

Saying that you still have a 2 in 3 chance is approximately as accurate as the following:

There are 4 cars in the lot: one each in brown, blue, grey, and red. You purchased the red car. Ergo, there is a 25% chance your car is red, because 25% of the cars in the lot were red.

See how that's in error? It's the same error as saying there's a 2/3 chance that your card that already shows a dot is one of the two cards that has dots on them.

 

[Frothingslosh]

Brian:

Sorry, I read

presumably some mad professor somewhere has used a bunch of students to prove that what happened before is still relevant to the result.

as 'previous draws'.

I guess I just get annoyed whenever I run into people who think the dice have memory, and it splashed onto you.

Sorry about that.

 

[SmallTomato]

There are 4 cars in the lot: one each in brown, blue, grey, and red. You purchased the red car. Ergo, there is a 25% chance your car is red, because 25% of the cars in the lot were red.

We all know this is the inverse of your Blood alcohol level^1-colorblindness factor/42. 42 being the car color constant.

 

[Frothingslosh]

We all know this is the inverse of your Blood alcohol level^1-colorblindness factor/42. 42 being the car color constant.

Dammit, I had the constant at 4.815162342! No wonder my car turned out to be chartreuse!