What are the chances ?

[ladyalex]

My family have been arguing about this all day so i have decided to ask it on here and see what AB makes of it.

You have 3 cards in a box. One is red on both sides, one is white on both sides and the third one is white on one side and red on the other side.
You draw one card out of the box showing only one side. The side showing is red.
What is the probability of the other side of that card also being red ?

[dachse]

50:50 or 1 in 2

[factor-fiction]

it can't be the one that's white on both sides. It could be either red-red or red-white. As you have seen one red face you know nothing more about the other side- I think each is equally likely so it's p=0.5.

[ladyalex]

Well...one member of the family agrees with you both.
Another answer that we came up with was 1 in 3.
The reasoning behind this was :

The card is showing red.
If the other side is also to be red, there is only one card of the original three which would fit that criterion. The chances of having picked that card are 1 in 3, so the chances of the other side being red are also 1 in 3.
We also had another answer, but I'll wait for further responses.

[dachse]

You are right that you had a 1 in 3 chance of picking the card originally but that is not what you asked. If you have apicked the card with a red face uppermost the chance of the other side being red can only be 1 in 2 as there are only 2 possibilities - the red/red card or the red/white card. The remaining 2 cards in the box are totally irrelavent at this point.

[gen2]

I would argue that the chances of the other side being red are 2 out of 3 or (66.7%)

Firstly let's distinguish between the three red faces by calling them R1, R2, R3.
R1 and R2 are back to back, R3 has a white back.

Now, when picking a card there is a 1/3 chance of picking any particular card and a 1/2 chance of picking a particular face on that card. Every face has an equal chance of being picked and that is 1/6

The 3 possible red picks are therefore R1, R2, R3 and each has an equal chance of being chosen.

So what is on the back of each of these?
R1 -- R2
R2 -- R1
R3 -- White

You can now see that there are 2 possibilities that the other side is red to a single chance of it being white.


The correct answer is therefore 2 out of 3 (or 66.7%).

[squarebear]

The card you have picked has a red side. This can only be the red/red or the red/white. So the chances are 50:50 or 1 in 2.

Any other combinations of sides are irrelevent as you have already picked a red side and so this a done deal.

[Gizmonster]

I'm with gen2 on this one:

A card is taken out and placed face up on the table showing a red face ... so you can ignore the white/white card .... so basically the puzzle now boils down to:

There are 2 cards, one red/red, the other red/white. I will now remove a card and place a red side facing up - what are the odds that the other side is also red ??

First impressions are it's 50/50 but look again ....

When the red side is revealed, we do not know whether it is the red side of the red/white, or if it's side 1 of the red/red, or side 2 of the red/red. Clearly, there are 3 different red sides that you can pick. If side 1 of the red/red is showing, or side 2 of the red/red is showing, then obviously the other side will be red (2 in 3 chance). If the red side of the red/white is showing, then the other side will be white (1 in 3 chance).

An even simpler explanation:

3 cards in a bag:

red/white: R/W
red/red: R1/R2
white/white: W1/W2

Each side is labelled accordingly and you now proceed to pull one card out, look at one face, note its face and replace it back in the bag and continue for say 6000 results.

All things being equal, the results will be (or close to):

R - 1000 observations
W - 1000
R1 - 1000
R2 - 1000
W1 - 1000
W2 - 1000

This is in a purely random case, which also includes results for white being the noted colour. We want to discard white results, so we now see results as

R - 1000
R1 - 1000
R2 - 1000

i.e. for every 3000 times a red side is seen, the red side of the red/white card will be seen 1000 times (1 in 3 times) and the other 2 sides of the red/red card will be seen 200 times (2 in 3 times).

[Gizmonster]

Dam typos lol ... that last line in my last post should read 2000 times (not 200).

[Marijn]

50:50

[gen2]

Go convince yourself notafish.

Set up your own practical experiment and count them.

Two Red-Red to every one Red-White

[factor-fiction]

sounds a bit like the montyhall problem where intuitive answer is wrong. Like with th montyHall problem I plumped for 50:50 initially but I was wrong . I think gen2 is right.

Suppose cards are;
R1, W1
R2, R3
W2, W3
If first face you look at is red then there's ann equal,chance you are looking at R1, R2 or R3. Each has p=1/3.
If you are looking at R1 you turn it over and see W1
If you are looking at R2 you turn it over and see R3
If you are looking at R3 you turn it over and see R2
So two times out of three you turn it over and see another red.

Look at it another way. If you started the exercise again and got a white I'd then say it was 2/3 likely the other side is white. This is logical - it means whatever you pick first, the chances are always 2/3 that you pick a double sided card. Which is obvious as the cards are RR, RW and WW-so the chances of a RW combination are only ever 1 in 3

[Marijn]

gen2, your response to me was a bit cheeky.
I agree with factor30's explanation.

[Marijn]

I mean factor30's original answer.

[factor-fiction]

Definitely 2/3.
As I stated above, when the cards are RR, WW and RW the chances of picking out a mixed card (i.e RW) is 1/3.
Whatever side you see first doesn't matter: the chances are always 2:1 in favour of having a matching other side. So if you see a red face the chances of the other side being red are 2/3. Similarly if you see a white face first there is a 2/3 chances that its other side is white.

[Marijn]

The first card has already been picked. We are already looking at a red side. What is the probability of the OTHER side being red?
If it is the red/red card, it will be red. If it is the red/white card, it will be white.
Therefore, I think it's a 50/50 chance.

[factor-fiction]

Hello Notafish.
Do you agree that before the game starts the chances of picking a double sided card are 2/3 and the chances of a mixed card are 1/3. Yes? Good.
Suppose I spoke colours in a foreign language you didn't know and I told you the first side was Ugggh. Then the chances are 2/3 that the other side would also be Uggh. So if a friend then revealed Ugggh meant red the chances are 2/3 that the other side would also be red.

[Marijn]

But we don't have to consider what happened BEFORE we chose the card that had red on at least one side.
We start from where we are already looking at a red side. The two sides of the card are not seperate entities. The two sides of the card are on one card.
Whether we are looking at side A or side B of the card with red on both sides doesn't make a difference.

[ladyalex]

Hello everyone, thanks for all your answers and all the time you have taken to think about and justify your replies.

My son, the cleverest one in our family by a country mile, agrees with gen2 that the chances are 2 in 3.
I was the one who thought it was 1 in 3, but have been convinced that 1 in 3 is only correct before the card has been picked, not after.
Thanks again for all your responses.

[factor-fiction]

I hold my hands up that I got it wrong with my early posting when I went straight for the intuitive answer of 0.5. But I read the arguments, thought it through and soon realised my mistake. The answer has to be 2/3 for the simple reason given- the chances of picking a card with two equal coloured sides is 2/3. So if I see a red face first then there is a 2/3 chance the other side is red. If I see a white face first then there is a 2/3 chance the other side is also white. There is no logical reason why seeing a red or white would make you switch your odds to 50-50.

Notafish should reconsider his argument. There are in effect 3 red faces and 3 white faces. If you see a red it rules out the double white option. It also rules out the other side being one of the 3 reds. That leaves 2 more red face options available but only one white option. So in this case the chances of second side being red is 2/3, and chances of it being white are 1/3.

Notafish is ignoring the fact that he could be looking at one of three things when he sees the red face:
(a) the opposite side of a white (b) the first side of a double red, or (c) the second side of a double red.
Notafish is ignoring the third option.

If you are not convinved Notafish, please can you explain what's wrong with my argument that the chances of picking a double sided card is 2/3 whilst the chance of a mixed card is 1/3.

[Marijn]

Hi factor30, I'm not a he.
You know you are right. Don't worry about what other people think.