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# Alex's birthday cards

### #1

Posted 31 October 2007 - 05:28 PM

with Alex? he said. He's sitting in the corner fumbling in his

pockets and talking to himself.

I think he's had a few too many, Jamie replied. Plus, a friend

gave him a birthday present - a puzzle - and it has him stumped.

You remember the friend, the one with two children, one

being a boy, who kept asking the probability the other's a girl?

Anyway, he's been there a while. Let's see if we can help.

So I've got three cards, Alex explained...

* a black card - black on both sides,

* a white card - white on both sides, and

* a mixed card - black on one side and white on the other.

I pull one at random from my pocket, and place it on a table,

like this. There. As you can see, the side facing up is black.

Now what are the odds that the other side is black, also?

After thinking a moment - and downing a couple O'Doul's,

Ian smiled and whispered his answer to Jamie.

What would you have said?

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #2

Posted 31 October 2007 - 06:24 PM

### #3

Posted 31 October 2007 - 06:26 PM

### #4

Posted 31 October 2007 - 07:54 PM

Spoiler for solution

What if it was side 2 that was black?

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #5

Posted 31 October 2007 - 08:09 PM

EDIT:

What if it was side 2 that was black?

I think I see what you're saying, but I don't think it's plausible. I think you're considering each side as a separate entity, in which case for 1 side to be black, you have only 2 cards which leaves you with 4 total sides. If one is black, that leaves 3 other card sides , two of which are black... which means that there is now a 66.66667% chance to get a black side. HOWEVER, this only works if the sides are independent of each other, which they're not.

Since the sides are dependent upon the other sides, you take 2 sides out of the equation... the BLACK side showing on this card, and the BLACK side that would've been showing on the other card had it been the one chosen. That leaves you with only TWO possible sides and a 50% chance to get a black one... I think that makes sense.

### #6

Posted 01 November 2007 - 03:34 AM

That's it, exactly.As long as there are no indications of which side is which (such as the mixed card having a 1 on the black side and a 2 on the white side and the other cards numbering their sides as well) then it seems there is a 50% chance that the other side is black. If you have a card with one side being black laying on the table, then you either have 1 of 2 cards.... the mixed or the all black which would mean that you have a 50% chance.

EDIT:I think you're considering each side as a separate entity ...

What if it was side 2 that was black?

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #7

Posted 01 November 2007 - 04:25 PM

The idea of putting numbers on the card faces makes it very clear that there are more than 2 possible outcomes in this case. I will apply that concept (labelling the items) in the future when I have to explain it to someone.

### #8

Posted 01 November 2007 - 05:11 PM

I buy a lottery ticket.

There are two outcomes - it's a winning ticket or it's not.

One outcome is favorable.

My odds of winning the lottery are 1/2. Ooops ... reality check!

The times that this is really fun is when the faulty result is feasible.

With the black and white card problem, you can easily make the cards

and do the experiment say 30 times.

A valid intuitive solution notes that 1/3 of the cards have opposite color on the other side.

The probability of opposite color is 1/3 and same color [black or white doesn't matter] is 2/3.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #9

Posted 01 November 2007 - 05:39 PM

*or*white showing as an unknown state. Once you know the state, the probability is 50/50. Once you set black as the first condition, then the only possbile outcomes are:

1 black outcome,

1 white outcome.

If you set it up saying that a given color was showing, then you would have

3 states to set the color with 2 possiblities = your 2/3.

2 states for backside outcome

### #10

Posted 01 November 2007 - 09:24 PM

Let's eliminate the all white card for brevity's sake.

Draw the numbers 1 and 2 on the faces of the black card.

Draw the numbers 3 and 4 on the faces of the mixed card (3 = black & 4 = white)

Knowing that you will be looking at ONE black face, you have 3 possible scenarios.

side 1 will be showing

OR

side 2 will be showing

OR

side 3 will be showing

Those are the only three scenarios. So, in TWO of those, you'll be able to flip the card and have a black side, and one of them you'll have a white side. That equals 2/3 probability that you'll get a black side.

I think the part I always have a problem with is the part leading up to that. I don't think you can count the sides independently of each other. For instance. He pulled the card at random from his pocket and there was at least ONE black side. There were only TWO cards in his pocket that had at least ONE black side, so there's a 50% chance that it will be the

**black&white**card and a 50% chance that it will be the

**black&black**card. Since the ONLY SCENARIO in which there could be 2 black sides would be if he chose the

**black&black**card, there is a 50% chance that the other side will be black because there is a 50% chance that he chose that card.

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