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# Alex's birthday cards

## Question

Ian waved to Jamie and approached the bar. What's wrong

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?

## 21 answers to this question

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We know side 1 is black, so there are only 2 possible states:

Side down is black

Side down is white.

Odds are 50/50.

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50-50 chances?

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We know side 1 is black, so there are only 2 possible states:

Side down is black

Side down is white.

Odds are 50/50.

What if it was side 2 that was black?

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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:

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.

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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:

What if it was side 2 that was black?

I think you're considering each side as a separate entity ...
That's it, exactly.

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

Draw the numbers 3 and 4 on the faces of the white card.

Draw the numbers 5 and 6 on the faces of the mixed card - say 5 is on the black face, and 6 is on the white face.

The numbers 1, 2, 3, 4, 5, 6 have equal likelihood of being visible.

Nothing favors one of these numbers over any of the others.

If 1, 2 or 5, is visible, you see black.

If 1 or 2 is visible, the opposite side is black.

The odds are 2/3.

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Thanks for that spoiler, bonanova. I have run into these sort of puzzles before and I know how to determine the proper odds, but I have always had trouble explaining the reasoning to others who are convinced that there are only 2 possibilities so the odds must be 50/50.

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.

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Intuitive probability fails when it compares favorable to possible outcomes that are not equally likely.

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.

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That is only true if it's black 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

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I can't stand these types of riddles because I usually don't agree with the probabilities. I think I sort of understand this one whether or not I agree.

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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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.
You make a valid point. I hope that doesn't sound condescending ...

The answer depends on the premises of the set-up.

The way you describe the set-up I agree the answer is 1/2.

But look back at the story. I created a narrative that presupposed

nothing except that a card was placed on the table.

And then it was observed that the top face was black.

It wasn't presupposed to be black.

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?

And if it doesn't matter which color it would have been, a card needn't

even be drawn; only to ask the question: if I pull a card at random

from my pocket and place it on the table ... what are the odds

that the other side will match?

Where the answer is clearly 2/3.

Presupposing the color that's visible [as in your analysis] makes the answer 1/2.

Part of the fun I get in constructing these stories is providing enough clues for an unambiguous answer.

Sometimes I succeed , Sometimes I get it wrong.

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I guess that sort of makes sense. I dunno... like I said, I've never agreed with the probability behind these types of puzzles.

<more below, but if you're done discussing it, skip it>

If he pulls a card out, at random, and slaps it down on the table and a black side just happens to be showing (without presupposing the color of the visible side or any side) there are only 2 cards he could've pulled out that would've given him the ability to lay down a card with a black side showing. He had a 2/3 probability of pulling one of THOSE cards out (given that there were 3 cards)... But, It doesn't matter if the color was presupposed, or which side it's on (if it's the black&black card). There are only 2 cards with a black side on them. AND, of those two cards, only 1 is black on the other side.... whatever. I guess I'll just agree to disagree.

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There are only 2 cards with a black side on them. AND, of those two cards, only 1 is black on the other side.... whatever. I guess I'll just agree to disagree.
If you reason that because the only cards that can show a B face [the BB and BW cards]

are drawn with equal likelihood the answer must be 50%, then try the experiment.

Do precisely what Alex did, ignoring the WW card for simplicity.

Make two cards: BB and BW. Place one from your pocket onto a table.

[1] If it shows B, add 1 to your possible outcomes total.

[2] If the reverse side is B, add 1 to your favorable outcomes total.

Repeat until you have 30 possible outcomes.

Now, are the favorables closer to 15? or to 20?

If it's 17 or 18, do another 30. And maybe another 30.

1/2 or 2/3 will eventually come into sharp focus.

And when it does, keep in mind that you drew the BB card 1/2 of the time!

If I were a betting man, and I'm not, I'd bet an entire donut on 2/3. Why?

Because of the equal likelihood requirement.

You see a B face 100% of the time when the BB card is drawn

but only 50% of the time the BW card is drawn.

2 cards x 2 faces = 4 equally likely events.

3 events show a B face.

2 events are favorable.

Basically, you count the BB card twice - because it can show a B face

two ways; and count the BW card once - it can show a B face only one way.

Does that make sense?

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To whoever says 1/2:

the three cards:

WW

BB

WB

we can immediately eliminate the WW to get 2 differnet outcomes. yes, two outcomes but they are NOT equally likely:

BB

WB

the black came from one of them. Since there are 3 blacks, 2 on the BB, 1 on the WB, it is twice as likely that the black came from the BB as opposed to the WB

the odds are 2/3

thats the easiest way I can think of explaining it

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For the fun of it I marked two quarters with 3 black sides and looked at 50 cases.

10 blacks -> 7 black on other side

10 blacks -> 10 black on other side

10 blacks -> 6 black on other side

10 blacks -> 6 black on other side

10 blacks -> 5 black on other side

========================

50 blacks -> 34 black on other side => 68%

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I think I get it now...

We both agree that you have a 50% chance to pull each card out of your pocket...

So, the possibilities and probabilities of each, are this:

BB - Black 1 Showing - 25% chance

BB - Black 2 Showing - 25% chance

BW - Black Showing - 25% chance

BW - White Showing - 25% chance

Now, that makes a total of 100%.

But, since we've already decided to eliminate the BW with white showing, we've eliminated HALF of the times that the BW card can be pulled out of the deck and used. So now the probabilites are affected thusly:

BB - Black 1 Showing - 25% chance (33% chance)

BB - Black 2 Showing - 25% chance (33% chance)

BW - Black Showing - 25% chance (33% chance)

-----------BW - White Showing - 25% chance------------ Eliminated

Since we've eliminated half of the BW's 50%, we have to adjust the percentages that are left to equal 100%. That changes everything to 33.33%.

Whew! Now I can sleep at night.

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Whew! Now I can sleep at night.

Great! What time zone you in?

I'm GMT-5, but it doesn't seem to affect when I sleep.

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Eastern here also (that's GMT -5 isn't it?).

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2 to 1 or 2:1

Odds for it being black =

(chances for it being black)

(chances against it being black)

[Probabiltiy it is black = 2/3

(chances for it being black)

(total chances) ]

Sorry -- didn't see this one when I posted my marble puzzle yesterday. :-)

Edited by xamdam

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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?

Just to make sure it's the other side of that same card right

I think the the two answers are correct but if use the 50% method you are most have a better chance of gusting the correct card look at it this way..

3 cards

W W

B B

W B

with the black side is facing up you know for sure that it can't be W W so you now know that is it has to be

B B

B W

this gives a 50% chance of being BB/BW dus u have a better chance of getting the correct card ..............but with math

it does give 1/3 because in math the commonsense of letting go the W::W is not logical in math....hard to explain

for example ,math problem (1+2+0= ?) commonsense will tell u that u can say 1+2 and get the correct ans but in math u must include "0"

for the record i only joined today........lol and now realized how old this post was

Edited by Zuh

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Like the Monty Hall problem and the puzzle bonanova alluded to in the OP, I enjoy when people say the logic does not follow the probablities or the math doesn't work, when that is precisely what happens. The application of probablilities is usually where the fault lies, but the numbers work.

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BY the way, meant to say good puzzle bonanova

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