## Black or White?

Posted by Chris on December 1, 2010 – 5:55 pm

I just found this one. I’m posting it as an extra special treat for Eketahuna who loves this sort of thing

A box contains three cards. One card is black on both sides, one card is white on both sides, and one card is black on one side and white on the other. One card is selected from the box at random, and one side is inspected. If this side is black, what is the probability that the other side of the card is black as well?

December 1st, 2010 at 6:00 pm

Thanks Chris I’d say 1 in 1,000,000 … that I’d get the answer right if I attempted it, so I’ll keep my answer to myself and let someone else answer it

December 1st, 2010 at 6:26 pm

Hi Eketahuna. It’s a disguised version of the double-headed coin problem.

December 1st, 2010 at 6:43 pm

Hi Chris, yep, I could work it out according to the ‘accepted’ logic, but steadfastly refuse to

Not to worry, beer-O’clock is rapidly approaching (though that’s relativity, not probability)

December 1st, 2010 at 7:36 pm

Hi Eketahuna. I found the (posted) problem during a futile attempt to find a web page with a simulator on it (for the other problem).

December 1st, 2010 at 7:42 pm

It’s the same probability again, I think? 1 in 3 chances?

December 1st, 2010 at 7:43 pm

I mean 2 in 3 chances this time, oops.

December 1st, 2010 at 8:03 pm

Hi Shelley. I see your confidence wavered for a moment. But have no fear, you got it right (2/3).

December 1st, 2010 at 10:45 pm

50/50 lol. Theres only two choices it could be: black or white.

December 1st, 2010 at 10:50 pm

Why would it be (2/3) if a black sided card was picked? there were only 2 cards with black sides so the white card is irrelevent to the end question. So there’s 2 cards in reality and only 2 color choices for the back to be so the chance would be (1/2) or 50/50. Does my logic make sense? :S Cause I cant see why it would be (2/3)…

December 1st, 2010 at 11:27 pm

hey i would believe it is 1 in 2 right?

December 2nd, 2010 at 12:18 am

I believe there would be a 50% chance the other side would be black.

reason:

Of the cards left you’ve eleminated the white/white card. So…that would leave the black/black and the black/white card. Since a black face has already been exposed the uninspected sides remaining would be black/black/white or 1/2. Is that correct?

December 2nd, 2010 at 12:21 am

sorry 66% chance

December 2nd, 2010 at 12:35 am

1/3 only one out of the three cards is a double black duh. lol

December 2nd, 2010 at 3:51 am

Wow I hadn’t expected so many responses. The logic is the same as used for the recent problem – “Heads or Tails? (yet again)”, http://trickofmind.com/?p=498

So a few of you need to reconsider their answers; that especially means Jaela Hicks.

December 2nd, 2010 at 7:51 am

I just want to ask even if my answer was wrong (i still believe differently), doesn’t my logic make sense, Chris?

December 2nd, 2010 at 8:19 am

here is the reasoning the answer is 2/3:

say the black/black card is labeled B/B’, the white/white W/W’, and we’ll use small letters for white/black w/b.

Here are the options for pulling out a card, with the first letter representing the face you can see:

B/B’

B’/B

b/w

w/b

W/W’

W’/W

There is a 50% chance that you will even pick a black card, but that is irrelevant, since we know it will be black

now out of the 3 choices where you see a black face, 2 of them have black on the back, and only one has white.

2/3 chance of the other side being black.

December 2nd, 2010 at 8:24 am

Thanks DP.

Bri, also take a look at http://trickofmind.com/?p=498

The W/W card is a red herring

Although there are two possibilities, they are not equally likely.

December 2nd, 2010 at 5:46 pm

i think 1 and 6 chances im not sure though after i read it thats what my head told me

December 2nd, 2010 at 8:32 pm

Interesting But I’m 99% positive its 2/3

December 3rd, 2010 at 1:31 am

It is one in 2 because the one that is white on both sides can be discarded imediatlly because we know one side is black that only leaves two cards the one that is black on both sides and the one that is black on one side and white on the other QED there is a 1 in 2 chance it is black on the other side

December 3rd, 2010 at 4:57 am

Hi Brian Burke. You’ve used decent logic to reject the W/W card, but you fail to continue to use it to realise that you are twice as likely to have picked the B/B card as the B/W card. DP (post 16) has explained it very well.

December 3rd, 2010 at 8:13 am

I’m quite confident that it’s 1/2.

December 3rd, 2010 at 8:19 am

Hi Jesse. I know that you are wrong. What is your reason for being confident?

If we ignore the W/W card (which the posted question makes possible), then the probability of drawing the B/B card is 1/2 as is the probability of drawing the W/B card. If you draw the B/B card then it is certain that the other side is B. If you draw the B/W card and you see B, then it is certain that the other side is W or if you see W then it is certain that the other side is B. But the question says that you see a B. As half of the times you draw the B/W card (it shows W) and you have to skip that go. So half of the times you draw the B/W card, it doesn’t count. So you are twice as likely to have drawn the B/B card as the B/W card, hnce the probability of drawing the B/B card is 2/3 and the B/W card is 1/3.

December 3rd, 2010 at 11:38 am

For those of you who still believe it’s 1/2 try the following. Replace the B/B card with a R/B card (Red/Black). I pretty sure that you’ll also say 1/2 for that case (and you’d be right this time). Will you assert that swapping the B/B for a R/B has no effect even though we are now having to throw away half of the R/B card selections – i.e. even though the problem has changed significantly?

NB for both problems the probability of flipping to see a W + the probability of seeing B (or R in new problem) must add up to 1.

December 3rd, 2010 at 2:06 pm

1/3 is wrong. the fact that there is a double white card in the hat is irrelevant, because you already have a black sided card. it is not asking the probability of drawing the B/B card out of the hat, it isasking the probability that the card you are holding is the double black. you are already holding a card with a black side. that means you are holding one of two cards, the B/B or the B/W, two cards, fifty percent. it is either goingto be white on the other side or black on the other side 1/2 chance of either

heres my number if you still want to be an idiot and argue it farther 2536775450

December 3rd, 2010 at 2:29 pm

Hi Anthony. 2/3 is the answer. Using your logic I can prove that the probability of throwing a 6 on a die is 1/2 (after all, either you throw a 6 or you don’t – 2 possibilities – 50% QED).

Please show me where the error is in the logic that DP and I (and others) have used. Then please show me how you CALCULATED 50% – but don’t simply say that it’s black or it’s white, therefore it’s 50% as they are NOT equally likely as you incorrectly assert.

Now why don’t you actually try following our logical arguments and calculations and pointing out exactly where we went wrong – personally I doubt that you can do that.

December 3rd, 2010 at 3:02 pm

there are two reasons you could be errored, one is that you are assuming all cards are in the hat and you are drawing one of the tree cards, the second is that, like in DP’s example, you are assuming all sides of the cards as different options, in DP’s example he rules out te sides starting with W and is left with the THREE OPTIONS B’/B B/B’ or w/b if yu were to stop there it would be a 2/3 chance because two of those optins have a B on the other side.

where you error in this is by the fact that Two of those options are of the same card. and the question is asking what the probability of the OTHER side of the card being black is. you are iether holdingthe B/B card or the B/W card their is no B’/B card because it is the same card. since only TWO cards hav a blck side andonly ONE has two black sides it is a 1 in 2 chance that the card you are holding isblack on the back

December 3rd, 2010 at 3:06 pm

all three cards have an equal chance of being picked. you do not have more of a chance of picking the B/B card than the B/W card because they are both only one card. their is also no more chance that you will be looking at the B/B card than the B/W card because again, they are both only one card.

December 3rd, 2010 at 3:36 pm

Hi Anthony. Thanks for coming back. I’ve acknowledged (and was fully aware at the time of posting the problem) that the W/W card can be ignored completely – it was only put there to confuse people (but almost no-one was taken in by it). From now I will now act as if the W/W doesn’t exist.

Your first paragraph (post 27) is fine.

The probability of drawing the B/B card and seeing black is 50%. The probability of drawing the B/W card and seeing black is 25%. If it shows white, we simply return it to the box, re-randomize, and have another go. So half the time you draw the B/W card you discard the trial, but you never discard the B/B card. So for the trials that count, you are twice as likely to be holding the B/B card as the B/W card. And that’s all there is to it.

December 3rd, 2010 at 3:42 pm

if you are looking at it as that, you are counting sides and saying there are three total black sides and one white side, looking at a black sides, two of the three remaning sides are black, however, that is like saying all the sides themselves are seperate sides. since you have two cards and are looking at a black side, only one card has a black side on the second side, tha other one has a white side on the back. so you are looking at a 50% chanc the side is black, and a 50% chance the side is white

December 3rd, 2010 at 3:58 pm

Hi Anthony. In my last version I carefully avoided looking at the sides as independent items. In essence I said that it is equally likely that, initially, it’s 50-50 that you draw the B/B or the B/W card. I don’t think anyone is arguing with that. But either card could be either way up. In the case of the B/B card you don’t know (or care) which way up it is. But for the B/W card, you do care which way up it is. If it’s B up then it’s good, if it’s W up then start over again.

Once you see that is what the question implies, then it’s not much of a leap to see that DP’s (and Shelley’s – see related coin flip problem) logic is in fact very clean.

December 3rd, 2010 at 4:05 pm

Hi again Anthony. I can see why you aren’t comfortable with the sides being independent. What’s really being said is that each side has an equal probability of being drawn. Obviously, that logic can only be applied because we are only drawing one card. It wouldn’t be so sweet if we were drawing two cards.

Addendum 10 Feb 2011. I am saying that it is equally likely that you could see any of the four faces (of the BB’ and bw card). So that’s a 3/4 probability of seeing black and a 1/4 probability of seing white).

December 3rd, 2010 at 7:50 pm

because both sides of the B/B can’t be an option at the same time. if the three optons are B’/B B/B’ and W/B and you are holding a known black side. the other side is either going to be the other black side or it is going to be the white side. there is no more of a chance that the black side you are looking at is of the B/B or the W/B

and reversely if you were looking at a white side, the question would ask for the probability that the other side will be white. you are implyng that you have less of chance ofdrawing the B/W card as the other two.

December 3rd, 2010 at 8:17 pm

Hi Anthony. I’ve been crystal clear that (before examing the face) that you are equally likely to have drawn the BB or the BW card. We do have a possiblility of dismissing the BW card, but we never dismiss the BB’ card. I have nowhere suggested that both sides of the BB card can be an option at the same time.

Having agreed that the WW card is a red herring, re-introducing it will only make the explanations longer, but add nothing new to the basic problem. But if the question had said that you are looking at a white side then the probability that the other side was white is 2/3 (and the BB card would be the red herring).

As you wish to keep using the BB’ notation I will go back to it. Actually put a little mark on the card so that the B’ side is visibly distinct from the B side. I’m dropping that code that the first letter is the displayed face). Now draw a card. The probability of drawing the BB’ card is 50%. The probability of drawing the BW card is 50%. If you draw the BB’ card then it is equally likely to be showing B or B’. Altogether the probability of it showing B is 25% and the probability of it showing B’ is 25%. If you draw the BW card then it is equally likely to be showing B or W. Altogether the probability of it showing B is 25% and the probability of it showing W is 25%. If it is showing W then he trial ends and we have another go.

Imagine we had done the trial 100 times, that means that for 50 trials we had selected (and kept) the BB’ card (25 times showing B and 25 times showing B’) and for only 25 trials we selected (and kept) the BW card (showing B) and for 25 trials we have discarded the BW card as it was showing W.

So that’s 50 times BB’ and 25 times BW => 2/3 of the admissable selecions was of the BB’ card.

That’s long enough for one response. I’ll write another after I’ve had a fag.

December 3rd, 2010 at 8:39 pm

Let the two cards of interest be BW and RG (red green).

Draw one and take a look. If it’s not white, what is the probability that it’s the RG card?

On the original draw it’s 50-50 that it was the RG or the BW card. If it’s the RG card it’s 50-50 that it shows R or G. If it’s the BW card it’s 50-50 that it shows W or B. If it’s W we abort and try again. So in 100 draws, we’ll get 25 R, 25 G and 25 B (and 25 W, but we discard those). So the probabiity of it being the RG card is 50/75 = 2/3. Now change R and G to B and B (or B’ if you prefer) and voila!

December 3rd, 2010 at 8:53 pm

Hi Anthony. I also suggest you read my post (24) which attempts to show an absurdity in claiming 1/2 is the answer.

It’s 4 am here in the UK, so I’m knocking off.

December 4th, 2010 at 4:19 am

Now that your up Chris…..? I am going to side with Anthony on this one, purely because of my new job role as Devil’s Advocate. However, I also feel that I would be wrong, and therefore, by extension the Devil would be wrong. There’s going to be hell to pay for this!

Sadly I actually feel that this one is obvious when you stop and think, and your explanations have been pretty clear – sorry Anthony!

The answer is indeed 28.4/42.6…..

December 4th, 2010 at 4:20 am

Oh, and thanks for posting the questions!

December 4th, 2010 at 7:12 am

by looking at a black side, you are already ruling out one of the black ide options, we already know you are lookin at a black side, therefore the other side of the card is either white or it is black.

just like in yor post 24, the color of the side you are looking at is irrelevent because it is theknown front side, the card you are holding is B/W or B/B. it cannot be either B/B or B/B because whicever black side you may be looking at doesn’t matter.

your error is in not canceling out the side you are looking at, it is the same probability in this instance as if you were laying both cards on a table with a black side showing, picking one of the cards at random t flip, you have the same probability of flipping the white side as you do the black side.

December 4th, 2010 at 8:10 am

2/9

December 4th, 2010 at 8:22 am

Hi Anthony. Re your post 39, para 1. True, the other side of the card is black or white (and the probability that it is black is 2/3 and the probability that is is white is 1/3 – but I can’t calculate that without a deeper analysis). (Remember the die).

Several means of seeing that the two black sides of the BB card are distinct (e.g. B’ as a sort of token white) have been tried. You seem to be suggesting that if both sides of the card are indistinguishable, then you are half as likely to have selected after you have observed it (but not before observing it – that’s magic, not reality).

Para 2. That both sides are the same colour (black) means that we never reject that card after drawing it. If they were a different colour, then we discard half the draws (when white shows) – and that’s why twice as many BB cards as BW cards are selected. This isn’t a sophisticated idea, it’s extremely straightforward.

Para 3. I don’t know what you mean by cancelling out the side I’m looking at. What do you mean?

None of your arguments are amenable to doing calculations with. You need to actually prove that the probabilities of the other side of the card being B or W are 1/2 – simply asserting it is no good (especially as it isn’t true).

December 4th, 2010 at 8:25 am

Hi Your name. If you are serious, would you explain how you obtained that result?

December 4th, 2010 at 8:46 am

Hi Athony. If you wish to pursue this, I ask you do do it by using my post 34 (and either para 3 or 4 or both) as a reference. Please tell me where you think I made a mistake.

But please don’t repeat that the other side is B or W and assert that makes the probability 1/2, because that is a incorrect assertion.

December 4th, 2010 at 8:49 am

but your missing where if you are looking at the B’ side of a B/B’ card, both B’ and B are not options of that card anymore, like i was saying earlier if you were looking at a white side, thequestion would be the same but with the double white card. it can be easily seen if you wipe the side you are looking at and it no longer exist (getting rid of a black face of both cards) now you are left with Two sides possible. one black, one white.

December 4th, 2010 at 8:49 am

and i think your name was not serious, their is absolutely no logic behind that one. lol

December 4th, 2010 at 9:03 am

Hi Anthony. I agree that 2/9 was probably a joke.

I don’t understand, at all, what you mean in your first sentence. And I certainly don’t understand the rest of it.

Of the 100 trials, on average, 25 times I see B (from B/B’), 25 times I see B’ (from B/B’), 25 times I see B (from B/W) and 25 times I see W (from B/W). As the problem states that I am actually looking at a black face, only 75 of the 100 draws are consistent with that. 50 of those are due to the B/B’ card and 25 due to the B/W card.

As for the rest, that’s just irrelevant, because I reject all cases where I see white. If I include the W/W card, it will simply mean more trials are to be rejected (in fact half of the trials would be fails, whereas by ignoring the W/W card only 25% are fails).

December 4th, 2010 at 10:05 am

Hi Anthony. This is probably going to be my last significant attempt at explaining the answer.

Let the three cards be WW’, BB’ and bw. The letters uniquely identify each face.

1. The probability of drawing a given card is 1/3.

2. It is equally likely that either face could be up.

3. Therefore, the probability of a given face being selected is 1/3 * 1/2 = 1/6.

4. In particular P(B) = P(B’) = P(b) = 1/6.

5. To comply wih the posted problem we must have seen B, B’ or b.

6a. The probability of picking B or B’ given that we have picked B, B’ or b is

(P(B) + P(B’))/(P(B) + P(B’) + P(b)) = (1/6 + 1/6)/(1/6 + 1/6 + 1/6) = 2/3.

OR

6b. The probability of picking b given that we have picked B, B’ or b is

P(b)/(P(B) + P(B’) + P(b)) = (1/6)/(1/6 + 1/6 + 1/6) = 1/3.

Before saying e.g. that picking B’ excludes having picked B and calling that a flaw, realise that we are only picking one card and only (initially) looking at it’s up face in any given trial. We only look at the other side if we see a black face up.

Aside. The probability of picking white = probability of picking black = 1/2 as you’d expect by symmetry. Also 1/2 = 1/6 + 1/6 + 1/6, so no obvious contradictions.

After each trial we return the card and re-randomize. There is no memory effect between trials.

Now, if you think I have made a mistake, please tell me which statement(s) is/are wrong and what is wrong with it. If appropriate tell me what value(s) is/are incorrect (and what they should be and why).

December 4th, 2010 at 11:34 am

Hi Chris. I see that you are trying hard to teach Anthony something but it my opinion this is pointless. He either has no idea of probabilities plus is extremely stubborn and self-assured or he is simply making fun of you and is trying to irritate you. In both the cases the dialog is just a waste of time

December 4th, 2010 at 12:28 pm

Hi slavy. I’m not that bothered. Although my patience is exhaustible, I quite enjoy coming up with different ways of explaining problems.

But I agree that I don’t think Anthony (or most people) actually understand what probability is.

December 4th, 2010 at 1:35 pm

In fact I found a web page that described how a student simulated this on a PC, and said it’s true, but only if you try it thousands of times – go figure.

December 5th, 2010 at 3:31 am

It is a 50/50 chance.

The first side is black so we have to eliminate the white-white card.

So the card is either the white-black card or the black-black card.

That’s two cards making it a 50/50 chance!

December 5th, 2010 at 4:26 am

Hi Geosam, try this – it’s your logic applied to a die:

So the die shows a 6 or not. That’s two possibilities making it a 50/50 chance [of throwing a 6]!

Can you see a problem with your argument?

Or perhaps you’re pulling my leg!

December 7th, 2010 at 4:49 pm

Hey Chris, what country do you live in.

Just curious about time zones affect on computer blogs.

7th grade problem.

December 7th, 2010 at 5:14 pm

Hi Spencer. I live in Wembley (North London), England.

December 7th, 2010 at 8:31 pm

Thanks Chris! I’m going to Wembley mid-July on vacation among other places. I know about the chaos theroy because I am an undeniably a NERD (And I am proud of it)!!!!

December 7th, 2010 at 8:45 pm

Nerds rule.

December 7th, 2010 at 8:54 pm

Hi Chris,

Can’t resist this one Re post 49 … understanding probability

I had a tree fall down in a storm and land on a metal gate, twisting the gate so that when still on the hinges at 90 deg. to the ground, the opposite end of the gate was 45 deg. to the ground.

What was the probability that I could fix the gate ?

December 7th, 2010 at 8:57 pm

I’d need more info than you’ve given to answer that question.

December 7th, 2010 at 9:02 pm

What tools do you have/ experiance do you have?

There is no finite possibility. There are infinite possibilities and even more branching off of those!

December 7th, 2010 at 9:12 pm

A brain, a computer, a copy of Mathematica, but I’ve only got that so I can make and idiot of myself trying to crack the Riemann Conjecture. I’ve also got a degree in Electronics.

Experience – that’s a long story as I’m 58. Definitely bed time for me.

December 7th, 2010 at 10:13 pm

1/3

December 7th, 2010 at 10:45 pm

I’m interested in electronics. I have modified my alarm clock to trigger several other applications such as a toaster. I’ve also studied anatomy and collage physics. GO NERDS!

December 8th, 2010 at 12:50 pm

It’s 3.66m in length, outside frame made of 25mm diameter metal tube welded together. But that’s irrelevant, as it’s 100% certainty that I’d fix the gate. It’s the amount of time that’s taken that is the only uncertainty.

Experience ? Never had to do it before, but I like to think I’m like Chris and have a brain (computer didn’t help, I tried throwing it at the gate when propped against a book (not Mathematica) but it just bounced off and got excessively scuffed).

My point ? (I’m sure I had one yesterday) Sometimes probability can be overthought and the simple approach is the correct answer.

Oh well, that’s me done – sorry for interrupting

February 9th, 2011 at 10:17 pm

Hi Lane

~~(Anthony?)~~The WW card is a red-herring. If there were 20 WW cards in the hat, the answer would still be 2/3. No matter how many cards there are in the hat, you are equally likely (a very important observation) to pick any of them. The probability of picking a WW card would be 20/22. The probability of picking the BB’ card is 1/22 and the probability of picking the bw card is 1/22 also. So the probability of picking the BB’ or the bw card is 2/22 = 1/11. Whichever card you pick, it is equally likely that you will see either of its faces. If you pick the BB’ card it is certain that you will see a black face (but you don’t know if it is the B or the B’ face). If you pick the bw card, the probability of seeing the black face is 1/2.

If you do not see a black face, you simply return the card into the hat, shuffle them about, and try again. The only time you look at the other side is when you see the black face first. You are therefore twice as likely to have picked and retained the black card as you are to have picked and retained the bw card.

On average, after 44 selections, you will have picked a WW card 40 times, the BB’ card 2 times and the bw card 2 times, and 41 times you will have seen a white face and 3 times you will have seen a black face. As you only inspect the other side when you see a black face, of the 3 picks that showed the black face, on inspecting the other side you will see a black face 2 times, and a white face 1 time. That makes the relative probability of having picked the BB’ card be 2/3 and of having picked the bw card be 1/3.

A slight variation is to remove the WW card(s). The probability of picking the BB’ card is 1/2, as is the probability of picking the bw card. The probability of picking the BB’ card and seeing a black face is 1/2 * 1 = 1/2. The probability of picking the bw card and seeing a black face is 1/2 * 1/2 = 1/4. So the probability of drawing a card and seeing a back face is 1/2 + 1/4 = 3/4, and he probability of drawing a card and seeing a white face is 1/4. Obviously the probability of drawing a card and seeing a black or a white face is 3/4 + 1/4 = 1. So the probability of drawing a card, seeing a black face, and the other side being black is (1/2)/(1/2 + 1/4) = 2/3. The probability of drawing a card, seeing a black face and the oher side being white is (1/4) / (1/2 + 1/4) = 1/3.

The key ingredient is to realise that the probability of seeing a black face on the BB’ card is 1, and the probability of seeing a black face on the bw card is 1/2. So you can also calculate the probability as 1/(1 + 1/2) = 2/3.

Addendum: I should have done the calculation directly. In the case of 20 WW cards, the probability of holding the BB’ card, given that you have seen a black face is: (1/22)/( 1/22 + 1/44 ) = 2/3

February 10th, 2011 at 6:03 am

Hi, Chris,

No, I’m not Anthony and have never corresponded with Anthony. I just happen to agree with him. After seeing the messages between the two of you, I don’t think either one of us will ever convince the other, so I didn’t even try.

February 10th, 2011 at 7:03 am

Hi Lane. I was only guessing as Anthony was the only one really pushing for the other solution.

It’s a shame that you aren’t persuaded by logic, but are persuaded by fallacious and seriously flawed (non)reasoning. You’re right, Anthony won’t convince me of his viewpoint because he hasn’t presented a logical argument. He only presents fallacious arguments (i.e. incorrectly assuming that all possibilities are equally likely, over and over again), arguments that are so poorly phrased that they are unintelligible (e.g. Anthony’s posts 39 and 44) and arguments that demonstrate a complete failure to understand other people’s reasoning.

See Anthony’s post 27. He says I’m in error by assuming that all three cards are in the hat and that I’m in error in believing that it is possible that I could pick any one of the cards. Does that seem to be a reasonable demonstration that I’ve used faulty logic? How many cards do you think that he thinks are in the hat. Why do you think that he thinks that I can’t pick any one of the three cards?

He then seems to find a problem with DP’s post, for saying that there’s only three options: B’/B, B/B’ or w/b – he doesn’t seem to have appreciated that the fourth option (b/w in order to be consistent with his post) has been rejected AFTER picking it and returning it to the hat because it was showing white. The fact is that DP was saying (first letter is the visible face, the second the hidden face) that initially you could have picked BB’, B’B, bw or wb and that you discard the wb because it showed white, and the posted question demands that you see black. i.e. Anthony has spectacularly failed to understand DP’s argument. I think Anthony believes that DP thinks there are three cards (BB’,B’B,wb) in the hat, whereas DP means that you could have picked the BB’ card showing B or showing B’ (i.e. either way up), but can only pick (and keep) the wb card if it is showng the b side. DP doesn’t mean (as I’m sure that Anthony believes) that you could pick BB’, and then B’B (without returning the BB’ card) – and then (under that understanding, quite reasonably) thinks that DP has goofed. DP hasn’t goofed, Anthony hasn’t understood what DP actually meant.

Anthony’s post 28: I can no longer tell if by three cards he means WW,BB’,wb or if he means BB’, B’B,wb. But I think he means that you are equally likely to pick the B/B card as the B/W card (that at least is correct). He hasn’t realised that if you pick the B/B card that you will see black, but if you pick the B/W card then it is only 50% likely that you will see black. It is a faulty argument, due to his complete lack of understanding of the problem.

I’ve just read one (post 39) where he changes the problem to having the two cards on the table, both showing black and correctly observing that if you turn one over that it is equally likely to show black or white – but that has nothing to do with the question. He is in effect saying that if you draw the bw card, you will see the black face. He should have placed two bb cards and one bw card (all black side up). Sadly, I didn’t point that out to him. I probably didn’t do that because the reasoning required is on a par (or harder) than the original problem.

Notice that I am able to pinpoint the error(s) in the fallacious arguments that people are presenting but not vice-versa; even though I have directly challenged them to do that (several times). That tells me a lot, and it should tell you a lot also. i.e. they are quite unable to do so (and that’s because I have not made a mistake). They just ignore the logic and obdurately stick to their guns. The few attempts at pointing out my errors have been crazy arguments; they hinge on putting words that I haven’t uttered into my mouth.

To accuse me of “insisting” on my answer, is the pot calling the shiny kettle black.

But despite that, it would be boring if we never made mistakes. And life is to short for me to hold a grudge over something as silly as this problem. But before you consider implying that I’m an obstinate unreasoning so and so again, just consider that you might be wrong (as is the case in this case). I endeavour to consider that possibility before disagreeing with someone elses argument.

In the case of this problem, the answer is 2/3 – and is easily verified by experiment and by the several arguments that I and others have presented on this page.

February 10th, 2011 at 7:59 am

Hi Lane and anyone else who still doesn’t get it; try the following thought experiment.

Sticking with all three cards, imagine trying the process 300 times. In what follows, I’ll be working with the average of that would occur if we repeated the 300 trials infinitely many times.

We would expect to pick the WW card 100 times, the BB’ card 100 times and the bw card 100 times. Of the 100 times we pick the WW card we will see a white face 100 times. Of the 100 times that we pick the BB’ card we will see a black face 100 times. Of the 100 times we pick the bw card, we will see a black face 50 times and a white face 50 times. So of the 300 picks, we will see a white face 150 times and a black face 150 times. Of the 150 times that we see a black face, 100 of them was because we had picked the BB’ card and 50 was because we had picked the bw card. So the relative frequency (i.e. the probability) of picking the BB’ card, given that we saw a black face is the number of times we saw a black face because we had picked the BB’ card divided by the number of times that we saw a black face because we had picked the BB’ card or the bw card (and saw a black face) = 100/(100 + 50) = 2/3.

That’s a very simple to follow argument; it uses the relative frequency definition of probability. I challenge anyone to find a fault with it.

Despite it’s simplicity, I do expect that people will still fail to accept the result. That’s because they do not understand what probability actually means. I can sympathise with that. I promise that two years ago, I would have made the same mistakes as others on this page have done. Less than two years ago, I definitely didn’t accept the result of the (in)famous Monty Hall problem.

It’s partly because of that, and partly because I’m a nice guy, that I am so keen to help others to understand how to correctly understand and solve problems like the posted one. I am quite sincere about this. It’s because of that, that I can be quite upset when people think I’m a stubborn and pig-headed and am accused of simply insisting that I’m right – that conclusion couldn’t be further from the truth.

Keeping an open mind is IMHO essential. A closed mind is a dead mind. I’ll add that (especially slavy) has caused me to learn new things, and I get a genuine intellectual thrill in doing so.

The recent problem “Girls, girls, girls” was a very simple problem and yet had so many aspects – it was a roller-coaster thrill ride for me. I just want to be able to give others that experience. I feel it’s one way that I can contibute to this site and this world.

February 10th, 2011 at 10:39 am

I just can’t stop thinking of other ways of seeing the result. So here’s another argument. I’m going to ignore the WW card as it’s irrelevant and only clouds the issue. So we simply have a BB’ and a bw card in the hat. Stick some labels on the four sides. Label B with 1, B’ with 2, b with 3 and w with 4.

Now obtain an assistant. Out of your sight, your assistant draws a card and tells you that it is not labelled with a 4. If he actually sees a 4, he puts the card back, shuffles them about and tries again (he will keep on doing this until he sees 1,2 or 3). The question now translates to “if he says it’s not 4, what is the probability that it’s 1 or 2?”

It is manifestly obvious that it is equally likely that he initially could have seen 1,2,3 or 4 (as each card is equally likely to have been drawn and neither face is preferred). The fact that he puts the 4 back in no way affects that it is equally likely that he saw 1,2, or 3. So the probability of him seeing 1 or 2 given that he saw 1,2 or 3 is 2/3.

Both 1 and 2 correspond to the BB’ card, and 3 corresponds to the bw card, so the probability of him picking the BB’ card given that he saw B, B’ or b is also 2/3 – because it’s exactly the same thing.

Just to be clear, if he saw 1 (B) then the other side is 2 (B’), if he saw 2 (B’) then the other side is 1 (B), or if he saw 3 (b) then the other side is 4 (w). It is equally likely that he saw one of 1,2 or 3 (i.e. he saw one of B, B’ or b) and so the other side of the card is equally likely to be 2,1 or 4 (i.e.the reverse sides for 1,2 or 3). But 2,1 are both black, whereas 4 is white i.e. 2 out 3 times the other side is black and 1 out of 3 times the othe side is white.

So it is twice as likely that the other side is black than it is white i.e. it is NOT equally likely that the other side is black or white as several of the posters assert (with not even the merest attempt at justifying their assertion). They simply say it’s black or white so 50%. That “logic” when applied to a fair die says that the probability of throwing a 6 is 50%.

February 10th, 2011 at 8:27 pm

LOL. Even my mate Lee gets this one. Lee doesn’t even know why a minus times a minus is a plus. He sent me this email:

Hi ChrisJust looking at your B&W problem – very good one.

That is hilarious.

Even I think I get it.

I have 3 B and 1 W.

I can only see 1 B.

This leaves 2B and 1W.

So chance of W on reverse is 1 in 3

If you had the BB and BW card in your hand, shut your eyes and randomly shuffle and turn over there is a 1 in 4 chance of seeing a W not 50-50 because of black and white.

TaLee

Maybe that wasn’t a model of clarity, but the gist of it was that (ignoring the WW card) when picking you will get the bw card white face up 1/4 of the time. Clearly, you will get the bw card black face up 1/4 of the time and the BB’ card black face up 1/2 of the time. So P(B or B’ given B or B’ or B) = (1/2) / (1/2 + 1/4) = 2/3.

February 11th, 2011 at 2:54 pm

LOL again. My mate Lee also caused me to realise that the folks who think it’s equally likely to be black or white, must believe that if you pick the wb card it must come out black side up. Why does that happen? Does it know that you require the answer to be 1/2 and so obliges and never comes out white side up!

OR perhaps the answer isn’t 1/2 after all!

February 12th, 2011 at 10:24 am

A slightly different way of seeing it is:

A pick is only a “success” if we pick a card and see a black face. i.e. we will only inspect the other face if the initial pick showed a black face.

If we pick the BB’ card, that is automatically a success. If we pick the wb card, then half the time it is a fail, as we will see a white face, and half the time it is a success because we will see a black face. So of the successful trials, 2 out of 3 were because we picked the BB’ card, and only 1 out of 3 were because we picked the wb card.

February 13th, 2011 at 5:51 am

I’ve just been re-reading some of the early posts, and noticed that there are two different reasons people believe the answer is 1/2. Nearly everyone is correctly realising that the WW card is irrelevant (so, curiously, they are intuitively using conditional probabilty correctly for that bit of the problem).

One group correctly realise that you are, before inspecting the card, equally likely to pick the BB or the BW card, and so say the probability is 1/2 of picking the BW card. They completely overlook that the question said you saw a black face first. They are not allowing that half the time you pick the BW card, that it will not be seeing the black face.

The other group are completely ignoring the probability of picking either card. They are simply saying there are two possibilities and without consciously realising it, incorrectly assuming that each possibility is equally likely, and so concluding the probability is 1/2.

Conjecture: I suspect that if there had been 100 BB cards amd 1 BW card, that a fair number of the first group would have said 1/101 and at least some of the second group would still say 1/2. (The correct answer is 200/201).

July 12th, 2011 at 10:44 pm

1/5

July 13th, 2011 at 2:13 am

LOL

May 16th, 2012 at 8:45 pm

hey, i have this question in my grade 12 data management book, the answer is 1/2

May 17th, 2012 at 12:16 pm

Hi xtina – the answer is 2/3.

August 28th, 2012 at 7:00 pm

Hi I just wanted to point out that the question in itself is wrong. The original question states that there are A NUMBER of cards in a hat, 20% of which are white on both sides, 50% of which are black on one side and white on the other and 30% of which are black on both sides. Then one card is drawn and only one side is inspected, and that side is black. So what is the probability that the other side is white? This makes it significantly more difficult.

August 29th, 2012 at 3:41 am

Hi John. You are describing a different problem. I’ll post it as a new problem. Thank you for providing it.