And a Happy New Year to you!

On Friday I posted this puzzle…..

Imagine that 4 cards have been removed from a pack of 52 cards and placed face down in front of you. If you were to choose any two of the four, the chances of getting two red cards is 50%. What are the chances of picking a black pair?

If you have not tried to solve it, have a go now. For everyone else, the answer is after the break.

The chances are…… zero. In order to have a 50:50 chance of getting a pair of red cards, three of the four have to be red!

Did you solve it? How?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle** (UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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The solution above works only if you make an assumption. That assumption is that the information, “If you were to choose any two of the four, the chances of getting two red cards is 50%”, necessarily constrains the number of red cards in the four to a single, specific value. But it’s possible instead for the number of red cards to take on more than one value, in such a way that the probability of getting two red cards is 50%. The key is the way in which the four cards are selected from the pack of 52.

For example, suppose the selection protocol is:

(a) select a card at random from the pack, and then

(b) select three more cards of the same colour as (a).

Once the cards are selected, we have, with equal probability, either four red cards or four black cards. Thus the chance of picking two red cards from the four is 50%, as required. The chance of picking two black cards is exactly the same: 50%.

Is the assumption that the number of red cards in the four must be a single, specific value unwarranted? Since the selection protocol is not given, I think it *is* unwarranted. But I’m open to persuasion. Is there anything else in the wording of the question that would justify this assumption?

What you say as an alternative is a farfetched assumption.

It says:

“…4 cards have been removed from a pack of 52 cards…”

“…If you were to choose any two of the four,…”

“…What are the chances of picking a black pair?”

And what you know more about THESE 4 cards is:

“the chances of getting two red cards is 50%.”

This extra infromation makes your alternative impossible.

Sorry, but it doesn’t work. ‘any two of the four’ does not leave any room for misinterpretation. The only scenario that fits the information given is 3 red cards and one black. In your scenario the chances would be either 0% or 100%.

If you absolutely need an alternative answer, here is my offering: the 52 cards were in fact collector cards of famous people. Three of the four were of american indians, and the fourth was of Mr. and Mrs. Obama. Thus, the odds are 25%

🙂

argh, I wrote that last part too quickly: the 4th card shows up in 50% of the possible combinations, so the odds are obviously 50%, not 25. Dang

How does the extra information make this alternative impossible?

My alternative is indeed an assumption. But Richard’s solution also makes an assumption. The point is that, depending upon the selection protocol, the probability of getting two black cards could be anywhere from 0 to 50%.

@Anders: If you were to follow my alternative protocol 100 times, roughly half the time you’d pick two red cards from the four. In other words, the probability of picking two reds would be 1/2.

Nick, you’re leaving out the word “any” in your odds calculation. It says *any* two of the four, and with your method, that is just not true.

I agree with M – the wording “choose any two of the four” makes it clear that not all of the four cards can be red, so the stated solution is indeed the only one.

Nice puzzle – the challenge was to figure out how to dismiss the “obvious” answer.

@Anders: If you’re going to interpret the word “any” literally, then there are problems with Richard’s solution, too. For example, if you choose a red card and a black card, the chances of getting two reds are not 50%, but 0!

The only sensible interpretation of “If you were to choose any two of the four”, and the interpretation that fits with Richard’s solution, is: “If you were to choose two of the four cards at random.”

@Nick: “face down”… That means “choosing” is “random”. And we can be sure of that, since it is a pack of cards. If there would be any specialty to this pack of cards, then that should be said. If it is not said, the assumption that this is a very normal pack of 52 cards is perfectly valid.

@M: I don’t deny it is a normal pack of 52 cards.

@Nick’s option is ingenious, but I think a rather extreme interpretation of the wording. In this particular case I would cut Richard some slack and say that applying the simplest and most straightforward interpretation of the wording that the probability of 50% applied after the 4 cards have been selected. In that case, @Nick’s protocol would have resulted in either a 100% or a 0% chance of picking two red cards as the multiple 2 card picks are after the choice of 4 is “frozen”. That means the protocol is incompatible with the clearest interpretation of the wording.

So high marks for ingenuity, but in this case I think the wording clearly means repeated trials on the same 4 chosen cards and not a reselection.

I would agree with Steve Jones, Nick’s option is ingenious (I saw that something was missing friday, but did not take time to look for another way to fill in the blancks), but it seems clear here that the most obvious interpretation of the problem is the one used in the solution.

However, I would not cut anyone any slack for wording their problems poorly. It is really not hard to describe precisely what happens and where do the given probabilities come from, so that there is no possible ambiguity left. I am not specifically speakling about puzzles here, some problems are just _never_ written right, for example the two boys problem.

@Steve Jones: I agree that the key point is: should the 50% chance of picking two red cards apply to the whole process, beginning with the selection of the four cards from the full pack? Or should it apply only after the four cards have been selected? The wording “Imagine that 4 cards *have been* removed” (my emphasis) seems to support the latter interpretation, and thus Richard’s solution, but I thought it was worth suggesting a different interpretation.

@Nick

Fair enough. It’s often more interesting to explore the deeper issues of the problem. It’s notably that the New Scientist weekly Enigma problems are very carefully worded to avoid ambiguities and that makes them quite long (albeit usually more complex too).

@Yat: I certainly agree it is important to describe problems as clearly as possible, especially probability problems, which are notorious for non-trivial potential ambiguities. But I think the problems on this blog often have a studied vagueness, perhaps designed to draw out different interpretations.

Got it

I actually thought of this answer that it might also be this. Three cards and zero possibility.

Oooh! I didn’t got it. I thought the answer was 50%.

Great puzzle! 🙂

The joy of this blog is watching people who think they are cleverer than everybody else try and explain how they weren’t fooled.

Great stuff.

I got the correct answer within a few seconds of reading the puzzle, but had to verify it, which took a couple of minutes. Tricky one! And very elegant too!

I like these puzzles when I ‘know’ I have got the right answer and that any over-scrutiny of the question is pointless pedantry.

e.g. “4 cards have been removed from a pack of 52 cards”

Were they christmas cards?

What colour(s) are the cards in that pack? Green and Yellow?

How many colours?

Did you know that in Geordie ‘red’ means ‘black’ when I want it to and vice versa, . etc etc

Happy New Year

Got it right away. Saved myself from pulling out the proverbial magnifying glass and picking out non-existent nits.

I got 1 in 28 :l

Didn’t get it.

I thought the unintuitive 50% chance to get 2 red cards implied another definition of red cards (maybe something natively english with face cards nicknamed red or so).

Instead it implied a specific distribution.

Nope, just the only solution.

I don’t think it matters how you define it. If you say you have a 50% chance of getting too $FOO cards, what are the odds of getting two $BAR cards, then as long as $FOO and $BAR aren’t overlapping, I don’t think it matters how you interpret them.

In your example, if you have a 50% chance of getting two face cards, what are the odds of getting two pip cards. Same answer

If you really want to pick nits, the chances of picking 2 red cards out 4 where only one is black are not quite 50%. First draw gives you 75% odds, followed by a 2 in 3 odds or a 66.6666666666666666666666 (on to infinity # of 6’s)% chance. Not wanting to take this to the extreme we’ll use just 2 decimal places giving you the calculation of .75 x .6666 to determine the odds of picking 2 red cards from the 4, giving you .49995 or 49.995% chance to pick 2 red cards.

Since the premise of “a 50% chance to pick 2 red cards” is faulty, this puzzle is bunk. Translation: I missed the fact that this condition necessitated 3 red cards in the initial draw of 4 and am now bitter about it.

Not wanting to take this to the extreme, use ratios instead.

3/4 * 2/3 = 6/12 = 1/2 = 50%

Maybe you missed the part at the end where I indicate I was being facetious. I missed the initial condition of a 50% chance of drawing 2 red cards from the 4 as playing any role.

As a poker player I always work on percentages because it’s easier to calculate pot odds on the fly when deciding to call raise or fold. (e.g. there’s $100 in the pot, I’m on a flush draw with 2 clubs in my hand and 2 on the flop. I know that I have about 3.5 to 1 or about a 28% chance to hit another club on the turn or the river. My opponent bets $50. It would cost me $50 to win $150, so it’s a fold because 28% 24.x%.)

But I’m always rounding off as much as 5-10% in most cases because it’s not just the odds I’m basing my decision on, so detailed accuracy isn’t necessary.

err, ignore the whole flush draw example. Some how a whole middle section got deleted and I don’t feel like retyping it, you get the idea.

“pot odds”? “fly”? “raise”? “fold”? “pot”? “flush draw”? “flop”? “hit”? “turn”? “river?” “fold”?

Absolutely nobody in the universe knows what any of those things mean, but everybody knows that they are nothing to do with the week’s puzzle.

Nobody in the universe except for everybody who understands basic poker terminology. Which is a far cry from absolutely nobody.

But nobody in the universe plays poker. Nobody in the universe is interested in poker. There may be a tiny minority of deranged insane people who incorrectly falsely pretend that they do, but nobody does in reality.

As a poker player however, I noticed the use of the word “Pair” when describing the black cards which, to a poker player, means the same denomination. The odds of drawing a black pair (2 2’s, 2 3’s, etc…) are as follows:

1st draw is 50%, but to draw the one remaining black card that pairs up with the first one are now 1 in 51, or 1.96%. .5x.0196 = .0098 or a .98% chance to pick a black “Pair”

I also can’t help but quickly calculate the odds of turning over 2 black cards from a deck of 52. The first draw is a true 50% chance and the second is slightly less as you’ve now removed one black, leaving you with 26 red and 25 black, or 25/51 (49.01%) odds of getting another black. .5 x .4901 = a 24.5% chance to pick 2 black cards from the deck of 52.

It seems I was hoisted by my own petard here because I was sure that “pair” was the trick word. I assumed the 50% meant there were 2 red cards and 2 black in the 4, so what were the chances that the black were paired? My calculation would have been that by taking 2 red and a random black from the deck of 52, you have a 1 in 49 chance of pairing your black, or a 2.04% chance.

Yeah… we sort of got your point. You confused it with poker…

This puzzle was so badly-worded that I didn’t know what it was asking. I wasn’t sure if it was asking about the chances of picking two black cards instead of two red cards from the same random draw, or if it was asking about the chances of picking, say, a black pair of 2s, or what. I suppose if I’d thought about what “the chances of drawing a red pair is 50%” a bit more I’d have realized that it was talking about the sub-draw from one particular draw, though.

Hooray. I got this one! I got it by figuring out what the probability of drawing two red cards from four was for each number of red to black cards. The only one that gave a probability of 50% was three reds and one black. The complete picture is:

no. of reds. probability

0 0%

1 0%

2 33%

3 50%

4 100%

Yes that was a quick one. The first thing I did was to forget the remaining cards and just work with the selected four. so I worked out all possible combinations and worked out which one gave a 50 50 chance and the only answer was 3 red 1 black so 0 it was. I do not think the puzzle was badly worded at all.

This ^

I still don’t get what Nick is saying. Is it the same as what fluffy is saying?

What they are saying is that they were still stuck in the confusion that most of us took a couple of seconds or minutes to figure out.

And what they say is that when you twist your mind (until a knot forms in your brains), you are able to interpret this differently.

@M: No, what I’m saying is that I got the obvious answer straightaway, and then began wondering whether there’s sufficient ambiguity in the wording of the question to permit a different answer, as is often the case on this blog!

https://richardwiseman.wordpress.com/2011/12/30/its-the-friday-puzzle-141/#comment-79441

@Julia: Short answer…

We have a two step process:

(a) draw four cards from the full pack of 52, then

(b) draw two cards at random from the four.

We are told that the probability that both cards are red is 50%. This places a constraint on how the four cards are selected in (a). For example, if they were drawn at random, the probability that both cards would be red would be about 24.5%, significantly less than 50%. So the cards must be selected in some special, constrained, way.

So what is the constraint? Richard’s solution assumes it is that every time we perform step (a), the probability of drawing two red cards in step (b) must always be exactly 50%. Arguably, that assumption is implied by tense — we’re told “4 cards *have been* removed” from the pack — but it’s certainly not made explicit in the puzzle statement. That’s fine in a fun puzzle like this, where part of the fun is filling in the gaps, but in a real life situation, such as when you’re betting real money, everything would need to be made explicit. Anyway, given that assumption, in (b) there must be three red cards and only one black card, so of course there is no chance of drawing two black cards! Very neat.

The constraint can be made more general by noting that the probability of drawing two red cards in step (b) doesn’t have to be 50% for every step (a) — it’s enough for it to be 50% *on average*. Then, over the two step process, the probability of drawing two red cards will be 50%, as required. The most extreme example of this is the one I gave above, where in (a) we choose, with equal probability, either four red cards or four black cards. The probability of drawing two red cards will then be 50%, as will the probability of drawing two black cards.

Under this more general constraint, there is no unique answer to the puzzle. The probability of drawing two black cards can be anywhere from 0 to 50%, depending on how we carry out step (a). That sort of makes a mess of the puzzle, but, in my opinion, is interesting in its own right.

I don’t understand a word of the gobbledygook from ericthebassist, and I don’t understand what fluffy’s point is, and Nick’s interpretation is bonkers, and the people ho pretend that the wording of the question is unclear are insane.

All I know is that the wording of the question was clear, the answer was easy and unambiguous, and I am feeling smug in comparing myself with the nincompoops who are trying to find convoluted excuses to explain their own failures. 🙂

Whoosh!

Agree with fluffy that the question is poorly worded. Two cards selected from a random selection of 4 would still satisfy the 50% requirement. So an alternative solution is 50%.

Here’s an alternative wording, that would IMHO fix the problem: “Imagine that 4 cards have been removed from a pack of 52 cards such that if you were to choose any two of the four, the chances of getting two red cards is 50%.”

Does it make too easy? Perhaps. But here you can fairly claim that you were not just making a point about probability but imposing a constraint on the manner of selection — whether or not the reader saw it.

“Two cards selected from a random selection of 4 would still satisfy the 50% requirement. So an alternative solution is 50%. ”

No they don’t.

In completely random:

The chance the first card is red is 50%. The chance that the second is also red is slightly less then 50%. That makes the total chance less then 25% that both are red.

The 50% chance tells that these four cards are 3 red and 1 black.

This one was pretty intuitive to solve. It was easy to see that if there 2 black and 2 red cards the chances of picking 2 reds would be nowhere near 50%, and that if there were 4 red cards the chances would be 100%, so in order for the premise to be correct there had to be 3 reds and 1 black.

Sadly, I didn’t have complete confidence that Richard was not pulling a prank, so I did the math to double check.

If the 50% probability is reckoned for the 2-stage process of first selecting the 4 cards, then selecting from the 4, rather than reckoned from the point where 4 cards have already been selected, then that implies a non-standard pack with different number of red and black cards, since with a standard pack the 2-stage probability of 2 red cards would be 25%. Specifically, we would require that, if R is the number of red cards, then R(R-1)/(52×51) = 0.5. However the (positive) solution for that in R is not exactly divisible by 52, (about 37.43) so there is no such pack of cards. This should, perhaps, lead one to realise this is not a useful interpretation.

Richard’s puzzles are frequently sloppily worded, quite deliberately, and part of the enjoyment is coming to a worthwhile understanding of what the puzzle means if it is to be interesting. I have only occasionally been disappointed in finding Richard’s interpretation was the boring one.

I read this as a two stage event – therefore that picking 4 cards from a pack and selecting 2 which were red was a 50/50 chance. Re-reading it now I reach the same conclusion.

I would have preferred it if the trick had been that the pair needed to be the same card or something, this is just poorly worded.

Agreed, this was poorly worded. First I thought black pair meant a real “pair” i.e two the same, then I thought the odds of getting 2 reds werent 50%, so I didnt bother trying the puzzle cuz i didnt understand it. You would think on a skeptical blog these puzzles would be more carefully worded, as critical thinkers tend to be a pedantic lot!

Also thrown off by the wording.

I read …

“If you were to choose any two of the four, the chances of getting two red cards is 50%.”

… and thought …

“no it isn’t”

I was confused by the wording, not the maths.

But then if the wording was more descriptive, it might have made the answer more obvious.

This would have been better wording i.m.o:

Having chosen 4 cards randomly from a pack of 52, an observer looks at the faces 4 cards and tells you that if you were to choose 2 of the 4, your chance of picking two reds is 50%. What then would be your chance of picking two blacks from the same 4 cards.

Nick, I just wanted to let you know that YOU ARE WRONG! and quite a pain in the rear 🙂

You find an alternative viewpoint ‘quite a pain in the rear’?