On Friday I posted this puzzle….
Let’s play a little game. Imagine taking 4 playing cards out of a deck. Two of them are red and the other two are black. We mix them up and place them face down on the table. I am going to ask you to turn two of the cards face up. If they are both the same (as in, if both of them are red cards or both of them are black cards) then you win. Otherwise, I win. Is that fair?
If you have not tried to solve it, have a go now. For everyone else, the answer is after the break.
Nope, it is not a fair game. Let’s imagine you start off by turning over a red card. Of the three remaining cards, two are black and one is red, and so you only have a one in three chance of winning.
Did you solve it? Any other answers?
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.
Richard Wiseman 
Yeeh, knew it. But was an easy one this time
Got it, but surprised by the result. At first glance intuition said 50:50, but of course it’s not quite that simple. Otherwise it wouldn’t be a puzzle, would it!
I got it.. but thought if you turn 2 cards over at the same time it is still 50/50 isn’t it ?
Nope…. you’ll have 6 possible combinations and in only 2 of them the cards will be the same.
Same for me
Oops, I wanted to post this below Martyn’s comment.
I need to get used to this new layout. It looks better, but it’s less user friendly.
Not fair. It clearly states not to post the answer and the plonker Richard has gone and done so.
Steady on, Richard, you’re going into dangerous Monty Hall territory there!
At first you think it’s fairbut than you look at the possible combinations of cards and you see it’s not!
Not convincing at all: Richard is cheating! The odds are exactly the same.
No, they’re not.
Let’s call the cards R1, R2, B1 and B2 (where R = red, and B = black)
Then you’ve got these six combinations:
R1 – R2
R1 – B1
R1 – B2
R2 – B1
R2 – B2
B1 – B2
So it’s 1 out of 3 you win.
Got it! Not too hard with a few math classes worth of knowledge 🙂
Oh and the new website looks… Hmm well different! Guess I was used with the old design. Change never hurts. Have a nice monday!
Imagine the cards are labelled A, B, C and D. There are six combinations you can take in total (AB, AC, AD, BC, BD, CD) of which only two of these will be the same colour… doesn’t matter if you turn them over at the same time or not.
Can’t see any similarities with the Monty Hall puzzle though Eddie.
It was just that Richard was taking about turning one card over and then discussing the probabilty of what the other was going to be. That’s a bit like the Monty Hall situation. Richard did originally say turn both cards at the same time. It was just a light hearted observation from me.
I also though it was 50:50 but then I realised if the cards are R R B B and I choose R then the options left for the second card are R B B so there is only a 33% change of picking another R.
Yay, I got it but I’m usually crap at probability questions so thought I’d probably be wrong ;-D
It’s because there is a limited pool. If I were to say “Flip a coin twice, if you get the same result twice you win” then the odds are genuinely even, because there is always a 50% chance, regardless of the first flip. However, with the cards, card A reduces the pool of cards that could be card B (it doesn’t matter whether you turn them over together or one after the other). However, many people will do the maths for the coin scenario, and forget about the new odds for card B.
It’s worth noting you can make this 50/50. After the first card is chosen, turn it back over, and reshuffle, before choosing the second card.
You can get any answer you want, Martin, if you change the question.
If I get $10 when I win and have to pay $1 when I loose, and I can play this as often as I want, I definitely want to play this game 🙂
Clever solution, rather than brute force.
But the site revamp seems to have been put in the hands of emotional challenged experts who are blind to the fact that color and shape form rewarding attachments for readers and should be retained when coding gremlins are making layouts more “efficient”.
RIP the cute warm brown colorful somewhat primitive tabs that characterized the heroic Wiseman’s site and made it seem the function of individual personality and genius that it is.
Is this really true? The solution, I mean. In the famous three doors TV challenge, where three closed doors hide different objects, one a Ferrari and the other two toilet rolls, say, the participant is asked to choose a door, the host opens another door, showing a toilet roll, then the participant is asked if he/she wants to change his/her door choice. Readers almost always says It makes no difference, the chances of a Ferrari remain 50/50, when of course it improves their chances if they switch.
In this case you choose two cards out of four, then the host turns over one. If you still have a chance to switch your choice of the second card, maybe you could improve your chances by switching, perhaps even to 50/50?
The math is far beyond me. It is on the Mad Hatter level. All I know is that the host opening the door in the TV challenge is governed by the fact he has to open a worthless door, so it is a choice based on information about where the Ferrari is, so it is an informed choice, not a random choice.
Here the host doesn’t know what the color of the other three cards are, once one is turned over, so I guess it doesn’t apply. It will be a random choice.
The problem you are thinking of is Monty Hall. This is actually quite different, though it does sound quite similar. In Monty Hall it’s all about the odds of the first choice, in this it’s all about the odds of the second choice. There’s no switching, it’s a separate card, and actually, this is far easier to grasp than Monty Hall.
very similar to the Monty Hall paradox.
simple probabilities, but the unfair/fair question is unanswered. I don’t see any subterfuge so I’d still say fair.
I’m astounded that there’s any question about how this result was arrived at, especially with -M- and Craig Jones both explaining it so clearly. So here’s my attempt to muddy things up.
There are only six different combinations of four cards, and of those six, only two will win for you. The other four lose.
This has NOTHING to do with the “Monty Hall paradox”. Knowing the color of the first card tells you nothing, because whether it’s red or black, you know exactly the same thing: Only one of the remaining three cards will match. You still don’t know which is which.
Here’s how it could BECOME a Monty Hall paradox: Say you know your first card is red. Say you pick your second card without turning it over, and Richard then turns over one of the cards you DIDN’T pick, showing that it’s black. Should you change your mind or stick with your first choice? Now the choices match the Monty Hall paradox. In that case, the odds favor you change your mind, because the odds of the second card you chose matching the first one are STILL only 1:3.
But that’s not what this puzzle was. It was simple random choice where there were six options total: two ways to win and four ways to lose, and the odds of each condition were 1:6.
i object to the use of the word “fair”. Black Jack is a fair game, even though the odds favour the house. To me, a fair game is one in which everything is known beforehand, you know the risks and chances, and there are no nasty surprises (lollapaloozas). To win “fair and square” means to win according to the rules of the game, not that the odds are 50-50
It also depends on the odds you give me. If you give me 1-1 odds then it’s in your favour (but still fair) and if you give me 10-1 odds it’s in my favour
A lot of fuss over one of the easiest puzzles presented.
Well, you would probably have said the same thing about Einstein when he asked, How does gravity work exactly, does it have anything to do with acceleration or where the observer is placed?
Anyhow, I think I am changing my mind about this easiest of all puzzles that you have seen.
There are four cards, right?
Then there are two pairs, one RR and one BB.
Both those are winners.
There are two other possibilities in terms of selection, BR and RB.
Both are losers.
Therefore the chances of getting a winning combination is fifty per cent if you randomly select two of the four cards.
So the game is fair.
What the heck is wrong with that reasoning?
Seems OK to me.
Does the fairness of the game depend on the path that one’s neurons take in reasoning it out, then?
I mean, I can see that the second choice is one among three of which two are going to be duds.
But my reasoning seems armor plated too. It is simply a matter of thinking in pairs.
Two wrong and two right.
Half good and half bad. A fair game.
Must be overlooking something.
Dormouse, you are indeed missing something. Consider that in picking up one card at random with your left hand, there are three other cards remaining for you to pick up with your right hand. Regardless of whether you pick them up together or one after the other the three right hand choices will only have one card which matches the card in your left hand, so the chance is 33.33% not 50%.
The simplest way to show this is to call the cards R, r, B & b and to write down all 12 possible combinations:
Rr RB Rb rR rB rb BR Br Bb bR br Bb
of which only Rr rR Bb bB (4 combinations) give a ‘Win’. 4 in 12 is not a fair chance of winning.
Sorry, the 12th combination should have read bB, as Bb had alrwady been shown.
Dormouse. I’d sure like to play poker with you!
safc4ever
I’m a big fan of Sir Alex as well.
;o)
Dormouse – as well as the number of possibilities you need to consider the frequency of each possibility…btw my wife and I visited Monty Hall lat year. It’s one of the finest English country houses I have ever seen.
Well, Dormouse, I see, upon rereading Richard’s puzzle, that your reasoning is supported. One must give credit, right? I assumed that one would NOT turn over two cards simultaneously. Have you ever seen anyone playing cards or performing a trick do that? I pictured turning over one card at a time. But Richard doesn’t say one at a time, does he? There always seems to be something questionable about Richard’s wording.
Makes not difference if the cards are turned over simultaneously or not, that’s the point.
He he good one Phyllis. I think we have something. Turn over two cards, any two of the four, and you have either a dud pair or a good pair.
Since there are only four cards, that means you win in two instances (RR, BB) or lose in two (RB, BR). FAIR!! Definitely fair.
In fact, Richard’s reasoning is up the creek. No one said you have to pause after the first card, and even if you do, so what? The chances are governed by the choice outcome possibilities, of which there are only four.
safc4ever, you are obviously a distinguished Mad Hatter but you are here just repeating what I said, which is Richard’s reasoning. Cactospiza, you too.
Yes, having chosen one card, you have then only a 1/3 chance of hitting on the pair. But that’s not what the probabilities as a whole consist of. They have to be worked out from the beginning, ie before the first card is chosen.
From before the first card choice, you have only four possibilities in your double choice outcome: BB, BR, RR, RB.
Two are good ie the chances of winning are fifty per cent.
So, Dormouse wins 1.
Wiseman Loses 1.
Mad Hatter Niels says “there are four other possibilities B1R1, B2R1, B1R2, B2R2.” In fact those are the only four possibilities.
Mad Hatter 14manga makes the same mistake as others by writing out all the possibilities as if B1R1 was different from R1B1, etc. Just like safc4ever.
There are only four cards, folks. If you choose two, you only have four possibilities, RR, RB, BR, BB, and two of them are winners, two losers.
Fair!
Of course, there is another way of looking at it, I admit.
Call the cards 1,2,3,4 and ignore the colors for a minute. The possibilities in choosing two are 1-2, 1-3, 1-4, 2-3, 2-4, and 3-4.
If 1 is Red, 2 is Red, 3 is Black, 4 is Black, a distribution which is effectively the same as distributing them any other way for the purposes of gauging the outcome of the selection of two, then we have
Win, Lose, Lose, Lose, Lose, and Win.
Chances are actually one is three, clearly.
But my question was, what is wrong with my earlier reasoning?
Nothing!
It is simply proof that different routes in reasoning from the same basic assumptions can lead to different outcomes, each perfectly valid.
Let’s play poker, guys!
Dormouse: There are four other possibilities, B1R1, B2R1, B1R2, B2R2. You forgot you have two cards of each color. So you have a total of 6 combinations, of which two are winning…
So yeah, you overlooked something.
Most people here say that there are six combinations, something like this:
R1-R2
R1-B1
R1-B2
R2-B1
R2-B2
B1-B2
But shouldn’t you include B2-B1 and R2-R1 as well? It is a wrong assumption to compute the odds after one card has been shows.
Anyone has a simulator or a pack of cards to check the results? I guess that the chances are the same.
If you are going to include B2-B1 and R2-R1 (which you shouldn’t) then surely by your same logic you should also include B1-R1, B2-R1, B1-R2 and B2-R2 in your list as well?
Ops, my poor neurons… I was totally wrong
Wow, this totally bamboozled me, and even when people asserted it’s 1:3 I still had to write every combination down.
Let’s say you have 4 cards, as previously explained, and call them R1, R2, B1 and B2.
Then the combinations are:
R1 – R2 *
R1 – B1
R1 – B2
R2 – R1 *
R2 – B1
R2 – B2
B1 – R1
B1 – R2
B1 – B2 *
B2 – R1
B2 – R2
B2 – B1 *
Only those 4 highlighted combinations meet the criteria. Therefore the odds are 4:12, or 1:3.
What a devilish puzzle. I am taking this to work tomorrow to ensure I don’t have to take back the dirty dishes.
It’s easier than this.
The first card is irrelevant. It only decides which color you need to match.
Regardless of which color you initially flip, you will only have a 1/3 chance of flipping that same color again, giving you a 1/3 chance of winning and the “dealer” a 2/3 chance of winning. i.e. Regardless of which card initially flipped, of the three cards left, there is only one that will allow you to win. (1/3 chance of winning)
e.g.:
Flip RED – Leaving one red and two black – only 1/3 chance of flipping another red.
Flip BLACK – Leaves one black and two red – again, 1/3 chance of flipping another black.
Sorry Niels I misstated that the possibilities you wanted to add were the same as my basic four. Erase that remark
Yet again I feel humbled by Richard’s complete mastery of the arcane science of probability.
For anyone in any doubt of his skill check out the following link:-
https://richardwiseman.wordpress.com/2011/06/27/answer-to-the-friday-puzzle-110/#more-4357
My son pointed out that the more cards there are, the fairer the game.
From before the first card choice, you have only four possibilities in your double choice outcome: BB, BR, RR, RB.
Two are good ie the chances of winning are fifty per cent.
Nothing wrong with that reasoning!
Do you still not understand that the odds are not 50-50 Dormouse? This really is probability 101 —
6 possibilities, 2 are winners, 4 are losers, therefore your odds os winning are 1/3.
Different routes in reasoning can lead to different outcomes indeed, but only one is correct!
Thanks Mike. My problem is that I know, as I wrote above, that there are six possibilities and four are losers.
(That’s clear if you follow the following logic:
Call the cards 1,2,3,4 and ignore the colors for a minute. The possibilities in choosing two are 1-2, 1-3, 1-4, 2-3, 2-4, and 3-4. That’s Win, Lose, Lose, Lose, Lose and Win).
What I can’t see is what is the flaw in my flawed reasoning!
Look, you have two red cards and two black cards. So if you turn over two of the four, it seems quite clear that you have only RR,BR,RB,RR as the possibilities, and therefore it’s fifty per cent chances of a matching pair.
So where are the other two possibilities?
They seem to be invisible. Vanished!
Where the heck are they?
They were eaten up by your faulty wording.
The possibilities are
R1,R2
R2,R1 You put these two together as RR
B1,R1
B1,R2
B2,R1
B2,R2 You put these four together as BR
Starting to see where they went yet?
R1,B1
R1,B2
R2,B1
R2,B2 Another four possibilities that you list as RB
B1,B2
B2,B1
So this is 12 possibilities, of which 4 win.
You have it sorted out well, Anders. thanks. I agree, I see you’re right, I missed out those possibilities.
But I think I can still say my way above is much simpler, the one which gives the outcomes Win, Lose, Lose, Lose, Lose WIn by simply numbering the cards 1,2,3 and 4.
Then you have 1+2, 1+3, !+4, 2+3, 2+4, 3+4, which are Win, Lose, Lose, Lose, Lose, Win if you have laid them out BBRR, say.
If you have laid them out BRBR, or BRRB, it gives the same result.
Its always two out of six, ie one in three.
But I don’t think your method is stated correctly in every respect, is it?
B1 B2 is the same as B2, B1 isnt it? Same two cards.
R1 R2 and R2 R1 are exactly the same cards too!
You cant see this? 🙂
I had better stop or you will think I am trolling.
But I think I can simplify your method:.
You agree that if you have four cards in a row, half red, half black, then the pairing possibilities are the same however the cards run ie BBRR would yield the same chances of a pair or not as BRBR or BRRB or RBBR, right?
So assuming that you have dealt say B1B2R1R2, for the sake of showing the solution, you have the following pair possibilities: B1 B2, B1 R1, B1 R2, B2 R1, B2 R2, and R1 R2.
That’s six different possibilities of which only two are good.
And none of them are repeats of the same two cards, as you have in your arrangement, no?
Anyway you certainly put your finger on what I was missing, thanks.
Easier to see it if you pick some real cards, eg. Queen of Hearts, Jack of Hearts and the Queen of Spades, Jack of Spades. Two the same colour is a win so QH & JH or QS & JS but the losing 4 combos are QH & QS, QH & JS, JH & QS, JH & JS OK this actually looks confusing, lol, if I had used different colours and heart& spade symbols it woulda looked better, but point is there’s TWO BR’s and Two RB’s along with a BB and RR to make 6 combos
Actually you only need the colors, dont you, not the rank, RRBB does fine, no? 🙂
Actually I think numbers without letters ie 1,2,3,4, is the quick easy way.
My point (after acceding to all the rest) was that B1R2 is the same pair as R2B1, right? Thats my threadbare tattered point rescued after the rest sank.
Also that the logic of the false answer (that only BB BR RB RR were possible choices, therefore half would be good, therefore the game was fair)
was in need of exploding, which you just did, too. Thanks.
The Monty Hall puzzle is a good one to catch those bulls that charge at a red banner waved in front of them without pausing to consider they might be wrong.
You choose a door out of three closed doors because you hope to pick the one with the Ferrari.
Monty hall opens a door to show a toilet roll is behind it.
Then you change your choice of door to the third one.
But why is it good to change doors?
Your first choice is one in three. Then Monty Hall opens a door to show you a toilet roll. Therefore you change doors. But why? Why does it make your first choice less likely than the door you switch to? Why is it more likely that the Ferrari is behind the one you didnt choose?
The answer seems to be that the chance of the Ferrari behind the third door (that you now switch to) is now one in two.
So that’s a lot better than one in three.
But that cant be it, surely. Can the chances differ in that way? The third door was one in three before. Suddenly it becomes one in two?
Wouldn’t the chances of your original door being a Ferrari rise to one in two also?
So why would there be any difference between either of the two doors that are still closed?
Seems wrong.
A lot of mathematicians get caught by this one so I know it is probably not my simple answer.
OK came across this later and see what the factor was. Monty knows which is which – which one has the Ferrari and which ones the toilet roll. He HAS to choose to open the door on a toilet roll. So he ADDS information, and because of this increased information the chances change from 1 in 3 to 1 in 2 if you change your selected door.
As to the card trick, still don’t see why B1R1 isn’t the same as R1B1. The choices are B1R1 B1R2 B1B2 R1B2 R1R2, which is five, not six.
Those who see six choices are counting the same card combinations twice, just because they are reversed. B1R2 is the same as R2B1 and B2R1 is the same as R1B2..
sure they are the same, in that the colour selection is the same, but numbering them that way shows you that it occurs twice in the pool of possible outcomes so it’s still 6, not 5
to spell it out using your numbering scheme, it would be B1R1, B1R2, B1B2, R1B2 (times two) R1R2 which is 6
This is one reason why so many probability puzzles have incorrect solution. One possibility is discarded because it is “obviously the same” as some other possibility. While true, it doesn’t mean it should be discarded. The probability is “number of winning combinations” out of “total number of possible outcomes” by definitions, and those totals need to be accurate, you can’t just discard possibilities and still expect to come up with a correct probability
If you made a fake die with the 5 replaced by a 6, what are the odds of rolling a 2? is it one in 5, since the two sixes are identical so they cancel out? Or is it still one in 6 because the six-side is counted twice? possibly a bad analogy, but the best I can think of this early in the morning