## Red and black brainteaser

By David Shaw, 12 June 2019

I have four playing cards: two of them are black and two are red. I turn them face down so you can’t see the colours, and shuffle them so you don’t know which is which.

If you pick two of these cards at random, are they more likely to be the same colour or different colours?

Scroll down for the answer!

## Brainteaser answer

Imagine you pick the cards one at a time. No matter what colour your first card is, there will be three cards left to choose from. One of those cards will match the card you’ve picked, and two will be different. So you’re more likely to pick two different cards!

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12 June, 2019 at 12:55 pm

Another way of solving this is to write down all the possible ways that you can choose two cards. We’ll name the card R1, R2, B1, and B2 (for Red/Black)

The left column is the first choice, the right column is the second choice

R1 R2

R1 B1

R1 B2

R2 R1

R2 B1

R2 B2

B1 R1

B1 R2

B1 B2

B2 R1

B2 R2

B2 B1

But choosing R1 first then R2 ends up with the same two cards as if we chose R2 first and then R1. So we can eliminate these “duplicates” and mark down if the two cards are the same or different.

R1 R2 = S

R1 B1 = D

R1 B2 = D

R2 B1 = D

R2 B2 = D

B1 B2 = S

Coming to the same conclusion as above.

12 June, 2019 at 1:13 pm

Good one!

Now, is there any way to have four cards and have a 50% chance of picking two the same colour?

13 June, 2019 at 8:44 am

Another way to see this; the probability of picking a red sock first is 1/2 and then with only 3 socks (2 black and one red) picking red is 1/3. The probability is the product 1/2 times 1/3 or 1/6. Repeat for black as a first pick (leaving 2 red and 1 black) and you get 1/6. Adding the probabilities 1/6 plus 1/6 you get 2/6 or 1 time out of three for a pair of the same color.

Still, view the probability of picking either a black or a red on the first pick is 1. Then the probability of picking the matching color on a second pick is 1/3. Multiply 1 times 1/3 and you get a pair, either color, 1/3 of the times.

Various ways to get to the same answer. Be sure to read the question carefully as in this case either a pair or red or a pair of black would count. The question could be for just red (or black) or just mismatched sets. Try to solve the mismatched set using probability as homework.

13 June, 2019 at 11:46 am

Put the card back?

13 June, 2019 at 11:54 am

Yup, that’s one way to get a 50-50 chance. There’s another way too, by changing the cards you start with…

14 June, 2019 at 12:55 pm

But the question said “pick 2” … not “pick one, then pick another one” … so what is the probability of getting 2 the same or 2 different if you pick 2 up at the same time? …

14 June, 2019 at 2:45 pm

Hi Andrea,

Good question!

What if you drew two cards at the same time, but only looked at them one at a time? Does the same argument work for you?

If you still don’t like it you could read Martin’s reply – he’s got a different path to the same solution, which you might find more satisfying.

14 June, 2019 at 3:08 pm

What if you don’t look at them? Would the chance be 50-50, like Schrödinger’s playing cards?

14 June, 2019 at 3:18 pm

That’s probably best answered by a philosopher, not a mathematician…

But I think that the answer might still be 33%. And if it’s not, it might not be a number at all.

14 June, 2019 at 9:52 pm

> Now, is there any way to have four cards and have a 50% chance of picking two the same colour?

If we allow the person to choose 2 cards and then allow them to change one of their cards, but the choice of swapping a card is done after an unmatched card from the remaining ones is removed, this would change the chance to 50%. This extra step in similar to the “game show doors” in which a contestant is shown 3 doors, they choose 1, the game show removes an incorrect door, and the contestant has the choice of staying with their original door or switching.

The following table shows a section of all the possible paths that could occur.

— The first column shows what the original 2 cards were

— The second column shows which of the remaining cards is removed

— The third column shows the results of changing the “second” card

— The last column depicts whether the remaining cards are the (S)ame or (D)ifferent

R1/R2 – B1 – R1/B2 – D

R1/R2 – B2 – R1/B1 – D

R1/B1 – B2 – R1/R2 – S

R1/B2 – B1 – R1/R2 – S

This section of the table can be extended to include all possible combinations but you will end up with a similar pattern — which is 50% chance of having the same colour cards.

18 June, 2019 at 6:36 pm

Hmmm … I can see why they say “there are 3 kinds of lies – lies, damn lies, and statistics” 🙂

Not being that great at maths, I am probably thinking more like a geneticist with punnet squares, or something – lol

As to the probability of one chose the cards one at a time but didn’t look? … maybe ask Shrödinger?

As to the probability of me getting 2 cards of the same, or different colours, however I picked them – Murphy’s law states that whatever I did – if I needed one particular combination or pair, I would most likely get some other 🙂

*thinks about it a bit more*

I guess another way to figure it out in my mind (and explain it) is to realise that to have 2 of the same colour chosen – each card only has one other card that it can be chosen with – the other one of that colour, whereas to have 2 different colours chosen – each card has 2 other cards it can be chosen with to achieve that – either of the 2 other coloured cards … If there is a different way to work out the same result/answer everyone else got by doing it the normal way – it will be me who did it the weird way – lol

So … now that I have untangled my brain slightly, time I did something else …

*wanders off, humming “2 out of 3 ain’t bad” …*

(sorry – for some reason, that song kind of popped into my head – lol)

18 June, 2019 at 6:39 pm

btw, yes, I am the same Andrea who commented earlier … not sure why it has picked up my profile picture from somewhere this time, but didn’t before … I typed in the same email address both times – lol