On Friday I set this puzzle….

From a full deck of 52 playing cards, how many cards do you need to draw to ensure that you have four of a kind in your hand?

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

Imagine randomly removing 39 cards from a deck. Amazingly, you remove all of the spades, diamonds and hearts. You would not have any four of a kind. However, the moment your remove an additional card you would have a four of a kind. And that’s the answer – 40 cards. Did you solve it?

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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If I can select the cards, I need to draw only 4. Nowhere specified that I have to draw the cards randomly.

I went straight to 40 as my answer then also considered the sneaky nature of some of the questions posed and reasoned 4 could also be an answer. I accept that “to draw cards” differs from selecting four cards but I’d prefer to cover both bases.

If you really want to misinterpret the puzzle, you might aswell say that 1 card has 4 corners and corners is the kind you’re looking for.

Or just hold the deck in your hand and say that you never removed any, but you now have 4 of a kind in your hand.

For the rest of us, Richard’s solution stands :)

“From a full deck of 52 playing cards, how many cards do you need to draw to ensure that you have 4 of a kind in your hand?”

That statement is not ambiguous. “Drawing” by definition does not allow you the freedom to “select” your cards.

With thinking like that, I could go and get another pack of cards (or even a few dice) and have four of a kind before I’ve drawn any cards. Nowhere does it specify that the four have to come from this pack of cards.

Were you like this when you did your 11+ exam?

I didn’t ‘solve’ it. I just answered the very easy question.

Don’t be smug, Eddie

I worked it the other way. If the pack is left with 13 cards one of them could be 2, stopping quad 2s; a 3 stops quad 3s etc. If 13 cards are left each of the values could be “stopped”:: draw one more and at least one four of a kind isn’t “stopped”.

I thought this was a good puzzle. I certainly did not solve it straight away but did so after a while. Thank you Richard.

The general formula seems to be : V x (T – 1) +1 (V = number of variants, here 13 ; T = number of types, here 4 => 13 x (4 – 1) + 1)

So, same question with 32 cards (8 of each color): 8 x (4 – 1) + 1 = 23

Other examples:

– If you have 2 colors of socks and want to be sure to take 2 of the same color : 2 x (2 – 1) + 1 = 3

– If you just want 4 cards of the same color: 4 x (4 – 1) + 1 = 13

Etc.

Not demonstrated, but seems valid: anyone has a counter-example?

Clever generalization! I wonder how one could prove it.

I think 8 x (4-1) +1 = 25 (not 23)

Ooops! You’re right, sorry and thanks for the correction

Got it, by the same logic. Not being into Poker I had to look up what exactly the hand means.

Interesting side questions would be:

What is the probability of four-of-a-kind when drawing

kcards? orHow many cards should one draw to have at least a 50% chance of getting four-of-a-kind?

I saw some discussion of that hand in 5-card and 7-card poker, but the cases there are not easily generalizable to

k > 7— one has to account for cases with more than one four-of-a-kind in the numerator. Messy, I think. The result would have to come out such that strictlyp(k)< 1 fork< 39 andp(k)= 1 fork≥ 40.I complain when it’s too easy (like today) and when it’s too difficult for me to solve. I should just be happy that Richard throws in an easy puzzle once in a while…

Ohhhh … so that’s what four of a kind means xD … my answer was 13, I’m sure you know why.

Close! What you were aiming for was a Flush, though generally that’s 5 cards all of the same suit, not 4 :)

I enjoyed the puzzle, even if it wasn’t too difficult this week. Few too many moaners on here deliberately misinterpreting puzzles for dull answers.

Many a poker player has lamented over a 4 flush…. glaring at the river card which failed to make their hand. :)

i remain a little confused with the ans. if i have a 100% chance of 4 of a kind with 40 cards surely if i have (for sake of aguement) 45 cards my changes of 4 of a kind is more than 100%. how so!?

A lot of types of gloves – wool gloves for instance – will fit either hand or both hands (but not at the same time).

i thank you all for your kind ans but i remain confused probably. two gloves may be the same colour but for opposite hands. does that effect the chances? sorry for spelling it changes before. i am learning.

No, if you draw 45 cards your chances (not changes) of having 4 of a kind is still 100%. Open a textbook and learn what is meant by a probability.

not sure if troll or just plain dumb

Imagine you have 2 colors of gloves, it’s dark and you want to be sure that you will take at least a good pair (i.e.: 2 gloves of the same color). You will have to take at least 3 of them: if you take only 2 you have risks to get 2 gloves of different colors, with 3, 2 will be of the same color, the third one will be of one color or the other.

The problem here is exactly the same, except numbers are higher.

I got 13. (3+3+3+3+1)

Because the “worst case” is: 3 spades + 3 hearts + 3 diamonds + 3 clubs taken, so only one card is needly to complete the set “four of a kind”

But besides that, your logic was spot on so you got the puzzle right in that sense :)

That’s not “four of a kind” as defined in poker. You’re thinking of a flush, sort of. In poker, a flush is all five cards of the same suit.

The other Matt: not 4 of the same suit, but 4 of a kind, which is the same “rank” ie, 1 2 3 J Q K etc

I got 40 but was worried there would be some kind of sneaky trick I’d missed!

If you simply pick up the pack of cards you will have all possible 4-of-a-kinds in your hand without having to draw any cards at all. Or does that count as drawing all 52?!

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