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?

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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.

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.

“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.

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 :)

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.

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?

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

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

Clever generalization! I wonder how one could prove it.

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!?

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.

not sure if troll or just plain dumb

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.

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.

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

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”

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.

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

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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