A cards problem

Probability Level 3

Suppose I have a normal 52-Card pack, jokers out, facing down. I start to flip the first, and then second, and so on until I get a combination. How many cards do I have to flip at least to ensure I get a combination?

Combos include: 5 of the same suit, 5 ascending cards, a straight flush, four of a kind and a full House.

The answer is 17.

2 solutions

Mark Hennings
Jun 4, 2018

After $16$ cards have been chosen, we could have picked $4$ cards of each suit, and not obtained a flush. We are bound to obtain a flush on turning over the $17$ th card.

It takes at least $27$ cards to obtain a full house (after $26$ you could have chosen $2$ of each kind of card). It takes at least $40$ cards to obtain $4$ of a kind (after $39$ you could have chosen $3$ of each kind). A royal flush is a special type of flush, and so does not need separate consideration. You can choose $11$ cards in a suit without getting $5$ in a row (for example, A2346789JQK), and so it could take $45$ cards to get $5$ in a row.

Thus you are sure of getting a combination after $\boxed{17}$ cards.

A flush = in 17 cards
A straight = in 45 cards
A straight flush = in 45 cards
A four of a kind = in 40 cards
A full House = in 27 cards

Does the probability of hands also the same in this order? P(flush) > P(full house) > P(4 of a kind) > P(straight) = P(straight flush)? It's hard to remember their order when you're only allowed the jets and helis cards to play with (no gambling in the house).

Saya Suka - 3 months ago

David Vreken
Jun 5, 2018

You can flip over 16 cards and still not have one of the combinations, as long as you had 4 cards of each of the 4 suits. (One example is the 2, 3, 4, and 5 of hearts and diamonds and the 7, 8, 9, and 10 of spades and clubs.) On the 17th card , though, you will add a 5th card to one of the 4 suits, so you will be ensured the combination of 5 of the same suit.

A cards problem

The answer is 17.

2 solutions

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