# Aces for Dinner probability problem

## Aces Pay or Aces Free?

You and a friend order a pizza and your friend says s/he will put 6 playing cards on the table, two of which will be aces. You can pick two of the cards and if you get one or both of the aces, then s/he will pay for the pizza and you can pay for the tip. When you hesitate your friend says. If you don't think I am giving you better odds, then you put 6 cards, two of which are aces and I will pick two of the cards and if I get one or two aces, then you can pay for the pizza and I will pay the tip.

Would it be better to pick two cards
or
to deal the cards and let your friend pick two?

Hint: Generate all the different possible combinations of cards that could be drawn. Try the aces in different positions.

Cards Card paired with others
Ace other other other other
Card 1 = Ace Ace, Ace Ace, other Ace, other Ace, other Ace, other
Card 2 = Ace
Card 3 = other
Card 4 = other
Card 5 = other
Card 6 = other

Cards Card paired with others
Card 2 Card 3 Card 4 Card 5 Card 6
Card 1 Card 1, Card 2 Card 1, Card 3 Card 1, Card 4 Card 1, Card 5 Card 1, Card 6
Card 2
Card 3
Card 4
Card 5
Card 6

Discussion:

As cards are paired the number of pairings decreases by one. Is there are four cards (1,2,3,4) the possible pairings are:

• 1,2
• 1,3
• 1,4
• _
• 2,3
• 2,4
• _
• 3,4