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?

 

 

 

 

 

bar

 

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          

 

Analysis diagram

 

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:

 

Dr. Robert Sweetland's notes
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