You find yourself playing a game of GreenJack with your friend.
It is played with a deck of only 16 cards, divided into 4 suits:
Red, Blue, Orange, and Green.
There are four cards in each suit:
Ace, King, Queen, and Jack.
All Aces outrank all Kings, which outrank all Queens, which outrank all Jacks, except for the Green Jack, which outranks every other card.
If two cards have the same face value, then Red outranks Blue, which outranks Orange, which outranks Green, again except for the Green Jack, which outranks everything.
Here's how the game is played: you are dealt one card face up, and your friend is dealt one card face down. Your friend then makes some true statements, and you have to work out who has the higher card, you or your friend. It's that simple!
Round 3:
You are dealt the Red Queen and your friend makes three statements:
My card could lose to a Blue card. Knowing this, if I am more likely to have an Ace or a King than a Queen or a Jack, then I have an Orange card. Otherwise, I don't. Given all of the information you now know, if I am more likely to have a Jack than an Ace, then I actually have a King. Otherwise, I don't.
Who has the higher card, you or your friend?
Hint
List all of the cards, and then eliminate some using (1).
Answer
Your friend.
Reasoning
You were dealt the Red Queen.
The possible cards, in order, are:
Green Jack
Red Ace
Blue Ace
Orange Ace
Green Ace
Red King
Blue King
Orange King
Green King
Red Queen (your card)
Blue Queen
Orange Queen
Green Queen
Red Jack
Blue Jack
Orange Jack
By (1), their card could lose to a Blue card (the Blue Ace), leaving:
Orange Ace
Green Ace
Red King
Blue King
Orange King
Green King
Red Queen (your card)
Blue Queen
Orange Queen
Green Queen
Red Jack
Blue Jack
Orange Jack
By (2), their card is not more likely to be an Ace or a King (6) than a Queen or a Jack (6), so their card is not Orange, leaving.
Green Ace
Red King
Blue King
Green King
Red Queen (your card)
Blue Queen
Green Queen
Red Jack
Blue Jack
By (3), their card is more likely to be a Jack (2) than an Ace (1), so their card is a King, leaving:
Red King
Blue King
Green King
Red Queen (your card)
All of which beat your Red Queen.
?
Puzzle 2
Yesterday I was asked to buy some stamps.
In the land of BrainBashers, stamps are available in denominations of 4p, 5p, 7p, 12p, and 19p (just the five different values).
For three denominations of stamps, I was asked to buy six of each. For the other two denominations. I was asked to buy ten of each.
Unfortunately, I forgot which I was supposed to buy six of, and which to buy ten of!
Luckily I had been given the exact money required to buy the stamps, £3.50, so the shopkeeper was able to work out the stamps I needed.
Reasoning There are 61 squares of size 1 x 1. There are 37 squares of size 2 x 2. There are 15 squares of size 3 x 3. There are 3 squares of size 4 x 4.
Giving a total of 61 + 37 + 15 + 3 = 116 squares.
??
Puzzle 4
How many people must be at a party before you are likely to have two having the same birthday (but not necessarily the same year)?
Reasoning
By likely, we mean greater than 50% chance.
With one person there is a 0 percent chance that you'll have two people with the same birthday.
With two people the probability that they won't share a birthday is 364 ÷ 365. The probability that they will share a birthday is therefore 1 − (364 ÷ 365).
With three people the probability that they won't share a birthday is the same as for two people, times 363 ÷ 365. So the probability that three people will share a birthday is 1 − (364 ÷ 365) x (363 ÷ 365).
Notice that with each additional person added, the probability that they share a birthday with one of the previous persons goes up, because there are fewer "free" days remaining.
We keep adding people until the percentage is greater than 50%.
