Francis defines a ☆ b as . What is (13 ☆ 12) ☆ 4? For a challenge, solve this problem without a calculator.
Solution to yesterday’s problem:
Let’s use the blanks method for this problem. However, as we did in this problem and its solution, we’ll put probabilities in the blanks instead of possibilities. We need three blanks for three draws:
____ ____ ____
Any of the cards can be chosen for the first draw; there are no requirements for the suit yet.
__52/52__ ____ ____
On the second draw, notice that there are only 51 cards to choose from. Also, the card can be any suit except the suit that we just drew. Therefore we exclude the twelve cards from that suit.
__52/52__ __39/51__ ____
On the last draw, there are 50 cards to choose from and twelve less cards that satisfy the conditions.
__52/52__ __39/51__ __26/50__
The total probability multiplies to become 52,728/132,600. Reducing this fraction, 8,788/21,100; 2,197/5,275. So the final probability (at least what I got) is 42%.
There was some debate today over whether we would need to divide the final answer to get rid of the repeats. For instance, clubs-hearts-spades would count the same as hearts-spades-clubs. However, I decided not to divide the number because the number of possibilities is still put over the total number of possible draws. For example, the probability of choosing a head and a tail is still 1/2; you don’t need to make heads-tails and tails-heads.
By the way, does anyone actually read my solutions? Please leave a comment or like if you read the entire post!