Brain Teaser 1: Warm Up with Monty HallYesterday, on the comments board, I posted a Brain Teaser, something I had remembered solving several years ago. Meepers_anonymous solved it correctly and DuchessofDork gave the correct answer but without calculating the exact odds of winning. Chaosandorder pointed out that this one is known as the "Monty Hall" problem, named for the old television game show where the contestent chooses a door hoping to win a car and avoid a donkey. Here is the problem the way I stated it, followed by a link to a good website describing the problem and solution. You can use this as a warm-up and preparation for Brain Teaser 2.
Waiting for a new video… here’s a brain-teaser for you (not a trick question, but requires careful thinking)…
A dealer shuffles a deck of 3 cards, an ACE and two KINGS, and places them face down on the table in front of you, asking you to pick the one you think is the ACE.
You take one card, without looking at it. The dealer looks at the other two cards, turns over a KING and keeps the remaining card face down.
Your goal is to get the ACE, and you have a choice… keep the card you originally picked, or trade it for the card the dealer has kept face down.
What should you do, (does it even matter) and what are your odds of getting the ACE?
Try to solve it first, then google “Monty Hall” or just look here:
Brain Teaser 2: Place Your Bet(a QtheC original, so don't bother googling)
A smart dealer offers a $50 bill if you can win her game by ending up with a pair of diamonds.
The dealer shuffles a deck of five cards including two diamonds and three clubs, and places them face down on a table, allowing you to take any two you choose without looking at them. The dealer then looks at the remaining three cards and discards a club.
The dealer tells you, "You can bet any amount of at least $1 you wish to try to win the $50 prize. Along with a bet amount, you must choose one of three options, to be performed by you without looking at any of the cards: keep the original pair of cards you have, trade one card from the original pair for one of my cards, or trade both of the original cards for both of my cards. If I know the odds are even or favor me, I must accept your bet, and if you end up with both diamonds cards, you will keep your bet and win the $50 prize. If I accept your bet and either of your cards turns out to be a club, you lose the amount you bet. However, if I reject your offer, you keep your bet but you must pay me a $25 penalty."
You have four $1 bills, two $5 bills, and three $20 bills. No change is available. If you decide to play, What is the best bet offer you can make?
Place your bets on the comments here or PM me on the comments board. For example, though I wouldn't recommend this solution, you might choose:
bet $67, trade one card