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Wednesday, May 16, 2007

☺ Brain Teasers - something to do while waiting for the next video

Brain Teaser 1: Warm Up with Monty Hall

Yesterday, 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

~ QtheC


  1. Clarification...
    The order of events is:

    You select two cards.
    Dealer discards one club.
    You make an offer (amount + option).
    Dealer accepts or rejects.
    If dealer accepts, then you perform the option and see if you have the two diamonds.

  2. ok, I'm going to give this my final try. If you choose to switch both cards or none the dealer knows who will win. If it's you he refuses your bet and you pay $25. If it's him you lose your bet.

    So you trade 1 card. That's the easy part. Now to have a chance on winning you need 1 D and 1 C. Chances for that are 3/5. But wait that doesn't matter. because if you don't have that the dealer already knows if you will win or lose and the same as before applies.

    the only option where your bit matters is having 1 D and 1 C. 4 trades are possible and one lets you win. That's 1 out of 4, to even the winnings you need to bet $12,5. But since that is not possible it will have to be $13.

    So my answer would be 13$, trade one card.

    But that would give you a 3/5 chance for even odds. And a 2/5 chance for certain loss. It's rigged.

  3. From a quick back-of-the-napkin run of the odds, I think the best bet is $1 without changing your cards. 9 out of 10 times you'll just lose the buck, and 1/10 you'll be penalized $25. Or as Brig suggests, don't bet on this rigged game.

    I will now run the numbers to see how much the dealer should OFFER ME to play this game... sound fair?

  4. Nomenclature:
    D=diamond, C=club, b=bet amount, W=expected winnings.

    First pick odds:
    DD CCC = 2/5*1/4 = 1/10=.1
    DC CCD = 2/5*3/4*2 = 6/10=.6
    CC CDD = 3/5*2/4 = 3/10=.3

    A) no trade
    A1) DD CC: (.1) dealer rejects bet (knows she's lost). W(A1)=-25
    A2) all others: (.9) dealer accepts bet (knows she'll win). W(A2)=-b
    Combined W(A)=-2.5-.9b

    B) trade 2
    B1) CC DD: (.3) dealer rejects bet (knows she's lost). W(B1)=-25
    B2) all others: (.7) dealer accepts bet (knows she'll win). W(B2)=-b
    Combined W(B)=-7.5-.7b

    C) trade 1
    C1) DC DC: (.6) dealer must consider bet amount to accept or reject. To win, player must pick the C in his hand (1/2) and trade it for the D in the dealer's (1/2) so the odds are 1/4 (3 to 1). So the bet must be at least $50/3 to be accepted (sorry Brig, you had this part wrong, the reason for this is in the winning payout wording: "you will keep your bet and win the $50 prize"). However, no bet is possible between $14 and $20 (we don't have a $10 bill), so 14 is rejected and 20 accepted. W(C1)=-25 if b<=14, =12.5-.75*b if b>=20.
    C2) all others: (.4) dealer accepts bet (knows she'll win): W(C2)=-b
    Combined W(C)=-15-.4b if b<=14, =7.5-.85b if b>=20

    To maximize W:
    A) b=1 W(A)=-3.4
    B) b=1 W(B)=-8.2
    C) b=20 W(C)=-9.5 (b=1 results in W(C)=-15.4)

    So, I was right before about option A: to lose the least, you pick the scenario where the dealer has the highest chance of a sure win! (and thus she has to accept your lousy buck). If you try to win the 50, as Brig said you're forced to trade 1 card, but you'll have to bet a 20 to convince the dealer. Sure you might get lucky and pick right and win the 50, but in the long run it'll cost you almost 10 bucks per game. Brig, turns out the choice of b=13 yields one of the worst expected winnings, as it's case C but the b<=14 formula, and W(C)=-20.2! Only a few worse than that:
    A) b=25 to 74: W(A)=-25.0 to -69.1
    B) b=20 to 25: W(B)=-21.5 to -25.0
    C) b=14 W(C)=-20.6

    Q, nice twist in the answer, although you made it too obvious (as you can see I got the right answer before running all the numbers). I would only bet in this game if they offered me a drink per game (gold label on the rocks please). Of course I'd only use the singles, and if I do get penalized, I think 4 gold labels for 29 bucks is not a bad deal, with a chance of getting them almost free. The dealer hopes I get drunk and greedy and start betting my 20's, or so unlucky that I get penalized twice.

  5. Great job on the analysis guys. And Deagol also helped me find a flaw in my original solution in the way I was calculating a "fair bet" for the trade 1 card case, so I think I need to rework the puzzle a bit.

    Also, out of that discussion came the idea of what happens with a series of bets with finite starting bills, and the possibility of winning or losing?

    So a next generation puzzle may emerge at some point.


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