Tuesday’s Challenge: A Card Logic Game

After the success of Tuesday’s afternoon logic puzzles (which were correctly answered in the comments in less than thirty minutes), I promised to post something a bit more challenging. So here we go.


You’ve just flown into Las Vegas and have decided to peruse the casino floor of the MGM Grand, your favorite casino. It’s mid-afternoon, and you haven’t had anything to drink, and the blackjack and poker tables are sparse. You don’t really want to get into those games until later in the evening when all your friends arrive. As you keep walking around the casino floor, you notice a card game which you’ve never encountered before. It’s called Lucky Strike.

Here’s how Lucky Strike works. The dealer has a standard deck of 52 cards (26 red cards, 26 black cards). The cards are shuffled prior to the game and all the cards are placed face down. It costs $1 to play the game.

The rules: the dealer draws one card at a time and shows you the card after each turn. For every red card that’s drawn, you win $1. For every black card you draw, you have to pay the dealer $1. After the dealer draws the card, it goes into the discard pile and isn’t seen/used again. You have the option to quit playing the game after each turn (i.e., when the dealer shows you a new card). Here’s the question: devise an optimal (rational) playing strategy to maximize your payoff in this card game. What is the expected payoff in this game?

Two thoughts to ponder:

1) Since this is a standard deck of 52 cards, and if you choose to see all 52 cards, your ultimate payoff will be $0 (all the $1 winnings from seeing red cards exactly offset the $1 losses you pay to the Dealer from seeing the black cards). But remember, you paid $1 to play the game, so the House has the advantage if you choose to see every single card. So this isn’t an optimal playing strategy.

2) There will be multiple optimal strategies on when to “cash out” in this game. The challenge in this game is to deduce your expected payoff after seeing one card, two cards, three cards, etc. Remember: the order of the cards drawn matters: if you draw five straight red cards, for example, I can guarantee you that it would have been more than optimal to walk away from the table, collect your $5, for a total winnings of $4. So you need to consider the order of how the cards are dealt and when the optimal strategy is to “cash out”.

I will revise this logic puzzle for clarity in case some things aren’t clear. One critical assumption to make is that you are rational. Needless to say, but you are also an expert at calculating probabilities on the fly.

Any questions?


To make things more interesting: if you’re in the Atlanta area and submit a detailed, correct response to this challenge question, I will buy you a drink at an establishment of your choice (ITP only, please).

Tuesday’s Logic Puzzles

A brief break from reading this afternoon to tackle two logic/math problems below. See if you can deduce the answer on your own. Leave a comment if you know the answer!

1) Consider an analog clock with both an hour hand and a minute hand. What is the first time after 6PM that the hour hand and the minute hand are exactly coincident (i.e., on top of one another)? NOTEYour answer should be in this format: HH:MM:SS.DDD, where HH = hour, MM = minutes, SS = seconds, and DDD is the 1/1000th of a second decimal equivalent. (HINT: the first thing that comes to mind, 6:30PM, is an incorrect response).

2) Consider a room  with a very large table on which stand 100 lamps, each with an on/off switch. The lamps are arranged in a straight line, and each one is numbered 1, 2, 3, …, 99, 100. At the beginning of the experiment, all the lamps are turned off.

This room has an entry door and a separate door for an exit. One hundred people are recruited to participate in this experiment. Each of the 100 participants is also numbered 1 to 100, inclusive. When participant number 1 enters the room, he turns on EVERY lamp, and exits. When participant 2 enters the room, he flips the switch for every second lamp (thus, turning off lamps 2, 4, 6, 8, 10, and so on because participant 1 has turned all the lamps on his turn). Participant 2 exits and then participant 3 enters. Participant 3 flips the switch on every third lamp (thus changing the on/off state of the lamps which are numbered 3, 6, 9, 12, and so on). This process continues until all 100 participants have taken their turn and passed through the room.

Assume each participant can properly count and doesn’t make any mistakes in changing the on/off state of the lamp(s) he’s assigned to change the state of. Here is the question: after the 100th participant completes his journey through the room, how many lamps are illuminated? And which of those lamps (i.e., reference by number) are they?


UPDATE (1:30PM): Both questions have been answered in the comments. To make up for these relatively simple questions, I’ll post a much more challenging logic question in the evening. It will have to do with a deck of cards.