Ayecarumba
Joined: Nov 17, 2009
• Posts: 6763
October 29th, 2019 at 10:37:56 AM permalink
Quote: Wizard

Quote: Gialmere

Dice Riddle
A dealer slams a shaker containing two dice down on the table, peeks under the lid and tells you that one of the dice is showing a 1. You then place a \$1 bet that snake eyes has been rolled. The dealer opens the shaker lid and reveals two 1s. Assuming true odds, how much money should you be paid?

If I may reword the question as "A dealer will slam a shaker containing two dice down on the table, peek under the lid until at least one of the dice is showing a 1. You then place a \$1 bet that snake eyes has been rolled. The dealer opens the shaker lid and reveals two 1s. Assuming true odds, how much money should you be paid?

Given that wording my answer is

\$10 + your original \$1 wager

As was noted, the same problem was debated here for months already. It is still a sore subject with me.

Hehe.... Happy Halloween Wizard! The undead are coming back...
Simplicity is the ultimate sophistication - Leonardo da Vinci
Gialmere
Joined: Nov 26, 2018
• Posts: 1075
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October 30th, 2019 at 9:40:04 AM permalink
Card Conundrum
While sitting at a card table you agree to an unusual wager. First, you are blindfolded so you cannot see. Next, eleven playing cards are placed in a pile before you. Finally, you are (correctly) told that seven of the cards scattered in the pile are faceup while the remaining four are facedown.

Using all eleven cards while blindfolded, the challenge is to make two groups of cards and each group must contain an equal number of facedown cards. Can you win this bet without literally resorting to blind luck?

Yes (How?)
No (Why?)
Insufficient Data
Have you tried 22 tonight? I said 22.
charliepatrick
Joined: Jun 17, 2011
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October 30th, 2019 at 9:58:12 AM permalink
Let's call the cards XXXXXXX YYYY, you need and equal number of Y's in both piles.

(i) Divide the cards into two piles, one with 7 and the other with 4.
This can result in N=0-4 Y's in the 4-pile (and 4-N X's in the 4-pile). Thus the remainder, 4-N Y's, are in the 7-pile.

(ii) Turn over all the cards in the 4-pile.
All Y's turn into X's, and all X's turn into Y's.
So now there are 4-N Y-cards in the 4-pile.

So both piles have equal number of Y's.
beachbumbabs
Joined: May 21, 2013
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October 30th, 2019 at 10:06:48 AM permalink
Quote: charliepatrick

Let's call the cards XXXXXXX YYYY, you need and equal number of Y's in both piles.

(i) Divide the cards into two piles, one with 7 and the other with 4.
This can result in N=0-4 Y's in the 4-pile (and 4-N X's in the 4-pile). Thus the remainder, 4-N Y's, are in the 7-pile.

(ii) Turn over all the cards in the 4-pile.
All Y's turn into X's, and all X's turn into Y's.
So now there are 4-N Y-cards in the 4-pile.

So both piles have equal number of Y's.

That's really well figured, CharliePatrick! Impressive.
If the House lost every hand, they wouldn't deal the game.
Gialmere
Joined: Nov 26, 2018
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October 31st, 2019 at 5:17:24 PM permalink
Note that if you turn over the 7-card stack you end up with an equal number of faceup cards. Also note that this will work with any number of cards in any up/down combination. Knowing this, the trick can be weaponized into an entertaining bar bet. Using a standard 52 card deck, have the mark cut the cards, turn one stack up and shuffle the stacks together. If you're told that there are 20 facedown cards, simply count the top 20 cards off the deck, turn them over and spread both stacks. You'll win every time.

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New Game (?)
At G2E you talk to some designer friends at their booth and they show you a new 2-player game they are developing. Nine cards numbered 1-9 are placed faceup on the table. Each card is worth its face value. To begin, the first player takes any card on the table. The second player then takes any one of the remaining cards. Play alternates back and forth with each player taking a remaining card. The winner is the first player to collect a combination of three cards whose sum is fifteen (such as 9-4-2).

You think about it and realize the game is flawed. Assuming both players utilize optimal strategy...

a) The player going first will win.
b) The player going second will win.
c) Neither player will win.

Bonus Question
The designers ask your opinion on what to name the game. What should you tell them?
Have you tried 22 tonight? I said 22.
charliepatrick
Joined: Jun 17, 2011
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October 31st, 2019 at 11:45:49 PM permalink
Quote: Gialmere

You think about it and realize the game is flawed. Assuming both players utilize optimal strategy...
a) The player going first will win.
b) The player going second will win.
c) Neither player will win.

There are only eight ways to have three numbers 1-9 add up to 15. Thus the objective is to prevent your opponent making any of these combinations.

Consider the following where you arrange the cards into a 3x3 perfect square, such as this...
 2 7 6 9 5 1 4 3 8

In essence you have to prevent your opponent from making any of the eight possible lines. If you can do this then the game will be a draw. So just use standard Noughts and Crosses strategy to pick your card and this will force a draw.
Gialmere
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November 1st, 2019 at 5:58:31 AM permalink
Correct. Well done again CP!

Although it's possible for either player to win, assuming optimal strategy, neither of them ever will. Essentially the designers have invented Tic Tac Toe.
Have you tried 22 tonight? I said 22.
Gialmere
Joined: Nov 26, 2018
• Posts: 1075
November 1st, 2019 at 5:49:40 PM permalink
Amazon Abstruseness
While shooting dice in a craps pit a software engineer standing next to you says that it reminds him of a test question he got while interviewing for a job at Amazon. You're given a pair of dice. One of them has the standard numbers of 1-6 on its faces. The other one is blank. Can you fill in each blank face with a number from 0-6 so that when the dice are rolled together they generate a number from 1-12 with equal probability?
Have you tried 22 tonight? I said 22.
beachbumbabs
Joined: May 21, 2013
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November 1st, 2019 at 10:10:51 PM permalink
Quote: Gialmere

Amazon Abstruseness
While shooting dice in a craps pit a software engineer standing next to you says that it reminds him of a test question he got while interviewing for a job at Amazon. You're given a pair of dice. One of them has the standard numbers of 1-6 on its faces. The other one is blank. Can you fill in each blank face with a number from 0-6 so that when the dice are rolled together they generate a number from 1-12 with equal probability?

The blank die would have to have 3 faces of 0 and 3 faces of 6.

You would then have 3 ways to make each number. A blank face + any number = 1-6, and a 6 + any number = 7-12. Each total has an equal probability.
If the House lost every hand, they wouldn't deal the game.
Gialmere
Joined: Nov 26, 2018
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November 2nd, 2019 at 4:05:36 PM permalink
Quote: beachbumbabs

The blank die would have to have 3 faces of 0 and 3 faces of 6.

You would then have 3 ways to make each number. A blank face + any number = 1-6, and a 6 + any number = 7-12. Each total has an equal probability.

Winner Winner Chicken Dinner!

An easy way to think of it is that you have 12 numbers and 36 possible outcomes. So each number must have 3 chances to come up. Because the only way to roll a 1 is 1+0, the blank die must have three 0s. To roll midnight, the blank die must also have three 6s. Somewhat surprisingly, doing this makes all the other numbers kosher as well.

This really was an Amazon interview question and it generated some minor buzz around ten years ago. As often happens here at WOV, the debate didn't center on how to solve the riddle but rather on how to correctly word the question. For example, here is one rather vague way I've seen of wording it...

Given a normal dice and a dice with blank faces, fill in the blank dice with numbers from 0-6 so that the probability of each number coming up, when you roll the two dice together is equal.

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NFL Nonsense
At a casino sportsbook you examine some futures bets on which NFL teams will make it into the playoffs. After thinking a while, you ask yourself two questions...

1) What is the minimum number of wins an NFL team needs to reach the playoffs?
2) What is the maximum number of losses an NFL team can suffer and still make the playoffs?