8 Table Games

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November 4th, 2019 at 7:38:57 PM permalink

Quote: Joeman
1. All intra-divisional games would have to end in a tie except one, and all inter-divisional/inter-conference games are losses for all 4 teams. Then you end up with:

Team A: 1-10-5
Team B: 0-10-6
Team C: 0-10-6
Team D: 0-11-5

And Team A makes the playoffs. (They probably will not have a bye!)

to 0 games. All divisional games could end in ties, leaving all teams with a record of 0-10-6!

13. If all teams in the division lose all their non divisional games, and split all their divisional games with each other, they will all end up with a 3-13 record. One of these teams will make the playoffs due to the appropriate tiebreaker.

Winner Winner Chicken Dinner!

1) The answer is 0 wins. As charliepatrick points out, this is sort of a trick question. Because regular season NFL games can end in a tie there are several ways for it to happen. For example, a team might finish the season 0-0-16 (which equates to 8-8-0) and manage to get in. (Under the current playoff rules Seattle once earned a spot with a record of 7-9-0.) The optimal solution, however, would be the worse possible record and Joeman is spot on by assuming a pathetic division where all four teams lose all their games outside the division and tie all their games inside it. The teams all end with winless records of 0-10-6 and the tiebreakers decide who gets the guaranteed playoff birth.

2) The answer is 13 losses. This solution is similar to #1 except you need to maximize the number of losses. Once again Joeman is absolutely correct to use the pathetic division where all four teams lose all their games outside the division. For the head-to-head games, instead of ties you now start handing out losses. Of course, for a team to lose a game the other team must win so you need to dole out an equal number of wins as well. You end up with all four teams sporting 3-13-0 records with tiebreakers once again deciding who reaches the postseason.

Note that a record of 0-10-6 is equivalent to a record of 3-13-0.

At first glance the NFL regular and postseason schedules have a lovely symmetry to them. Look closer though and you see some clunky rules that can allow very bad teams to reach the playoffs. (Go Bengals!) There have been calls to rectify the situation. An obvious fix would require a divisional winner to at least have a winning record to claim the automatic playoff birth. Otherwise the spot would convert to a wildcard.

--------------------

Flip Flummoxed
You and a friend want to quickly whip up a game to compete in this year's Cutting Edge table game competition. You decide that using a coin flip mechanic would be the simplest thing to do with punters wagering on various sequences coming up before a certain number of tosses. As you sit down to work on the math your friend asks you if the expected number of coin flips until you get two heads (HH) in a row is the same as the expected number of coin flips until you get a heads followed by a tails (HT). You answer...

A) Yes. The number of expected coin flips for both is equal.
B) No. The number of expected coin flips for HH is greater than for HT.
C) No. The number of expected coin flips for HT is greater than for HH.
D) Uhhhhhhh...

Last edited by: Gialmere on Nov 4, 2019

Have you tried 22 tonight? I said 22.

yogitagodara

yogitagodara

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November 4th, 2019 at 11:59:49 PM permalink

Hello,

Tricky question, definitely gets my up vote for your opinion of considering it as a puzzle.
I am also curious to know about the responses about this question.

Bingonice

Jessica

charliepatrick

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November 5th, 2019 at 1:01:23 AM permalink

Maybe I'm missing something but my understanding is the bet would be resolved on the toss following the first Head. So there can be any number of consecutive Tails, but as soon as there's a Head then on the next toss there will either be Head (in which case HH has happened) or Tail (in which case HT has happened). Since these are equally likely, the expected values are the same.

Gialmere

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November 5th, 2019 at 8:05:01 AM permalink

Quote: charliepatrick
Maybe I'm missing something but my understanding is the bet would be resolved on the toss following the first Head. So there can be any number of consecutive Tails, but as soon as there's a Head then on the next toss there will either be Head (in which case HH has happened) or Tail (in which case HT has happened). Since these are equally likely, the expected values are the same.

Unfortunately you are indeed missing something. Note that it's not an either/or question.

Have you tried 22 tonight? I said 22.

Gialmere

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November 6th, 2019 at 9:42:39 AM permalink

CP's logic is close but misses a possible path to sequence completion. Consider each sequence separately.

Have you tried 22 tonight? I said 22.

beachbumbabs

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November 6th, 2019 at 9:49:48 AM permalink

Still not seeing the twist.

Seems like it has something to do with the bet being HOW MANY flips in total before seeing the winning sequence, not just whether it happens. But that seems like equal probability to me.

If the House lost every hand, they wouldn't deal the game.

unJon

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November 6th, 2019 at 11:12:25 AM permalink

Quote: Gialmere
CP's logic is close but misses a possible path to sequence completion. Consider each sequence separately.

Greater number of flips expected for HH than HT. For HT, once you flip the first H, you are assured of getting the sequence the next time T flips. So you lock in the first step of sequence and never need to start from scratch. On HH, the first time you flip a H, you would still need to go back to scratch if a T flips next. So you might have to restart.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

Gialmere

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November 6th, 2019 at 3:44:56 PM permalink

Quote: unJon
Greater number of flips expected for HH than HT. For HT, once you flip the first H, you are assured of getting the sequence the next time T flips. So you lock in the first step of sequence and never need to start from scratch. On HH, the first time you flip a H, you would still need to go back to scratch if a T flips next. So you might have to restart.

Winner! Winner! Chicken Dinner!

unJon's version is ready for field trials!

The answer is B "The number of expected coin flips for HH is greater than for HT."

If you flip a coin two times, the four possible sequences generated are HH, HT, TH and TT. Each sequence has an equal 25% chance of occurring. This is a rather intuitive concept. If, however, you specify one of those sequences occurring in a longer sequence, strange things start to happen.

An easy way to think of it is to imagine that both HH and HT are at a starting line, both of them a minimum of two flips away from completion. If the next coin flip is tails then they both remain at the starting line since T is not the first needed result in their sequences. If the following flip is heads then they both move forward to a midpoint between start and finish, each now a minimum of one flip away from completion.

If you bet on HH, and the next flip is heads, HH crosses the finish line by completing its sequence. If, however, the next flip is tails, then HH must move back to the starting line and revert to a minimum of two flips from completion.

Now consider HT at the midpoint. If the next flip is tails, HT crosses the finish line by completing its sequence. If, however, the next flip is heads, then HT stays at the midpoint--still a one flip minimum from completion--because although H is not the second needed result in its sequence, it is still the first needed result.

In other words, from the midpoint HH will either complete its sequence or end up with a two flip minimum to completion while HT will either complete its sequence or stay at a one flip minimum to completion. Because of this, the expected number of coin flips to achieve HH is 6, while the the expected number of coin flips to achieve HT is only 4. (Note that HH=TT and HT=TH in terms of probability.)

--------------------------

Weird Words
With your coin flip game finished it's time to go to the Cutting Edge competition. While making that loooong drive across the desert to Vegas you pull over at a rest stop to attend to the needs of the body. In your stall (amid all the phone numbers and middle school witticisms scrawled everywhere) you read this...

There are are five things
wrong with this sentence;
only geniuses will be able to
to spot all of the mitstakes

Is this statement true or false?

[Try to hurry with this one. It smells really bad in here.]

Have you tried 22 tonight? I said 22.

charliepatrick

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November 6th, 2019 at 4:19:53 PM permalink

Neither as it is a paradox.

Initially ignoring the "there are five things wrong" phrase, there are four things wrong (1) are are (2) to to (3) mitsakes (4) lack of full stop.

Thus at this stage there aren't five things wrong, so the phrase "there are five things wrong" is also wrong.

Thus at this stage the phrase "there are five things wrong" becomes correct; and now the sentence only has four things wrong.

Reduce ad absurdum.

This is a version of the card which has on the front: "The sentence on the other side of this card is TRUE." and on the back "The sentence on the other side of this card is FALSE."

ThatDonGuy

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November 6th, 2019 at 4:33:15 PM permalink

Quote: charliepatrick
Neither as it is a paradox.

Initially ignoring the "there are five things wrong" phrase, there are four things wrong (1) are are (2) to to (3) mitsakes (4) lack of full stop.

Thus at this stage there aren't five things wrong, so the phrase "there are five things wrong" is also wrong.

Thus at this stage the phrase "there are five things wrong" becomes correct; and now the sentence only has four things wrong.

Reduce ad absurdum.

This is a version of the card which has on the front: "The sentence on the other side of this card is TRUE." and on the back "The sentence on the other side of this card is FALSE."

(5) it's a run-on sentence - either that, or two sentences separated by a semicolon instead of a period.

gordonm888
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November 6th, 2019 at 4:54:55 PM permalink

Answer #1
The so-called sentence is actually two statements. The first is declarative: "There are are five things etc....." The second one is a prediction about the future: "Only geniuses will be able to to spot all the mitstakes."

It is impossible to declare a prediction about the future to be true or false. Also, the two statements need not be 'True, True' or 'False, False.'

Answer #2
Five things wrong with the sentence.

1. Sentence is a run-on.
2. Mitstakes (spelling)
3. are are
4. to to
5. There are only four things wrong with the sentence.

Given the fifth defect of the sentence, the truthfulness of the sentence is a logical loop. In order to be true, it must be false, in which case it is true -so it is false.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

Gialmere

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November 7th, 2019 at 5:17:29 PM permalink

Quote: charliepatrick
Neither as it is a paradox.

Initially ignoring the "there are five things wrong" phrase, there are four things wrong (1) are are (2) to to (3) mitsakes (4) lack of full stop.

Thus at this stage there aren't five things wrong, so the phrase "there are five things wrong" is also wrong.

Thus at this stage the phrase "there are five things wrong" becomes correct; and now the sentence only has four things wrong.

Reduce ad absurdum.

This is a version of the card which has on the front: "The sentence on the other side of this card is TRUE." and on the back "The sentence on the other side of this card is FALSE."

Quote: gordonm888

Answer #1
The so-called sentence is actually two statements. The first is declarative: "There are are five things etc....." The second one is a prediction about the future: "Only geniuses will be able to to spot all the mitstakes."

It is impossible to declare a prediction about the future to be true or false. Also, the two statements need not be 'True, True' or 'False, False.'

Answer #2
Five things wrong with the sentence.

1. Sentence is a run-on.
2. Mitstakes (spelling)
3. are are
4. to to
5. There are only four things wrong with the sentence.

Given the fifth defect of the sentence, the truthfulness of the sentence is a logical loop. In order to be true, it must be false, in which case it is true -so it is false.

Two Winner Winners!! Two Chicken Dinners!!

The classic answer is the sentence falls into paradox, but give ThatDonGuy (and gordonm888 - answer #1) credit for thinking outside the box. Speaking of boxes...

------------------------------

Chip Chase
Well, viva Las Vegas. You've made it to Cutting Edge and are attending a meet-and-greet for designers. To your surprise, your name is called out as a possible door prize winner. You are taken to a table that has five boxes (numbered 1-5) placed in a line and told that one of the boxes contains a $5000 casino chip.

Also on the table is a bag containing ten keno balls (numbered 1-10). You need to draw a ball from the bag to indicate how many guesses you get at which box contains the five grand cheque. The catch is, after each incorrect guess, you must turn your back and the chip is moved to an adjacent box before you can guess again.

Fortunately, being a game designer, you've seen this sort of mechanic before and realize that the game is beatable. You just need to get lucky with the draw.

What is the lowest numbered keno ball you can draw that guarantees you'll find the $5000 chip?

Have you tried 22 tonight? I said 22.

charliepatrick

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November 8th, 2019 at 1:38:50 AM permalink

I'm not sure this is minimum but it's an idea.

On turns 1 and 2 go to 4. This picks up the chip if it started in 5 (5 4) or (4). Otherwise the chip has now moved to 1,2 or 3.

On turns 3 and 4 go to 2. This picks up the chip if (after 2 moves) it was in 1 (x x1 2) or 2 (x x2).
If the chip is still not found then it started in 3 after turn 2, moved to 4, and so is sitting in 5 or 3 (x x3 45) (x x3 43).

On turn 5 go to 4, this picks up the chip if it was in 5 (as it has now to move back to 4 (x x3 45 4) ) and also picks up 3 and went to 4 (x x3 43 4)
If the chip is still not found then it is in 2 (x x3 43 2), it is therefore going to either 3 or 1.

On turn 6 go to 3, this picks up the chip if it went to 3 (x x3 43 23). Otherwise it is sitting in 1 (x x3 43 21).

On turn 7 go to 2, this picks up the chip as it has to return to 2 having been in 1 (x x3 43 21 2).

beachbumbabs

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November 8th, 2019 at 9:20:36 AM permalink

Turn 1. Pick 4. If chip, you win. If not, go again.
Chip in 1. Must move to 2.
Chip in 2. Moves to either 1 or 3.
Chip in 3. Moves to either 2 or 4.
Chip in 5. Must move to 4.

Turn 2. Pick 4. If chip, win. If not, go again.
Chip could not have been in 5 or 4 on turn 1 or you would have won.
Chip 1. Moves to 2.
Chip 2. Moves to 1 or 3.
Chip 3. Moves to 2 or 4.

Turn 3. Pick 4. If chip, win. If not, go again.
Chip 1. Moves to 2.
Chip 2. Moves to 1 or 3.
Chip 3 (previous)must have moved to 2, or you would have won this turn. So no chip can move higher than 3 now.

Turn 4. Pick 3. If chip, win. If not, go again.
Chip1. Moves to 2.
Chip 2. Moves to 1 or 3.

Turn 5. Pick 3. If chip, win. If not, go again.
Chip 1. Must move to 2.
Chip 2 (previous) must not have moved to 3 or you would have won this turn, so chip was in 1 and must move to 2.

Turn 6. Pick 2. Win.

So you must have at least 6 chances to guarantee a win. 5 lotto balls 6 or higher, 5 lotto balls 1-5, so you have 5/10 chance of guaranteeing a win, but you have 10/10 chances of getting an opportunity to win.

If the House lost every hand, they wouldn't deal the game.

charliepatrick

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November 8th, 2019 at 11:38:13 AM permalink

I'm not sure the method of working your way down and hoping to trap the chip works. Essentially the chip can swap positions with your guesses. I'm not sure your statement "no chip can move higher than 3 now" works.

Let's presume each hour you are let into the room to make a guess. You make your guess on the hour. If you are correct you win. If you are wrong then you leave the room, the chip is moved at half past before your next guess.

Let's assume your "Turn 1" was as 12 noon, and "Turn 2" at 1pm. I agree that at 1:15pm the chip is in 1, 2 or 3. By 1:45 it can be in 1, 2, 3 or 4.

For instance suppose (at 2pm) you are looking in box #4 and on the next turn planning to go to #3; then the chip might currently (2pm) be at #3. Your guess at #4 is wrong, so the chip will be moved (at 2:30pm) to #4 before your next look (at 3pm) into box #3. Thus it has got away from your net.

Essentially if you can get to a point where you can determine the only box the chip is in is odd (1,3,5) then you know it can only be in 2 or 4 next time. By guessing 2 you now know it's in 4, so it has to move to 3 or 5 next time. By guessing 3 you now know it's in 5, so it has to move to 4. Thus you guess 4 and win.

By starting out picking 4 for the first two rounds, you know (before moving for round 3) the Chip is in box 1, 2 or 3.
By turn 3 the chip might have moved to 4, so can be in 1, 2, 3 or 4. So you pick 3. This leaves the chip in 1, 2 or 4.

By turn 4 the chip cannot be in box 4, but could be in 1, 2, 3 or 5. By guessing box 2, you now know the chip is in 1, 3 or 5. hey presto.

	B O X	B O X	B O X	B O X	B O X	Comments
Time	1	2	3	4	5	Boxes one to five
12:00				PICK
12:15	1	2	3		5	Possiblities remaining after picking
12:30						Chip is moved to adjacent box
12:45	1	2	3	4		Possible positions after one space movement
13:00				PICK
13:15	1	2	3			Possiblities remaining after picking
13:30						Chip is moved to adjacent box
13:45	1	2	3	4		Possible positions after one space movement
14:00			PICK
14:15	1	2		4		Possiblities remaining after picking
14:30						Chip is moved to adjacent box
14:45	1	2	3		5	Possible positions after one space movement
15:00		PICK
15:15	1		3		5	Possiblities remaining after picking
15:30						Chip is moved to adjacent box
15:45		2		4		Possible positions after one space movement
16:00		PICK
16:15				4		Possiblities remaining after picking
16:30						Chip is moved to adjacent box
16:45			3		5	Possible positions after one space movement
17:00			PICK
17:15					5	Possiblities remaining after picking
17:30						Chip is moved to adjacent box
17:45				4		Possible positions after one space movement
18:00				PICK
18:15						Possiblities remaining

beachbumbabs

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November 8th, 2019 at 3:08:37 PM permalink

It might be a question of my phrasing, Charlie, but I think my method works, and the chip can't slide past. You can do a mirror of it by starting at 2, but assuming

1. the chip MUST move each time, and

2.MUST be to an adjacent square,

you can trap it by choosing 4,4,4,3,3,2 (if you need that many).

The reason is, as I said, by looking at the chip's past history when you don't find it at the 3rd 4 (and the 2nd 3), it isn't in a position where it can reach beyond you.

I could be wrong, no question. But I think that's the trick.

If the House lost every hand, they wouldn't deal the game.

charliepatrick

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November 8th, 2019 at 3:20:56 PM permalink

This is my understanding, so I might be wrong.

If the chip starts at position 1 and initially just trundles up one each time:-
Round 1, Chip starts at 1. Guess 4. Chip moves to 2.
Round 2. Chip starts at 2. Guess 4. Chip moves to 3.
Round 3. Chip starts at 3. Guess 4. Chip moves to 4, (This is where the chip escapes.)
Round 4. Chip starts at 4. Guess 3. Chip moves to 5.
now chip can flip between 4 and 5.

beachbumbabs

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November 8th, 2019 at 3:27:01 PM permalink

Quote: charliepatrick
This is my understanding, so I might be wrong.

If the chip starts at position 1 and initially just trundles up one each time:-
Round 1, Chip starts at 1. Guess 4. Chip moves to 2.
Round 2. Chip starts at 2. Guess 4. Chip moves to 3.
Round 3. Chip starts at 3. Guess 4. Chip moves to 4, (This is where the chip escapes.)
Round 4. Chip starts at 4. Guess 3. Chip moves to 5.
now chip can flip between 4 and 5.

That's where i started. And perhaps should have stayed. I had a glimmer in there, though, that while this seems right, there's something else that has to happen, and now I can't recapture it.

Maybe you have to do one more 4, and the chip is forced backwards or you find it.

If the House lost every hand, they wouldn't deal the game.

charliepatrick

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November 8th, 2019 at 3:55:54 PM permalink

Quote: beachbumbabs
...Maybe you have to do one more 4, and the chip is forced backwards or you find it.

I'm not so sure as I think however many consecutive (>1) 4's you start with the only thing you know is while you were making your guess the chip was somewhere in 1, 2 or 3. In your scenario as soon as you change your guess from 4 to 3 it doesn't cater for the chip which will have been in 3 and be moved to 4.

Gialmere

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November 8th, 2019 at 5:29:45 PM permalink

Wow! Reading these responses makes me feel like a Gumby...

BBB is correct about the number of picks (6) but incorrect on the logic.
CP is incorrect about the number of picks but has almost flawless logic.

I was going to call it a split-decision (and probably still will) but will let it ride 'till tomorrow in case anyone wants to try the full solve.

Have you tried 22 tonight? I said 22.

unJon

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November 8th, 2019 at 6:50:45 PM permalink

This is a great puzzle. Did it on my fingers on train so may have missed a trick but think this works:

Box 4, Box 4, Box 2, Box 2, Box 3, Box 3, Box 4, Box 4

ETA: ugh see that there�s a method to do it in less picks.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

Gialmere

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November 8th, 2019 at 6:56:08 PM permalink

One trick is to reduce the number of boxes to three and solve for both odd and even.

Have you tried 22 tonight? I said 22.

unJon

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November 8th, 2019 at 8:38:19 PM permalink

Got it:

box 4, box 4. If you didn�t get it, it must now be in 1, 2 or 3. You can eliminate 1 and 2 by going Box 2, Box 2 (which will win if it was Box 1 or 2 at start of round 3). That means it must have been in Box 3 and went to Box 4. So it is now either in Box 3 or 5. Move Box 3. If it�s a miss, it was in Box 5 so now moved to Box 4. So an always win in 6 moves is:

Box 4
Box 4
Box 2
Box 2
Box 3
Box 4

Last edited by: unJon on Nov 8, 2019

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

Gialmere

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November 8th, 2019 at 9:29:39 PM permalink

Quote: unJon
Got it:

box 4, box 4. If you didn�t get it, it must now be in 1, 2 or 3. You can eliminate 1 and 2 by going Box 2, Box 2 (which will win if it was Box 1 or 2 at start of round 3). That means it must have been in Box 3 and went to Box 4. So it is now either in Box 3 or 5. Move Box 3. If it�s a miss, it was in Box 5 so now moved to Box 4. So an always win in 6 moves is:

Box 4
Box 4
Box 2
Box 2
Box 3
Box 4

The rakish dealer running the game smiles and shrugs as your final pick is an empty box. Suppose the chip starts in...

x Box 2 ... You pick Box 4
>Box 3 ... You pick Box 4
>Box 4 ... You pick Box 2
<Box 3 ... You pick Box 2
<Box 2 ... You pick Box 3
<Box 1 ... You pick Box 4 (D'oh!)

Hint: Note that the chip must always alternate between odd and even numbers every turn.

Have you tried 22 tonight? I said 22.

charliepatrick

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November 9th, 2019 at 2:17:20 AM permalink

Phase one chase down the chips that started in an even position.
Round 1 - pick 4 - this gets rid of a chip starting in position 4.
Round 2 - pick 3 - this gets rid of a chip starting in position 2 that then moved to 3.
Round 3 - pick 2 - this gets rid of a chip starting in position 2 that moved to 1 and is now forced to move to 2.

At this stage only the chips that started in 1, 3 of 5 have survided. In round 4 they must be moved to positions 2 or 4. So you can use the same logic to chase them down.

In theory you could repeat the logic for Rounds 1 thru 3 but the mirror logic of 2 3 4 also works.

	B O X	B O X	B O X	B O X	B O X	Comments
Time	1	2	3	4	5	Boxes one to five
12:00				PICK
12:15	1	2	3		5	Possiblities remaining after picking
12:30						Chip is moved to adjacent box
12:45	1	2	3	4		Possible positions after one space movement
13:00			PICK
13:15	1	2		4		Possiblities remaining after picking
13:30						Chip is moved to adjacent box
13:45	1	2	3		5	Possible positions after one space movement
14:00		PICK
14:15	1		3		5	Possiblities remaining after picking
14:30						Chip is moved to adjacent box
14:45		2		4		Possible positions after one space movement
15:00		PICK
15:15				4		Possiblities remaining after picking
15:30						Chip is moved to adjacent box
15:45			3		5	Possible positions after one space movement
16:00			PICK
16:15					5	Possiblities remaining after picking
16:30						Chip is moved to adjacent box
16:45				4		Possible positions after one space movement
17:00				PICK
17:15						Possiblities remaining

beachbumbabs

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November 9th, 2019 at 7:43:21 AM permalink

I think you have it, Charlie! Good job.

If the House lost every hand, they wouldn't deal the game.

unJon

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November 9th, 2019 at 8:16:28 AM permalink

Quote: beachbumbabs
I think you have it, Charlie! Good job.

Agree

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

Gialmere

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November 9th, 2019 at 5:32:33 PM permalink

Quote: charliepatrick

Phase one chase down the chips that started in an even position.
Round 1 - pick 4 - this gets rid of a chip starting in position 4.
Round 2 - pick 3 - this gets rid of a chip starting in position 2 that then moved to 3.
Round 3 - pick 2 - this gets rid of a chip starting in position 2 that moved to 1 and is now forced to move to 2.

At this stage only the chips that started in 1, 3 of 5 have survided. In round 4 they must be moved to positions 2 or 4. So you can use the same logic to chase them down.

In theory you could repeat the logic for Rounds 1 thru 3 but the mirror logic of 2 3 4 also works.

B O X
B O X
B O X
B O X
B O X
Comments

Time
1
2
3
4
5
Boxes one to five

12:00
PICK

12:15
1
2
3
5
Possiblities remaining after picking

12:30
Chip is moved to adjacent box

12:45
1
2
3
4
Possible positions after one space movement

13:00
PICK

13:15
1
2
4
Possiblities remaining after picking

13:30
Chip is moved to adjacent box

13:45
1
2
3
5
Possible positions after one space movement

14:00
PICK

14:15
1
3
5
Possiblities remaining after picking

14:30
Chip is moved to adjacent box

14:45
2
4
Possible positions after one space movement

15:00
PICK

15:15
4
Possiblities remaining after picking

15:30
Chip is moved to adjacent box

15:45
3
5
Possible positions after one space movement

16:00
PICK

16:15
5
Possiblities remaining after picking

16:30
Chip is moved to adjacent box

16:45
4
Possible positions after one space movement

17:00
PICK

17:15
Possiblities remaining

Ding! Ding!

Winner! Winner! Chicken Dinner!!

Correct. The Patterns 2, 3, 4, 4, 3, 2 -or- 4, 3, 2, 2, 3, 4 solve every time.

[Note that if there were seven boxes, then use 2, 3, 4, 5, 6, 6, 5, 4, 3, 2 etc.]

CP shoots ... AND SCORES!! with an assist from BBB. Unfortunately...

-------------------------------------

Box Boggled
D'oh! Just your luck. You knew how to win the $5k but drew keno ball #1 and your lone pick missed. Your host says you still have a chance at a consolation prize. Your led to another table with three numbered boxes on it. (Oh great, you think, more boxes. But at least there's only three.) You're told that one of the boxes contains a $500 dollar casino chip and you get one pick to find it. Each box has a statement written on it and you're told that only one statement is true...

Box #1: THE $500 DOLLAR CHIP IS IN THIS BOX.
Box #2: THE $500 DOLLAR CHIP IS NOT IN THIS BOX.
Box #3: THE $500 DOLLAR CHIP IS NOT IN BOX #1.

Which box do you choose?

Last edited by: Gialmere on Nov 9, 2019

Have you tried 22 tonight? I said 22.

beachbumbabs

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November 9th, 2019 at 7:44:13 PM permalink

Box 2.

If chip is box 1, statements 1 and 2 are true.
If chip is box 2, only statement 3 is true.
If chip is box 3, statements 2 and 3 are true.

If the House lost every hand, they wouldn't deal the game.

Gialmere

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November 10th, 2019 at 12:08:51 PM permalink

Quote: beachbumbabs
Box 2.

If chip is box 1, statements 1 and 2 are true.
If chip is box 2, only statement 3 is true.
If chip is box 3, statements 2 and 3 are true.

Winner! Winner! Chicken Dinner!!

------------------------------------

Dealer's Dilemma
Woohoo! You've won $500 dollars and the night is still young. Three other designers invite you up to a comped suite for a relaxing game of bridge. (By "relaxing" they mean that each player is wagering $500 on winning with partners splitting $1000.) Life in the fast lane, you think as you agree to play and hopefully do some networking. You head to the elevators.

It's a friendly game with no cheating. Since all players are designers (i.e. rule makers), however, strict table etiquette is observed. As things progress, the game is very close and it's your turn to deal. Then, in the middle of distributing the cards, you get an important cell call from the company delivering your CE booth supplies. You immediately excuse yourself from the table. There's a shipping holdup, but it can be quickly resolved if you grease a few union palms. You sigh and agree to the new terms hoping you win the bridge game to cover the new expenses.

You return to the table and (horror of horrors) realize you forgot where you left off with the deal. The other players (who have been drinking and talking shop) don't remember either. How embarrassing. Is there a way for you to correctly finish the deal without prolonging the awkward moment by counting all the cards dealt or that remain in the deck? Should you declare a misdeal?

Have you tried 22 tonight? I said 22.

charliepatrick

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November 10th, 2019 at 1:21:40 PM permalink

Co-incidentally I was playing teams of eight today (don't ask!)

If the players are happy you start from yourself and deal backwards (i.e. counterclockwise).

If the players want the same cards they would have got, then you deal the cards, one at a time, onto a pile and then deal that as above. For instance you were due to get the 52nd card, the the last card would now be at the top of the pile. East would get the 51st card, now 2nd on the pile, etc.

Gialmere

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November 11th, 2019 at 9:07:31 PM permalink

Quote: charliepatrick
Co-incidentally I was playing teams of eight today (don't ask!)
If the players are happy you start from yourself and deal backwards (i.e. counterclockwise).

If the players want the same cards they would have got, then you deal the cards, one at a time, onto a pile and then deal that as above. For instance you were due to get the 52nd card, the the last card would now be at the top of the pile. East would get the 51st card, now 2nd on the pile, etc.

Winner! Winner! Chicken Dinner!!
(Combined your responses are close enough.)

Since bridge is a four player game where all the cards are distributed, the dealer will always receive the last card. Knowing this, you can quickly rectify the situation by dealing the first card at the bottom of the deck to yourself (the dealer) and continue dealing from the bottom of the deck backwards around the table. Doing this will give each player the cards they were meant to receive, show which player you left off at, and make you look smart all at the same time (assuming you're not accused of cheating).

---------------------------------

Calendar Cubes
Well done! You've won the bridge game and now have two $500 dollar chips. You decide to cash them in at the cashiers cage before you go to bed so you'll have money on hand for tomorrow's early deliveries. As you leave the cage you pass by a gift shop and see a block calendar in the window. It occurs to you that, properly themed, such a calendar would make a clever advertising giveaway if you ever design a dice game.

Heading up the elevator you wonder about the numbers on the blocks. Off hand it doesn't seem like there's enough faces. Hmm... You decide to figure it out before bed. What numbers must be placed on each cube?

Cube #1 numbers = ?
Cube #2 numbers = ?

Have you tried 22 tonight? I said 22.

unJon

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November 11th, 2019 at 9:23:22 PM permalink

Fun. Guess based on an assumption.

Both cubes have to have 0 to get through 01 to 09. Both cubes need a 1 and 2 for 11 and 22. That leaves 7 numbers and only 6 spaces. So let�s assume the side with 6 can do double duty as a 9 when upside down? If that works then:

Cube 1 can have 0-5. Cube 2 can have 0-2 and 6-8 with the 6 serving as a 9 sometimes.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

Gialmere

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November 12th, 2019 at 4:46:36 PM permalink

Quote: unJon
Fun. Guess based on an assumption.

Both cubes have to have 0 to get through 01 to 09. Both cubes need a 1 and 2 for 11 and 22. That leaves 7 numbers and only 6 spaces. So let�s assume the side with 6 can do double duty as a 9 when upside down? If that works then:

Cube 1 can have 0-5. Cube 2 can have 0-2 and 6-8 with the 6 serving as a 9 sometimes.

Winner! Winner! Chicken Dinner!!

Absolutely correct. On a side note, the two-cube calendar was originally created, not as a commercial product, but simply as a math/logic riddle.

-------------------------------

Mate Mystery
With the cubes solved you try to get some sleep but, between the excitement and worry, it's only for a few hours. Oh well. You have to be there at the crack of dawn to receive the delayed shipments anyways. You go down to the booth area, set up what you can and start what you hope will be a short wait.

On a nearby table you see a newspaper and a few magazines. One of these is a chess periodical and you flip to the problems section hoping to pass the time. You decide to start with an easy problem and see a "white mates in one". You stare at it ... then sit down and stare some more.

White Mates in One...

Have you tried 22 tonight? I said 22.

charliepatrick

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November 13th, 2019 at 2:53:12 AM permalink

It's actually an impossible board as all 32 pieces are still on the board, for two pawns to be in the same column one of them must have taken something.

That was a hint!

beachbumbabs

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November 13th, 2019 at 10:18:03 AM permalink

I don't speak chess nomenclature (sorry for my ignorance, Ed!), but if the white bishop on the dark squares moves northeast to take the black pawn, all the king's moves are covered, and he is in check.

White knight covers move N. Same bishop covers move NE and SW. Current position is in check. All other possible black king squares are occupied, and he can't take his own piece.

ETA. OK, I see why my solution doesn't work. But I also don't understand the arcane abbreviations and specialized move referenced, so I'll bow out.

Last edited by: beachbumbabs on Nov 13, 2019

If the House lost every hand, they wouldn't deal the game.

gordonm888
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November 13th, 2019 at 10:56:17 AM permalink

The only way this can be solved is if the black pawn on E5 has just moved forward two spaces. In that case P(D5) X P(E5) en passant. The pawn on D5 goes to E6 uncovering the white Bishop's attack on the Black King, Checkmate. Thank You CP for hint.

Last edited by: gordonm888 on Nov 13, 2019

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

charliepatrick

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November 13th, 2019 at 11:01:44 AM permalink

^ The writing for your move is B x e5 ch (Bishop, takes piece on, e5; check). While the King cannot move Black's reply is to take the Bishop using the Pawn at d6: P x e5 (i.e. the Pawn, takes the, Bishop).

If only you could get rid of that Pawn!

Joeman

Joeman

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November 13th, 2019 at 11:31:44 AM permalink

d5 x e6# e.p. -- Assuming Black's last move was pawn to e5, white can take it en passant for a discovered checkmate by the black bishop. This one had me stumped until that last hint.

ETA -- I see that Gordon is smarter than me as he only needed one of Charlie's hints to solve it.

Last edited by: Joeman on Nov 13, 2019

"Dealer has 'rock'... Pay 'paper!'"

Gialmere

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November 13th, 2019 at 5:19:32 PM permalink

Quote: gordonm888
The only way this can be solved is if the black pawn on E5 has just moved forward two spaces. In that case P(D5) X P(E5) en passant. The pawn on D5 goes to E6 uncovering the white Bishop's attack on the Black King, Checkmate. Thank You CP for hint.

Winner! Winner! Chicken Dinner!!

(Joeman places and Grandmaster CP gets the assists.)

This problem caused quite a stir at the site I saw it at (which wasn't a chess site). Some loved it while others panned it siting some of the reality issues mentioned here. In the designer's defense, it's meant more to be a logic problem where the solution actually leads you back to the preceding turn (although you obviously need to know the rules of chess).

------------------------------------------

Direct Delivery
You put the chess magazine down as you're greeted by three other designers. The four of you sip coffee and talk in your booth area. Just as you start to explain your missing package problem, a passing delivery guy stops, stares directly at you and asks if you're John Smith. [For purposes of this thread narrative, you are.] You smile and sign for the delivery.

Neither you nor the other designers at your booth are wearing name tags yet. How did the delivery guy guess you were John Smith?

Have you tried 22 tonight? I said 22.

gordonm888
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November 13th, 2019 at 6:42:55 PM permalink

Quote: Gialmere

Direct Delivery
You put the chess magazine down as you're greeted by three other designers. The four of you sip coffee and talk in your booth area. Just as you start to explain your missing package problem, a passing delivery guy stops, stares directly at you and asks if you're John Smith. [For purposes of this thread narrative, you are.] You smile and sign for the delivery.

Neither you nor the other designers at your booth are wearing name tags yet. How did the delivery guy guess you were John Smith?

The other 3 designers were all female????

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

charliepatrick

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November 13th, 2019 at 6:53:00 PM permalink

The obvious answer is the rest were female.

Another option is you were wearing something different from the others - e.g. a company colour which matched the "John Smith, Big Blue Company"; a dog collar which matched the Reverend John Smith; etc.

Another option is you were the nearest to the door and hence were lucky enough to be the first to be asked.

Another option, assuming one is upposed to pick up hints in the question, is you were "explaining your missing package" which the delivery man overheard.

I have a feeling a melted block of ice in the Australian desert might be raisjng it's head soon!

Gialmere

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November 14th, 2019 at 5:07:07 PM permalink

Quote: gordonm888
The other 3 designers were all female????

Winner! Winner! Chicken Dinner!!

It'll be interesting to see how long this type of riddle (relying on gender assumptions) lasts in the 21st century.

------------------------------------

Ferris Fizzbin
One of the women designers you're talking to is none other than Heather Ferris. Being a big Star Trek fan, she has designed a casino version of the game called Fizzbin which appears in the original TV series. This is a nonsensical card game that Capt Kirk makes up on the spot to confuse and distract some guards so that he, Spock and and McCoy can get the jump on them. Many fans have made home versions of the game over the years.

For Ferris Fizzbin six community cards are dealt to the middle of the table à la Hold-em. The number of pocket cards a given player receives varies from 0-6 based on several silly factors such as the day of the week, where you sit at the table, if it's day or night etc. The highest ranking hand is a Royal Fizzbin. This consists of one queen, four clubs, two jacks (aka a half fizzbin), three aces and two diamonds.

What is the minimum number of hole cards you'll need to have a shot at a royal fizzbin?

Have you tried 22 tonight? I said 22.

unJon

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November 14th, 2019 at 5:35:33 PM permalink

Quote: Gialmere
Quote: gordonm888
The other 3 designers were all female????

Winner! Winner! Chicken Dinner!!

It'll be interesting to see how long this type of riddle (relying on gender assumptions) lasts in the 21st century.

------------------------------------

Ferris Fizzbin
One of the women designers you're talking to is none other than Heather Ferris. Being a big Star Trek fan, she has designed a casino version of the game called Fizzbin which appears in the original TV series. This is a nonsensical card game that Capt Kirk makes up on the spot to confuse and distract some guards so that he, Spock and and McCoy can get the jump on them. Many fans have made home versions of the game over the years.

For Ferris Fizzbin six community cards are dealt to the middle of the table à la Hold-em. The number of pocket cards a given player receives varies from 0-6 based on several silly factors such as the day of the week, where you sit at the table, if it's day or night etc. The highest ranking hand is a Royal Fizzbin. This consists of one queen, four clubs, two jacks (aka a half fizzbin), three aces and two diamonds.

What is the minimum number of hole cards you'll need to have a shot at a royal fizzbin?

Assuming four clubs and two diamonds means four cards of the club suit (and not the four of clubs), etc, and that the game is played with a normal 52 card deck:

You have to have a club in your hand. The rest can be on the board.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

Joeman

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November 15th, 2019 at 4:54:30 AM permalink

I agree with unJon, but the odds against getting a Royal Fizzbin are astronomical! :)

A Piece of the Action Might be my favorite Original Series episode. Now get me some of them fancy heaters! ;)

"Dealer has 'rock'... Pay 'paper!'"

gordonm888
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November 16th, 2019 at 9:46:01 AM permalink

I would have explained it differently.

You must have 4 clubs + 2 diamonds so that is six cards. But you also must have three aces, which means at least one ace that is not a club or diamond. So, you must have a minimum of 7 cards, which is one more card than the six on the board. So , one hole card is the minimum.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

Gialmere

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November 16th, 2019 at 1:48:58 PM permalink

Quote: unJon
Assuming four clubs and two diamonds means four cards of the club suit (and not the four of clubs), etc, and that the game is played with a normal 52 card deck:

You have to have a club in your hand. The rest can be on the board.

Quote: gordonm888
I would have explained it differently.
You must have 4 clubs + 2 diamonds so that is six cards. But you also must have three aces, which means at least one ace that is not a club or diamond. So, you must have a minimum of 7 cards, which is one more card than the six on the board. So , one hole card is the minimum.

Two Winner! Winners! Two Chicken Dinners!!

--------------------------------------

McNugget McNumbers
You finish building your booth setup just as the doors open and the public enters. It's a busy morning. Although your game is not getting much placement interest, the spectacle of punters wagering on a dealer flipping a coin is unusual (or perhaps absurd) enough to draw a lot of attention. By mid-afternoon, however, the crowds start to thin. With the initial excitement wearing off, you find yourself exhausted from little sleep and even less food intake. You decide to take a break and head to the food court for a quick bite.

You get a burger and fries at McDonald's and sit down to eat. At a nearby table, a crowd of CE attendees from the Wizard of Vegas website is hooting and hollering. From what you can tell, they are bidding on who can eat the most Chicken McNuggets. The highest bidder purchases the required number and proceeds to gorge on them.

You glance at the menu and see that this food court McDonald's sells McNuggets in boxes of 6, 9 and 20. So if you want 32 McNuggets you would buy a box of 20 and two boxes of 6. Of course, if you wanted 16 McNuggets you would have to buy 18 and have 2 extra. You wonder if these extras effect the rules at all.

At this particular McDonald's, what is the highest number of McNuggets that CANNOT be exactly bought with a combination of the available boxes?

Have you tried 22 tonight? I said 22.

charliepatrick