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charliepatrick
charliepatrick
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November 4th, 2019 at 9:48:37 AM permalink
Clearly if every team tied then it would be decided by, something like, points scored. So you don't need any wins to make the playoffs.

With 2) you need to clarify what the rules are.
I'm assuming that
(i) teams are divided into two halves (conferences) who will occasionally play each other.
(ii) On each side there are four groups of four teams and to qualify you must
either (a) be the winner of your group
or (b) have one of the best two records (presumably W-T-L then ties split by something else).
(iii) Each side plays each other side within their group twice
(iv) Each side (do they?) plays all the other teams within their half (conference) - I don't think this can be true as it would lead to too many games played: 6 within your group and 12 with the other teams.

My feeling is the winners of the division win every match and the 2nd/3rd/4th teams have the same record but are split by other factors.

However it is complicated because teams in one half (conference) could lose all their inter-conference games, and we don't know how many games that leaves.


There are 16 teams and each team plays 6+12 matches = 18 matches.
This means there are 16*18/2 matches in total = 144.
The leaders in each group win all their matches, so that's 4 * 18 gone = 72.
The second place in one of the group win all their matches except against the leaders = 18-2-1-1-1 = 13.
So there's 59 matches to share between the other 11 teams.
So at least one team has to win 6 matches and can lose 12.

Personally I'd use the same logic whatever the actual rules are.

Assume winners win everything, one second place wins as many as possible, all other teams lose their inter-conference matches, determine the number of matches left and divide by 11 rounding up. That's how many the other playoff place wins and hence you know how many it loses.
Joeman
Joeman
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Gialmere
November 4th, 2019 at 11:08:55 AM permalink
Quote: Gialmere

1) What is the minimum number of wins an NFL team needs to reach the playoffs?


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!


Quote:

2) What is the maximum number of losses an NFL team can suffer and still make the playoffs?

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.


Quote:

What does your brain answer back?

So, you're saying the Jags still have a chance!
"Dealer has 'rock'... Pay 'paper!'"
Gialmere
Gialmere
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MrCasinoGames
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
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
Gialmere
Joined: Nov 26, 2018
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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
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
Administrator
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
unJon
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Gialmere
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
Gialmere
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MrCasinoGames
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.

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