My Math Forum Expected Amount of Rounds the Game 4 Corners Takes

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 January 10th, 2019, 01:39 PM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 644 Thanks: 85 Expected Amount of Rounds the Game 4 Corners Takes I watched video of the game 4 Corners. One person stands blindfolded in the middle. Everybody else picks a corner numbered from 1 to 4 to go to. The blindfolded person picks a corner, and anybody at that corner is out. In the second round, everybody who is left gets a chance to go to any corner including staying put. Every round after Round 2 goes has the same rules. The game continues until 1 person is left and wins. There is no maximum amount of rounds the game takes because when there are few enough people left, the person in the middle could keep choosing corners with nobody there. Assume that people pick corners independently of each other with one exception which is that there will never be a time when everybody picks the same corner. That means that if 2 people are left, the probability of the game ending is 0.5 because picking either corner with a person will end the game. If there are 3 people left, the only way the game ends that round is if 2 people pick the same corner and the person in the middle picks that corner. Any amount of people can be eliminated in any round, and any round could be the last round. For example, with 10 people, hypothetically 9 people can pick the same corner and the person in the middle picks that corner. This is extremely unlikely, but possible. For amounts of players from 2 to 10 (excluding the person in the middle), can the expected amount of rounds it takes to get a winner be calculated? This requires 9 answers, one for each number of people from 2 to 10. If there is a formula that can calculate the expected amount of rounds necessary to get a winner that can take any amount of people as an input, that would be great.
 January 10th, 2019, 05:05 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,452 Thanks: 1337 once a corner has been chosen by the It person is that corner then out of play?
 January 11th, 2019, 06:49 AM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 How many are not blindfolded?
January 11th, 2019, 08:46 AM   #4
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Quote:
 Originally Posted by JeffM1 How many are not blindfolded?
That's going to have to be a parameter (unless OP picks a number).
Web suggests it can be played with many people, (and more than 4 corners as well)

January 12th, 2019, 05:31 AM   #5
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 Originally Posted by romsek once a corner has been chosen by the It person is that corner then out of play?
No. If a corner was out of play, it would always end within 4 times. After a corner is picked, everybody gets to stay or move. An alternate way of thinking about it is everybody picking one of four equal sections of a spinner and the people who picked the section that is spun are out. I could estimate using random numbers from 1 to 4.

Quote:
 Originally Posted by romsek That's going to have to be a parameter (unless OP picks a number). Web suggests it can be played with many people, (and more than 4 corners as well)
One person is blindfolded. One corner is picked at atime You may not be understanding me. Tell me what difference it makes in how many people are blindfolded and it may help me explain.

Using more than 4 corners would make the math more complicated and would make it more likely to have corner get picked with nobody there.

Last edited by EvanJ; January 12th, 2019 at 05:37 AM.

January 12th, 2019, 07:29 AM   #6
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 Originally Posted by EvanJ No. If a corner was out of play, it would always end within 4 times. After a corner is picked, everybody gets to stay or move. An alternate way of thinking about it is everybody picking one of four equal sections of a spinner and the people who picked the section that is spun are out. I could estimate using random numbers from 1 to 4. One person is blindfolded. One corner is picked at atime You may not be understanding me. Tell me what difference it makes in how many people are blindfolded and it may help me explain.
Not how many are blindfolded.. how many are not.
You're saying that the number of folks going to the corners at the start of the game doesn't matter??? This might be so but it's not obvious.

So we've got 1 blindfolded person, some unknown number of hiders, and all 4 corners are available during the entire course of the game. Check.

Stay tuned.

 January 12th, 2019, 12:59 PM #7 Senior Member     Joined: Sep 2015 From: USA Posts: 2,452 Thanks: 1337 How do you want to treat the cases where all the hiders are eliminated, i.e. there is no winner?
 January 14th, 2019, 06:23 AM #8 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 644 Thanks: 85 I want separate answers for each amount of hiders from 2 through 10. I said to make the assumption that all the hiders will never hide in the same corner, so there will always be exactly one winner.
January 14th, 2019, 07:26 AM   #9
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 Originally Posted by EvanJ I want separate answers for each amount of hiders from 2 through 10. I said to make the assumption that all the hiders will never hide in the same corner, so there will always be exactly one winner.
you'll have to explain in detail the process of hiders choosing corners then.

do all of them save one choose and then the last one has a reduced choice of corners?

January 15th, 2019, 04:30 AM   #10
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 Originally Posted by romsek you'll have to explain in detail the process of hiders choosing corners then. do all of them save one choose and then the last one has a reduced choice of corners?
The last one only has a reduced choice if everybody else picked the same corner, in which case the choice would be reduced by 1. As long as at least 2 corners were chosen before the last person, he can go anywhere. With many people, the probability that everybody except the last person picks the same corner is tiny. For example, with 6 people hiding the probability that the first 5 pick the same corner is 1/(4^4) = 1/256. It's 1/(4^4) rather than 1/(5^4) because the latter would be that the first 5 picked the same specific corner rather than the same any corner. P(first 5 pick Corner 1) = P(first 5 pick Corner 2) = P(first 5 pick Corner 3) = P(first five pick Corner 4) = 1/(5^4), and the sum if 1/(4^4). When there are 3 people hiding, the last one has a 3/4 chance at being able to pick any corner because the first 2 picked different corners and a 1/4 chance at having a choice of 3 because the first 2 picked the same corner. When there are 2 people hiding, the last one always has a choice of 3.

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