Jury problem (birthday paradox twist)

A bailiff who works in the court system asks 12-person juries, when they are first sequestered, to break the ice, to play the following game. The jurors sit around a circular table that could be thought of as a clock, with the fore-man (fore-woman) sitting at position 12, and the other 11 jurors, in clockwise order, sitting in positions 1, 2, …, 11. Each juror is asked to write the month of his/her birthday with a marker on a sheet of paper and put it face-down on the table. The bailiff first turns over the sheet of paper with the fore-person's month on it. The jurors proceed around the room, in order, turning over their sheets of paper until a month's name is repeated. The juror who turns over the sheet where a repeat first occurs wins a prize. The only way the fore-person can win is if the jurors make it all the way around the table with no repeats (i.e. all 12 months show up exactly once). The bailiff asks the jurors, before they play, which position is most likely to win. Here are some of the answers which he has received over the years. Do you think any are correct? How would you explain?

(a) All 12 positions are equally likely to win.

(b) Position 1 is the best, and chances decrease monotonically from there.

(c) Position 6 is the best because it is in the middle of the first 11.

(d) Position 3 is best.

(e) Position 4 is best.

Is there any posible way to doit on R?

Hi,

I hope I got the whole setup correctly, but this would be a simulation:

library("ggplot2")

winner = sapply(1:10000, function(x){
  
  #Give the jury a random birthday month
  jury = data.frame(pos = 1:12, 
                    month = sample(month.abb, 12, replace = T),
                    sameMonth = F)
  
  #Loop fast over all jurors
  jury$sameMonth[-1] = sapply(2:12, function(y){
    jury$month[y] %in% jury$month[1:(y-1)]
  })
  
  #If no-one has same month for-man wins
  if(!T %in% jury$sameMonth){
    jury$sameMonth[1] = T
  }
  
  #Return position of first person that won
  which(jury$sameMonth == T)[1]
})

#Adjust the position of the winner to the clock position around the table
winner = winner - 1
winner[winner == 0] = 12

#Plot results
barplot(table(winner))

Notes

  • Either position 3 or 4 seems to win most often, depending on randomness

Hope this helps,
PJ

UPDATE: Thanks to @Yarnabrina for pointing out earlier confusing notation in my code and graph which have been corrected.

Please take a look into our homework policy. Homework inspired questions are welcome but they shouldn't be verbatim

@manuel2308ab

If this is indeed part of a homework, I agree with @andresrcs and suggest that in future you note this in your posts so we can answer accordingly by giving pointers instead of answers.

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