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  • Mathematical Conundrum or Not? Number Five

    I think one of the best ways to envision probability is to imagine what would happen if you did the event 10,00 times, so that is what I have done

    n = 10000 #Number of flips
coin <- sample(c("Heads", "Tails"), n, rep = T) #The coin flip
MondayHeads <- 0
MondayTails <- 0
TuesdayTails <- 0 
#Loop to count the outcome
for (i in coin) { 
  if (i  == "Tails")
    MondayTails <- MondayTails+ 1}
for (i in coin) { 
  if (i  == "Tails")
    TuesdayTails <- TuesdayTails + 1}
for (i in coin) { 
  if (i  == "Heads")
    MondayHeads  <- MondayHeads  + 1}
Tails <- sum(coin == "Tails")
Tails #Number of tails
Heads <- sum(coin == "Heads")
Heads #Number of Heads

MondayHeads #Number of Monday and Heads
MondayTails #Number of Monday and Tails
TuesdayTails #Number of Tuesday and Tails

    Here is the output:

    [1] 5037 - Number of Tails
    [1] 4963 - Number of Heads

    [1] 4963 - Number of Monday and Heads
    [1] 5037 - Number of Monday and Tails
    [1] 5037 - Number of Tuesday and Tails

    You can see that the 1/3 argument does in fact lead to a 33% split and the 1/2 argument does lead to a 50% split; however, you should go with the 1/2 argument, as 4963/10000 is better odds than 4963/15037, therefore you have a better chance of being correct.
    Jeremiah,  
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