Here is a problem I thought of:
- Suppose I am watching someone flip a fair coin. Each flip is completely independent from the previous flip.
- I watch this person flip 3 consecutive heads.
- I interrupt this person and make the following offer: If the next flip results in a "head", I will buy you a slice of pizza; if the next flip results in a "tail", you will buy me a slice of pizza.
My Question: Who has the better odds of winning?
I wrote the following simulation using the R programming language. In this simulation, a "coin" is flipped many times ("1" = HEAD, "0" = TAILS). We then count the percentage of times HEAD-HEAD-HEAD-HEAD appears compared to HEAD-HEAD-HEAD-TAILS:
#load library library(stringr) #define number of flips n <- 10^7 #flip the coin many times set.seed(1) flips = sample(c(0,1), replace=TRUE, size=n) #count the percent of times HEAD-HEAD-HEAD-HEAD appears str_count(paste(flips, collapse=""), '1111') / n 0.0333663 #count the percent tof times HEAD-HEAD-HEAD-TAIL appears str_count(paste(flips, collapse=""), '1110') / n 0.0624983
From the above analysis, it appears as if the person's luck runs out: after 3 HEADS, there is a 3.33% chance that the next flip will be a HEAD compared to a 6.25% chance the next flip will not be a HEAD (i.e. TAILS).
Thus, could we conclude: Even though the probability of each flip is independent from the previous flip, it becomes statistically more advantageous to observe a sequence of HEADS and then bet the next flip will be a TAILS? Thus, the longer the sequence of HEADS you observe, the stronger the probability becomes of the sequence "breaking"?