Obtaining the Same Outcome in Repeated Runs - Flipping a Coin-CodePudding

What is the frequency of getting the same face of a coin in a row consecutively when you flip a single coin 1000 times? What are the relative frequencies of this experiment ending at 2, 3, 4, and 5 tosses?

I started with these codes but I don't know how to continue:

coin <- c("heads", "tails")

num_flips <- 10000

flips <- sample(coin, size = num_flips, replace = TRUE)

freqs <- table(flips)

CodePudding user response：

The secret here is to use run length encoding (rle), which will tell you the length of consecutive flips of the same result.

set.seed(1) # Makes example reproducible

coin <- c("heads", "tails")

num_flips <- 10000

flips <- sample(coin, size = num_flips, replace = TRUE)

RLE <- rle(flips)

If we examine the RLE object it will show us the number of consecutive heads and tails:

RLE
#> Run Length Encoding
#>   lengths: int [1:4912] 1 1 2 1 3 2 5 4 7 1 ...
#>   values : chr [1:4912] "heads" "tails" "heads" "tails" "heads" "tails" "heads" ...

If we table the lengths element, we will get the number of runs of each length:

runs <- table(RLE$lengths)
runs

#>    1    2    3    4    5    6    7    8    9   10   11   12   14   16 
#> 2431 1224  581  332  180   84   45   17   10    1    1    2    2    2

If we want to know the proportion of runs that are of each length, we can do:

runs / sum(runs)

#>            1            2            3            4            5            6 
#> 0.4949104235 0.2491856678 0.1182817590 0.0675895765 0.0366449511 0.0171009772 
#>            7            8            9           10           11           12 
#> 0.0091612378 0.0034609121 0.0020358306 0.0002035831 0.0002035831 0.0004071661 
#>           14           16 
#> 0.0004071661 0.0004071661

However, this is not the same as the estimated probability of any single flip belonging to a run of a particular length. To get this we need to multiply each element of runs by its associated length to get an absolute value for how many flips belong to a run of each length:

results <- runs * as.numeric(names(runs))
results

#>    1    2    3    4    5    6    7    8    9   10   11   12   14   16 
#> 2431 2448 1743 1328  900  504  315  136   90   10   11   24   28   32

Now if we want the proportion of flips that belonged to a run of one particular length, we can do:

results / num_flips
#> 
#>      1      2      3      4      5      6      7      8      9     10     11 
#> 0.2431 0.2448 0.1743 0.1328 0.0900 0.0504 0.0315 0.0136 0.0090 0.0010 0.0011 
#>     12     14     16 
#> 0.0024 0.0028 0.0032

This is the estimated probability that any single flip will belong to a run of the given length.

^{Created on 2022-09-26 with reprex v2.0.2}

CodePudding user response：

You can simulate your 10000 coin flips and count the frequency of X consecutive 1s and X consecutive 0s with rle function.

Example

# Your 10000 flip results
set.seed(1)
flips <- sample(c(0, 1), 10000, replace=TRUE)

# Consecutive 1s
nbConsecutive1s <- rle(flips==1)
table(nbConsecutive1s)

Output

The TRUE column corresponds to consecutive 1s.
The FALSE column corresponds to consecutive 0s.
As you can see, the number of consecutive 1s and 0s are close, whatever the length (this makes sense).

       values
lengths FALSE TRUE
     1   1232 1199
     2    588  636
     3    283  298
     4    183  149
     5     88   92
     6     46   38
     7     21   24
     8      8    9
     9      6    4
     10     0    1
     11     0    1
     12     0    2
     14     1    1
     16     0    2

If you want to converge towards an estimated frequency, then you can launch the code k times and aggregate the results.
Launching it only once will give you results with variability around your estimated frequency.

Don't forget to use a seed if you want to make the example reproducible.

rle function and example

https://www.r-bloggers.com/2009/09/r-function-of-the-day-rle-2/