Probability CAVS win the series


#1

# no packages needed 
# Assign a variable 'n' as the number of remaining games.
n <- 6


# Assign a variable 'l' to a list of possible game outcomes, where 0 indicates a loss and 1 indicates a win for the Cavs. 
l <- list(0:1) 


# Create a data frame named 'possibilities' that contains all possible outcomes for the remaining games.
possibilties <- data.frame(expand.grid(rep(l, n)))   


# Create a vector named 'results' that indicates whether each row in the data frame 'possibilities' contains enough wins for the Cavs to win the series.
results <- c(rowSums(possibilties == 4))   


# Calculate the proportion of 'results' in which the Cavs win the series. Print the outcome to the console.
mean(results) 
#> [1] 0

#2

I tried to make this a REPREX not sure if I did it right.

What I am trying to do: Two teams, say the Cavs and the Warriors, are playing a seven game championship series. The first to win four games wins the series. The teams are equally good, so they each have a 50-50 chance of winning each game.

My desired output:

If the Cavs lose the first game, what is the probability that they win the series?


#3

I don't really see the connection between your reprex and your question.

If you are more broadly interested in how to solve math probability questions, and answer those questions with simulations, I'd google around for examples and code vignettes on the topic.
With no assurance of quality, this appears to be a nice example with R,
http://math.slu.edu/~speegle/TempWebPageFolder/schedule/schedule.html
The Jan 20 and 22 lecture notes hit your topic.

For this specific question, the way you set it up I don't think you need to run simulations at all. If each team has a 50% chance of winning, and that's independent of whether or not one team won a previous match or not. So if a game is a coin toss, and the cavs winning is heads, it's like asking "what is the probability of getting 4 heads in 7 tosses, given the first toss landed heads". Or, what's hte prob of getting 3 heads in 6 tosses.

There are 36 ways to toss a coin 6 times. How many of those toss combos ends up with 3 or more heads?

image
6 fair coin flips: probability of exactly 3 heads

So you'd do this calc for (6,3) + (6,4) + (6,5) and (6,6), all the sets in which Heads was >= 3.


Here's a nice intro to probabilities with a 4-toss version of this question.
https://www.fourmilab.ch/rpkp/experiments/statistics.html


#4

# no packages needed 
# Assign a variable 'n' as the number of remaining games.
n <- 6


# Assign a variable 'l' to a list of possible game outcomes, where 0 indicates a loss and 1 indicates a win for the Cavs. 
l <- list(0:1) 


# Create a data frame named 'possibilities' that contains all possible outcomes for the remaining games.
possibilities <- data.frame(expand.grid(rep(l, n)))   
possibilities
#>    Var1 Var2 Var3 Var4 Var5 Var6
#> 1     0    0    0    0    0    0
#> 2     1    0    0    0    0    0
#> 3     0    1    0    0    0    0
#> 4     1    1    0    0    0    0
#> 5     0    0    1    0    0    0
#> 6     1    0    1    0    0    0
#> 7     0    1    1    0    0    0
#> 8     1    1    1    0    0    0
#> 9     0    0    0    1    0    0
#> 10    1    0    0    1    0    0
#> 11    0    1    0    1    0    0
#> 12    1    1    0    1    0    0
#> 13    0    0    1    1    0    0
#> 14    1    0    1    1    0    0
#> 15    0    1    1    1    0    0
#> 16    1    1    1    1    0    0
#> 17    0    0    0    0    1    0
#> 18    1    0    0    0    1    0
#> 19    0    1    0    0    1    0
#> 20    1    1    0    0    1    0
#> 21    0    0    1    0    1    0
#> 22    1    0    1    0    1    0
#> 23    0    1    1    0    1    0
#> 24    1    1    1    0    1    0
#> 25    0    0    0    1    1    0
#> 26    1    0    0    1    1    0
#> 27    0    1    0    1    1    0
#> 28    1    1    0    1    1    0
#> 29    0    0    1    1    1    0
#> 30    1    0    1    1    1    0
#> 31    0    1    1    1    1    0
#> 32    1    1    1    1    1    0
#> 33    0    0    0    0    0    1
#> 34    1    0    0    0    0    1
#> 35    0    1    0    0    0    1
#> 36    1    1    0    0    0    1
#> 37    0    0    1    0    0    1
#> 38    1    0    1    0    0    1
#> 39    0    1    1    0    0    1
#> 40    1    1    1    0    0    1
#> 41    0    0    0    1    0    1
#> 42    1    0    0    1    0    1
#> 43    0    1    0    1    0    1
#> 44    1    1    0    1    0    1
#> 45    0    0    1    1    0    1
#> 46    1    0    1    1    0    1
#> 47    0    1    1    1    0    1
#> 48    1    1    1    1    0    1
#> 49    0    0    0    0    1    1
#> 50    1    0    0    0    1    1
#> 51    0    1    0    0    1    1
#> 52    1    1    0    0    1    1
#> 53    0    0    1    0    1    1
#> 54    1    0    1    0    1    1
#> 55    0    1    1    0    1    1
#> 56    1    1    1    0    1    1
#> 57    0    0    0    1    1    1
#> 58    1    0    0    1    1    1
#> 59    0    1    0    1    1    1
#> 60    1    1    0    1    1    1
#> 61    0    0    1    1    1    1
#> 62    1    0    1    1    1    1
#> 63    0    1    1    1    1    1
#> 64    1    1    1    1    1    1

# Create a vector named 'results' that indicates whether each row in the data frame 'possibilities' contains enough wins for the Cavs to win the series.
results <- rowSums(possibilities == 4)   
results 
#>  [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#> [36] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

# Calculate the proportion of 'results' in which the Cavs win the series. Print the outcome to the console.
mean(results) 
#> [1] 0

#5

the first two lines of meaningful code I'm good with and the "possibilities" dataframe looks ok to me as well. my call to mean(results) is good too. trying to construct results <- rowSums(possibilities == 4) as a vector which checks the "possibilities" df for enough wins for the cavs to win the series, which would be 4. having trouble figuring out how to check for the right value.