The word field is puzzling. (Maybe a faulty translation into English.) I will try
to guess what you are trying to do. If I guess incorrectly, please edit your Question to explain more clearly and try to give your own solution; even if your try is
not exactly correct, it will may give someone a clue toward a more useful answer.
If you believe each of two teams has an equal chance of being chosen, and you make 100 random choices, then the split will not always be 50 for Team A and 50 for Team B.
In $n = 100$ random choices the number $X$ of times Team A is chosen has
the distribution $X \sim \mathsf{Binom}(n=100, p=1/2).$ Here are results, sampled in R, of $10\,000$ such counts for Team A among 100 choices.
set.seed(121)
x = rbinom(10^4, 100, 1/2)
summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
30.00 47.00 50.00 49.99 53.00 69.00
The most likely value of $X$ is $50,$ and half of the time the number of choices
for Team A will be between 47 and 53. But you should not rule out getting as
few as $35$ or as many as $65.$ Here is a histogram (blue bars) of my simulated
results. Red dots indicate the the theoretical binomial probabilities for each
value of the random variable $X.$ [Even with as many as $10\,000$ iterations the practical results do not exactly match the theoretical binomial probability model, but the model still provides a useful guide for what
results to anticipate.]
hist(x, prob=T, br=cp, col="skyblue2", main="50-50 Random Choices")
k = 0:100; pdf = dbinom(k,100,.5)
points(k, pdf, col="red")
By contrast, if you believe Team A will be randomly chosen $70\%$ of the time
(a 70:30 split), then the appropriate binomial distribution for random choices of Team A out of 100 is
$X\sim\mathsf{Binom}(n=100, p=0.7).$ Results for $10\,000$ such choices are shown below. On any one choice the probability of choosing Team A is $\frac{70}{70 + 30} = \frac{70}{100} = 0.7 = 70\%.$
set.seed(2021)
x = rbinom(10^4, 100, 7/10)
summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
50.00 67.00 70.00 70.06 73.00 87.00
