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BruceET
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The number $X$ of 6's in 4 trials has $X \sim \mathsf{Binom}(n=4,p=1/6)$ and he number $Y$ of 6's in 8 trials has $Y \sim \mathsf{Binom}(n=8,p=1/6).$

In R, $P(X \ge 1) = 1 - P(X = 0) = 0.5177.$

1-dbinom(0, 4, 1/6)
[1] 0.5177469

By contrast, $P(Y \ge 2) = 1 - P(X \le 1) = 0.3953,$ smaller than above.

1 - pbinom(1, 8, 1/6)
[1] 0.3953231

By simulation of a million games of each type:

set.seed(1120)
x = replicate(10^6,  sum(sample(1:6, 4, rep=T)==6))
mean(x >= 1)
[1] 0.517721     # aprx 0.5177

y = replicate(10^6,  sum(sample(1:6, 8, rep=T)==6))
mean(y >= 2)
[1] 0.395072      # aprx 0.3953 +/- 0.001
2*sd(y >= 2)/1000
[1] 0.0009777328  # 95% margin of simulation error
BruceET
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