Seeing two of those hands in only 10 hands
We use a binomial approximation:
[
\binom{10}{2} p^2 (1 - p)^8
]
Since p is very small:
[
\approx 45 \times (0.00024)^2
]
[
\approx 2.6 \times 10^{-6}
]
0.00026%
Approximately: 1 in 380,000 sessions of 10 hands.
We use a binomial approximation:
- p = 0.00024 per hand
- n = 10
[
\binom{10}{2} p^2 (1 - p)^8
]
Since p is very small:
[
\approx 45 \times (0.00024)^2
]
[
\approx 2.6 \times 10^{-6}
]
0.00026%
Approximately: 1 in 380,000 sessions of 10 hands.