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It is the first time I heard about bingo game and I would like to learn more on this game by mathematical analysis. To make it simple, I consider the American BINGO with 75 balls used and each game will at maximum draw 50 balls. The winning pattern on each card is the center cross (the column pass the free and the row pass the free) and two diagonals. I want to know what's the probability for not winning the game within 50 draws for a single board. Here is my math

the probability of winning a single pattern is $P_1=C(50,4)/C(75,4)=0.189477$, since there are 4 patterns to win, the total winning probability is $P_4=4P_1=0.757909$. The probability of nothing winning should be $1-P_4=0.242901$. However, by computer simulation, I've got that the probability of not winning the game is 0.414359. I think there might be something wrong in my math. Any comment will be appreciated.

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Define event $W_i$ = "Line $i$ wins" for $i = 1,2,3,4$.

Then the probability of none of the $4$ lines winning is $1 - P(W_1 \cup W_2 \cup W_3 \cup W_4)$.

Because more than one of the $4$ lines could win on any given draw, $P(W_1 \cup W_2 \cup W_3 \cup W_4) \neq P(W_1) + P(W_2) + P(W_3) + P(W_4)$.

This is why, while $P_1 = C(50,4)/C(75,4)$ is correct, multiplying that by $4$ is not.

Instead, we use the inclusion-exclusion principle:

\begin{eqnarray*} P(\mbox{no winning lines}) &=& 1 - P(W_1 \cup W_2 \cup W_3 \cup W_4) \\ && \\ &=& 1 - \sum_{i=1}^{4}{P(W_i)} + \sum_{i \lt j}{P(W_i \cap W_j)} - \sum_{i \lt j \lt k}{P(W_i \cap W_j \cap W_k)} \\ && \qquad + P(W_1 \cap W_2 \cap W_3 \cap W_4) \\ && \\ &=& 1 - \binom{4}{1} \binom{50}{4} \bigg/ \binom{75}{4} + \binom{4}{2} \binom{50}{8} \bigg/ \binom{75}{8} \\ && \qquad - \binom{4}{3} \binom{50}{12} \bigg/ \binom{75}{12} + \binom{4}{4} \binom{50}{16} \bigg/ \binom{75}{16} \\ && \\ &\approx& 1 - 0.7579086 + 0.1909348 - 0.0185883 + 0.0005759 \\ && \\ &=& 0.4150138 \end{eqnarray*}

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  • $\begingroup$ Thanks Mick. I see why my math is wrong now :) $\endgroup$ Aug 24 '14 at 19:51

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