Probability in a Card Game
In a game, a standard deck of 52 cards is shuffled and two cards are drawn without replacement. A player wins a prize if both cards are of the same suit. The game organizer claims that the probability of winning is about 25%. A player suspects the probability is different and decides to simulate the game 100 times, recording the number of wins. The simulation results in 30 wins. The player wants to know if this provides evidence that the organizer's claim is incorrect. Use a 5% significance level.
1. Calculate the theoretical probability of winning (both cards same suit). Show your reasoning.
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There are 4 suits, each with 13 cards. The probability that the second card matches the suit of the first card is 12/51, because after drawing the first card, there are 12 remaining cards of that suit out of 51 total. So probability = 12/51 = 4/17 ≈ 0.2353, or about 23.5%. Alternatively, count favorable outcomes: choose a suit (4 ways), choose 2 cards from that suit (C(13,2)=78), total ways to choose 2 cards from 52 is C(52,2)=1326. So probability = 4*78/1326 = 312/1326 = 4/17. So theoretical probability is 4/17 ≈ 0.235, not 25%.
2. Perform a hypothesis test at the 5% significance level to determine if the simulation result (30 wins in 100 games) is significantly different from the theoretical probability. State the null and alternative hypotheses, check conditions, compute the test statistic and p-value, and state your conclusion.
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Null hypothesis: ( p = 4/17 \approx 0.2353 ). Alternative: ( p \ne 4/17 ). Sample size ( n = 100 ), number of wins ( X = 30 ). Check conditions: ( np = 100*(4/17) ≈ 23.53 \ge 10 ) and ( n(1-p) ≈ 76.47 \ge 10 ), so normal approximation is reasonable. Test statistic: ( z = (\hat{p} - p) / \sqrt{p(1-p)/n} ), where ( \hat{p} = 0.30 ). Compute ( p(1-p) = (4/17)*(13/17) = 52/289 ≈ 0.17993 ). Standard error = ( \sqrt{0.17993/100} = \sqrt{0.0017993} ≈ 0.04242 ). ( z = (0.30 - 0.2353)/0.04242 ≈ 0.0647/0.04242 ≈ 1.525 ). Two-tailed p-value = ( 2 * P(Z > 1.525) ≈ 2 * 0.0636 = 0.1272 ). Since p-value > 0.05, we fail to reject the null hypothesis. There is not enough evidence to conclude that the probability is different from 4/17. The simulation result of 30 wins is not statistically significant at the 5% level.