Is the roulette wheel fair? Goodness of fit with unequal expected counts

Question

An American roulette wheel should land red, black, or green with probabilities $\frac{18}{38}$ , $\frac{18}{38}$ , $\frac{2}{38}$ . Over $380$ spins you observe $[172, 188, 20]$ (red, black, green).

Test whether the wheel matches its theoretical distribution. Give the statistic, degrees of freedom, and conclusion.

Accepted Answer

Expected counts are n 	imes pi, and here the categories are not equally likely: 380 	imes \frac{18}{38} = 180 for red, 180 for black, and 380 	imes \frac{2}{38} = 20 for green. So E = [180, 180, 20]. Statistic. \chi^2 = \frac{(172-180)^2}{180} + \frac{(188-180)^2}{180} + \frac{(20-20)^2}{20} = \frac{64}{180} + \frac{64}{180} + 0 = 0.356 + 0.356 = 0.71. Degrees of freedom. The probabilities are fully specified by the wheel's design (nothing estimated), so 	ext{df} = k - 1 = 3 - 1 = 2. Decide. The 5\% critical value for \chi^22 is 5.991. Since 0.71 < 5.991, we fail to reject: the spins are …

Is the roulette wheel fair? Goodness of fit with unequal expected counts

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