A fair or a loaded die, chosen by a coin

Question

You flip a fair coin. On heads you roll a normal fair die. On tails you roll a loaded die that shows a $6$ half the time, and shows each of $1, 2, 3, 4, 5$ with probability $\tfrac{1}{10}$ otherwise.

What face value should you expect?

Accepted Answer

Two stages: the coin picks the die, then the die is rolled. Condition on the coin and use the tower rule: E[	ext{face}] = E\big[\,E[	ext{face} \mid 	ext{die}]\,\big]. Step 1, average each die on its own. - Fair die: E[	ext{face} \mid 	ext{fair}] = \frac{1+2+3+4+5+6}{6} = 3.5. - Loaded die: it shows 6 with probability 	frac12 and each of 1 through 5 with probability 	frac{1}{10}: E[	ext{face} \mid 	ext{loaded}] = 6 	imes 	frac12 + (1+2+3+4+5)	imes	frac{1}{10} = 3 + 1.5 = 4.5. Step 2, blend by the coin, which is fair: E[	ext{face}] = 0.5 	imes 3.5 + 0.5 	imes 4.5 = 1.75 + 2.25 =…

A fair or a loaded die, chosen by a coin

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