Make me a market on heads in 100 coin flips

Question

Interviewer: "I am about to flip a fair coin $100$ times. Make me a market on the number of heads. I may trade either side, and neither of us will know the outcome in advance."

Give a bid and an ask, and explain how you chose the width.

Accepted Answer

Step 1: Fair value. The count of heads is Binomial(100, 0.5), with mean E[	ext{heads}] = np = 100 	imes 0.5 = 50. Your fair is 50, and this time it is not a guess. Step 2: Uncertainty. The standard deviation is \sigma = \sqrt{np(1-p)} = \sqrt{100 	imes 0.5 	imes 0.5} = \sqrt{25} = 5. So a "typical" outcome is 50 \pm 5, and about 95\% of the time it lands in 40 to 60. Step 3: Quote around fair, width from the standard deviation. Because neither side knows the outcome, information is symmetric: there is no better-informed counterparty to pick you off, so you can quote fairly tight. Still, a…

Make me a market on heads in 100 coin flips

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