Forty-card deck with five wilds, first wild position

Question

You shuffle a 40-card deck that contains exactly 5 special "wild" cards (the other 35 are ordinary). You turn cards face up one at a time from the top.

On average, at what position does the first wild card appear?

Accepted Answer

Use the same symmetry idea, adapted to this custom deck. Set up the picture. There are 5 wilds and 35 ordinary cards. Lay the 35 ordinary cards in a row: the 5 wilds split that row into 6 gaps (one before, four between, one after). By symmetry the wilds are equally likely to fall in any gap, so each gap holds the same number of ordinary cards on average. Ordinary cards before the first wild. The 35 ordinary cards spread evenly across 6 gaps, so the first gap holds on average: \frac{35}{6} 	ext{ ordinary cards.} Position of the first wild. Add one for the wild itself: EV = \frac{35}{6} + 1 = …

Forty-card deck with five wilds, first wild position

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