A wand that turns two coins over
Four fair coins are tossed, and you are paid 1 dollar for each head. After seeing the coins, you may use a magic wand, for free and as many times as you like. Each use picks any two coins and turns both of them over (a head becomes a tail and a tail becomes a head).
By how much does the wand raise the expected payoff? That is the fair price of the wand.
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
How can one use of the wand change the number of heads?
Something about the number of heads never changes, however many times you use the wand.
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s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. Turning two coins over changes the number of heads by , or , so its parity never changes. With an even number of heads you can reach 4 heads, with an odd number only 3.
Step 1: the invariant. If both coins were heads, the count drops by 2; if both were tails, it rises by 2; if one of each, it stays the same. So the parity of the number of heads is fixed.
Step 2: the best you can do. With an even number of heads (0, 2 or 4), turn over pairs of tails until all four are heads. With an odd number (1 or 3), you can reach 3 heads but never 4.
Step 3: the probabilities. Four fair coins show an even number of heads with probability So with the wand the expected payoff is .
Step 4. Without the wand it is . The wand is worth
A different wand. If the wand instead re-tosses two coins, the parity is no longer fixed, and you can always reach 4 heads eventually: that wand is worth 2 dollars.