5 coins are tossed simultaneously. it costs nothing to play. if 1,2,3,4 heads occur, you will be paid 1 dollar for each head. if all heads or all tails occurs, you lose 20 dollars. is it possible to adjust the pay amounts to make the game "fair" to both the dealer and the challenger if the expected payoff is 1.093?

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5 coins are tossed simultaneously. it costs nothing to play. if 1,2,3,4 heads occur, you will be paid 1 dollar for each head. if all heads or all tails occurs, you lose 20 dollars. is it possible to adjust the pay amounts to make the game "fair" to both the dealer and the challenger if the expected payoff is 1.093?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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sure, charge that much to play.
probability of all head or all tails is 1/32+1/32=2/32 P(1 H) = 5/32 P(2 H) = 10/32 P(3 H)= 10/32 P(4 H)= 5/32 expected value is 1*5/32 +2*10/32 +3*10/32+4*5/32-20*2/32
but question is not clear since obviously you can adjust your payoff in infinite number of ways. numerator of that fraction is 5+20+30+20-40 and when you divide by 32 you get 1.093 rounded. fair would mean the numerator was 0, so perhaps the easiest thing to do it so write 75-2x = 0 and solve for x as what you lose if you get all heads or all tails. this would make the game fair.

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in other words instead of losing $20 you would lose $37.5 for all heads or tails. then game would be fair.

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