I'm reading about an expected value problem in a philosophy book and the way the author explains it is so convoluted and overcomplicated that I was hoping people who actually deal with math would be able to explain it way better

"The Allais Paradox.

Number 100 cards consecutively from 1 to 100. The cards are shuffled, and a card is drawn. The possible outcomes are grouped as three mutually exclusive possibilities.

L: A“low” card is drawn, numbered below 90 (probability 0.89).

N: Card 90 is drawn (probability 0.01).

H: A“high” card is drawn, numbered 91 or higher (probability 0.1).

ALLAIS FIRST GAME

A: $500,000 if a high card (H) or card N is drawn (probability 0.11). Otherwise, nothing happens (probability 0.89).

B: $2,500,000 if a high card (H) is drawn (probability 0.1). Otherwise, nothing happens (probability 0.9).

ALLAIS SECOND GAME

F: You get an outright gift of $500,000 (free!).

G: $2,500,000 if a high card (H) is drawn (probability 0.1). $500,000 if a low card (L) is drawn (probability 0.89). Otherwise, nothing happens (probability 0.01).

Which choice would be correct According to expected value in each game? Answer: B in Game 1 and G in Game 2(So far so good)

Edit: The paradox is that at first glance most people will choose B in Game 1 but not G in Game 2( even though G technically has a greater expected value than F)

(Here is where the disaster comes) The Allais paradox uses money. Can we use something like Daniel Bernoulli’s diminishing marginal utility to beat the paradox? Or use some other way of fiddling with the utilities and probabilities?

Answer: No. The dollar utilities in the paradox were $5 million and $1 million. But in fact the paradox is quite general. Instead of $5 million and $1 million, let the utilities, in utiles, be x and y. Show that so long as

x > (1 [Pr(N)/Pr(H)])y,

there is no consistent way to maximize expected utility by preferring A to B, while preferring G to F. Why? We see that

Exp(A)=Pr(L)0 Pr(N)0 Pr(H)x.

Exp(B)=Pr(L)0 Pr(N)0 Pr(H)y.

Since x must exceed y by [Pr(N)/Pr(H)])y,

Pr(N)0 Pr(H)x >Pr(N)0 Pr(H)y.

and hence Exp(A) > Exp(B). But now add Pr(L)y to the above inequality:

Pr(L)y Pr(N)0 Pr(H)x> Pr(L)y Pr(N)0 Pr(H)y.

On the left-hand side we have Exp(F), and on the right-hand side we have Exp(G), so Exp(F)> Exp(G)."

How in God's name did he derive that "Show that so long as x > (1 [Pr(N)/Pr(H)])y"

Why is Exp(A)=Pr(L)0 Pr(N)0 Pr(H)x. There is a chance of getting money when we draw N in the "A" scenario. Why is he multiplying N by 0 in this case?

I'm quite lost with his explanation(Or is this guy simply mistaken, and I shouldn't ever trust a philosopher to explain algebraic manipulations of expected value equations). If anyone could help with showing the working out, I'd be very grateful.