PS–Just for the record, I do think the expected value of (total payoff after n games of Game 1) / (total payoff after n games of Game 2) probably doesn’t exist for any n in this case, I’m just claiming it’s not obvious based only on the fact that it doesn’t exist for n=1.
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OK, I see where you’re going, Cabbage, thanks. You’re saying, play Game (1) and Game (2) exactly n times (where n is some fixed integer, yet to be determined) and then sum all the winnings from both games, and Player 3 gets the ratio of the sum of the winnings… is it possible that Game (3) has a determinable (finite) expected value?
I don’t want to think that hard.
But… suppose there is such a fixed n.
We enumerate all possible outcomes for Game 3 (the ratio of the sums), and the probability for each outcome, as you have started. Then we sum all the expected values.
I’m going to try to find an unbounded sub-series of the sum, as follows:
One set of outcomes is obviously:
Game 1: Payout $3, $3, $3…, $3
Sum of payouts = 3n
Game 2: Payout $6, $6, $6,…,23^m where m ranges over all the integers m = 1, 2, …
Sum of payouts = 3*(n-1)+23^m = 3^(n-1)[1 + 23^(m-n+1)]
For each m, (3) = Ratio =(2)/(1) = [1 + 2*3^(m-n+1)]/3
Now we need to multiply this by the probability of that outcome, for each m… but the probability which is something like (1/2)^n or perhaps a sum of 1/2^n, I presume. My point is that (I think) the probability doesnt depend on m, just depends on n.
So multiplying the ratio of an outcome by the probability of that outcome may introduce a large denominator with 2^n, but n is fixed, so that denominator is finite, and eventually 3^m in the numerator will overtake it.
Hence, the sum of ratios has an unbounded subseries. Since everything in sight is positive, this means that the series itself must be unbounded, QED.
I think I’m basically trying to do this by induction. We’ve shown it’s true for n = 1, and assume it’s true for fixed n, then adding one more play won’t change the unboundedness of the expected value series.
Man oh man. Now I got a headache.
Actually, I think I’ve found a fairly simple way around it. Instead of trying to find the expected value of the ratio after n rounds (A1 + … + An)/(B1 + … + Bn), find the expected ratio of just one Game 1 payoff, to the total of n Game 2 payoffs: A1 / (B1 + … + Bn). If this is unbounded, it seems fairly clear to me that the former is unbounded as well. Now it’s pretty easy to get an unbounded subseries:
Game 1 payoff: $3
Total Game 2 payoffs: $6n
Ratio: 3/6n
Probability of this: 1/2^(n+1)
Add to the expected value: (3/6n)*{1/2^(n+1)}
Game 1 payoff: $3^2
Total Game 2 payoffs: $6n
Ratio: 3 * (3/6n)
Probability of this: (1/2) * {1/2^(n+1)}
Add to the expected value: (3/2) * (3/6n)*{1/2^(n+1)}
And so on, increasing the Game 1 payoff each time, where I’ve highlighted the fact that this is a geometric series with ratio 3/2.
…ebius sig. This is a moebius sig. This is a mo…
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