>>I'm not sure the analysis is totally correct, but agree that the possible winnings is theorically infinite, but gets there less rapidly - my analysis is
>>
>>50% chance losing everything
>>25% chance winning $1
>>12.5% chance winning $3
>>6.25% chance winning $7
>
>Well, and how much does this add up to? (25% * $1 + 12.5% * $3, etc.).
>
>These terms should be added to get the average value of the game: 0.25 + 0.375 + 0.4375 + 0.4688 + 0.4844 + 0.4922 + 0.4961 ... (the terms rapidly approach 0.5).

This is where the analysis is not complete, I think, I'll come back to it later.

>
>>Betting $1 would give me a 50% chance of winning at least $1, so in the long run I'm likely to profit. 50% of the time I lose, 25% of the time break even, 25% of the time profit by at least $2, so the odds are in my favour.
>>
>>If I bet $3, I would only have a 25% chance of winning at least $3, 50% of the time I would lose $3, 25% of the time I would lose $2, 12.5% of the time I would break even & only 12.5% of the time would I be in profit. At this level the odds are very much in your favour.
>
>It seems to me that your are confusing "average" with "median" here. If you get less than $3 in half the games, but much more than $6 on the other half, you are still in a favourable position. The game is worth more than $3, on the average.

I'm not looking at averages/medians at this point, simply probabilities of winning.

There appear to be 3 ways of viewing the problem, from an individual point of view, from the advertisers point of view & from the person offering the bet's point of view.

I was taking the individual point of view, playing a small number of games, I am simply interested in the probability of me winning. As I said, with a $1 bet, there is 50% chance of me losing, a 25% chance of me not losing anything & 25% chance of me being in profit (regardless of the amount). With a $3 bet, there is a 75% chance of losing, 12.5% chance of not losing anything & 12.5% chance of being in profit. With a $7 bet, I'm down to 1 in 16 chance of being in profit & so on. Averages are not involved as I'm playing so few games.

From the advertisers point of view, you want as many people to play as possible so you emphasis the winnings, which is where your analysis comes in, showing the average pay out over a large number of games - always positive & always increasing with the number of games played.

From the person offering the bet's point of view, we also need to involve the bet placed in the game, & calculate the average winnings that are paid out, now the calculation becomes (for a $3 bet) :

(50%*-$3 + 25%*$-2 + 12.5%*0 + 6.25%*4 + 3.75%*12 ....
(-1.5    + -0.5    +  0.0    +  0.25   +  0.375   ....
(-1.5      -2.0      -2.0      -1.75     -1.375   ....

(negative values are profit to the person offering the bet).

With an exponential payout, of course you are likely to lose out in the very long run, but with a $3 bet you have around a 95% chance of not losing out. With a $5 bet you are up to 99% likely not to lose out, in the very long run.

Of course you could equally lose a large amount on the very first go, but you could console yourself with the fact that because of the very high interest rates in your country, it won't be worth that much at all to the winner tomorrow.