A fair price for this game?

I'd just like your opinions on this. Consider the following game:

The player pays a fixed number of L$ to play. The player then has a series of turns, each of which consists of touching either of two white prims. One of the prims turns red when touched, the other green. Each turn it's completely random which is the red one. (OK it would be the same with one prim that either turns red or green, 50% chance of either). The game lasts until the player picks the one that turns red. When that happens, the game ends, and pays out depending on turn number, as follows:

Turn 1 - player gets L$ 2
Turn 3 - player gets L$ 4
Turn 3 - player gets L$ 8
Turn 4 - player gets L$ 16
Turn 5 - player gets L$ 32
Turn 6 - player gets L$ 64 ... and so on.

What price would the player have to pay for it to be fair (no advantage to house or player)?

You can work it out yourself on paper using a branching diagram. You have a starting point and from that 2 branches; win and lose. From the win branch you branch again to win or lose etc.

Pick a number of people to start with (e.g. 100). If your game is 50/50 then half win and half lose. Put these numbers on the branches and for the losing branch write how much would be paid out (e.g. if losing on the first round gets you $2, you pay $100 for the 50 that lose). From the 50 that won, 25 lose on the next turn (paying out 25x$4) and 25 move to the third round. Keep going until you run out of people, then add up the losing payouts.

I think the system you described has the property that it'll pay $1 for every turn in it. So if you have 6 turns you should be charging $6 to play in order to break even. But check it for yourself.

The method above also allows you to try out different odds of winning and losing, different payout schemes etc.

Lucius

Ah, it's a little more complex than that. Think about it. Rather than paying an extra L$ 1 for every turn you stay in the game, the payments increase exponentially. The game is not necessarily balanced if 50 people win and 50 lose, since the income from the losers maybe much less than the payout to the winners. One lucky winner could potentially wipe out all the house gains. If you were lucky enough to last 12 turns you would net over L$ 4000. Since all players get L$ 2 back, a price of L$ 6 would mean you would need 1000 unlucky players to cover that one lucky win.

As I said, I think you'd need to charge $1 for each level, that is assuming you have a maximum number of times that you can win. If the game allows people to continue indefinitely then I'm not sure.

However, the odds of winning 12 turns is 2 to the power 12, or 1 in 4096. So only one player in over 4000 should get the amount you mention (I think), on average.

Maybe instead of working from a maximum number of turns you could guess the number of people that would play it, and work from that.

Either way you're taking a gamble yourself.

This is the St. Petersburg paradox. See this page for much discussion.

Lucius and Seifer are both correct. You would have to set a maximum amount of turns to calculate a particular price. Lets take your 6 turns for example.

You said that the probability of a "win" was 50% and hence a "loss" 50%. Hence statistically, after enough trials have been completed, your numbers should approach the following:

Turn 1: Probability is 1/2 of a win. Hence 50 people win and 100L is paid out (50 people times 2L each).
Turn 2: Probability is 1/2 that these 50 remaining people will win. So 25 people will win and 100L is paid out (25 people times 4L each).
Turn 3: Probability is 1/2 that these remaining people will win. Obviously we can't have 12.5 winners, but let's assume that thousands of people have played your game so 12.5 actually represents 125, 1250, etc. So 12.5 people will win and 100L is paid out (12.5 times 8L each).
Turn 4: Probability is 1/2 that these remaining 12.5 people will win. So 6.25 people will win (same explanation as above) and 100L is paid out (6.25 times 16L each)
Turn 5: Probability is 1/2 that these remaining 6.25 people will win. So 3.125 people will win and 100L is paid out (3.125 times 32L each).
Turn 6: Probability is 1/2 that these remaining 3.125 people will win. So 1.5625 people will win and 100L is paid out (1.5625 times 64L each).

So the total payouts are Turn 1 + Turn 2 +... + Turn (N-1) + Turn N, where Turn N is Turn 6 in your example and Turn 1 + Turn 2 +... + Turn (N-1) + Turn N = 600.

Since we have 100 players in our example, a reasonable amount to charge would be 600/6 = L6 to play. Do note as Seifer said that if you allow an infinitesimal amount of turns, there is no suggested price as for example lets take Turns = 100

Then: Turn 1 + Turn 2 +... + Turn (N-1) + Turn N = 100N = L10,000 . This would suggest that games cost 100L to play, however the odds over 100L are actually 1 in 128 (2^7). So why is the estimated amount required to pay so high? Well there is a 1/2^100 chance you may end up paying out 2^100L which would bankrupt anyone in the game instantaneously (2^100 = 4^50 = 16^25 = 256^12.5 = 65536 ^ 6.25 = 4294967296 ^ 3.125). Now my Ti 89 can't even properly calculate exactly how much this is, so I tried my best to elucidate it. I couldn't imagine cubing 4294967296... In short, as Seifer said, be very careful how many turns you have because the Expected payouts can get extremely high. If you have infinite turns, the estimated payoff is actually infinity!

Hopefully this has helped, as said earlier, 1L per turn would be a reasonable amount to charge. Maybe charge 1.2L per turn or something like that to make a little profit for that hard work

Edit: Just remember to not script it such that it removes only 1L for turn 2 if you get to turn 2 for example, that would be eerily incorrect and cost you big time. If you have 6 turns for example, everyone would need to pay 6L to play, whether they lossed at turn 1 or at turn 6.

Seifert is a clever-clogs. This is indeed the St Petersburg paradox. I was wondering how people would react to it if presented as a SL game ...

Sorry

I had suspected as much, but if you had actually reinvented it, a reference to the standard name would have been very useful.

No need to apologize - I was also speculating as to if anyone would recognize it!

For the benefit of others, the problem is that the average win on the game for the player is

2 * 1/2 + 4 * 1/4 + 8 * 1/8 + 16 * 1/16 ....

which is

1 + 1 + 1 + 1 ...

to infinity. So the average win, mathematically, is infinite. So it would be worth paying any sum of money to play. Which is obviously not the case.

In the SL case, as Reece points out, you don't have to go very high in the sequence before, for practical reasons, the sequence crashes out. 2 ^ 20 is over a million, so if we assume that a payout of two million is not practical and the sequence must be capped at 20, the fair price becomes L$20.