I use both, but in some cases a simulation seems less likely to make a mistake, or a way to confirm you haven't made a mistake.

I can see that your strategy is optimal in the bounded case and in all bounded cases of this type.

H = 1/2 (1,3)
TH = 1/4 (3,9)
TTH = 1/8 (9, 27)
TTTH = 1/16 (27, 81)
TTTT = 1/16 (81, 243)

see a 1 --> switch to 3
see a 3 --> (2/3) switch to 1 + (1/3) switch to 9 = 3 2/3
see a 9 --> (2/3) switch to 3 + (1/3) switch to 27 = 11
see a 27 --> (2/3) switch to 9 + (1/3) switch to 81 = 33
see a 81 --> (1/2) switch to 27 + (1/2) switch to 243 = 135
see a 243 --> don't switch

in the unbounded case it appears to me without thinking too much about it that

stay < stayexcept1 < stayexcept1,3 < stayexcept1,3,9 < etc.

I did use probability (as opposed to simulation) in one post and you refused to look at it.

no 243,1 are the extremes not a possible pair.

The rules are that as i described above once you hit 4 tails and more you only award 81,243.

Every time you create correlation like that you restrict the problem.

I only proposed the bounded case because it has finite EV and therefore it can be reliably simulated with clear convergence. If my EV calculation wont convince you the simulation will (that switching unless you hit max is optimal). I even allow a deterministic simulation that offers both strategies the exact same "luck".

You can of course make the upper prize as big as you like definitely make it large enough to give room to the strategy to prove its superiority. I think 243 is ok.

I have already simulated it and it gives the switch method an edge as expected even including the stay method to abort 1 as needed.

You must not run them together correlated because it is a sampling issue. But you can feed both identical flip sequences (basically all given to them is identical).


Also notice that if the prize was not 3^t but say 1.8^t i would not be suggesting switch.


Ok good i see you are doing the finite ev examples now.

Yes I did that. I don't see any need to simulate it.

I simulated it anyway

So now what is the point. So far this is just confirming my intuition about not trusting the conclusion "because EV says we should switch then we should switch".

Because what I see going on here is kind of like a long chain of lotteries.

#1 We invest 1 to get 3.
#2 We risk 3 and erase (or reinvest) winnings in #1 but we get paid off when we hit 9.
#3 We risk 9 and erase (or reinvest) winnings in #2 but we get paid off when we hit 27.
etc.

So long as this chain is finite... it works.

But if we "reinvest" all our winnings we never actually get paid off and somehow AlwaysSwitch degenerates to AlwaysStay.

The point was that switching is the solution something you were not so accepting of earlier.

Now since i arrived to that conclusion using my EV/probability/Bayesian logic arguments they will also hold for the unbounded case.

In any case in our world there is finite money so you cannot have an infinite bankroll/prizes case closed. Switch always unless max.

All you can do in the theoretical infinite EV case is calculate the EV not for the entire strategy (its infinite) but for every single 3^k number received. This will tell you what strategy to choose. The general strategy for any 3^k will follow as a result.

I asked this earlier. If you have to choose a strategy in advance and stick with it prior to looking at the number, do you choose
a) always switch
b) something else
c) doesn't matter


And this is the part I don't necessarily agree with.

Suppose you can purchase a lottery ticket. You must pay $1 out of your pocket to play. You are guaranteed to win 2X your money. You can then reinvest your winnings, but it costs you another $1 out of your pocket. Once you quit, you can't play anymore. If you reinvest a finite number of times you win big. But if you always reinvest, you get a stream of -1$ and nothing else. You shouldn't always do what each individual ev calc tells you to do.

It seems the only way you get paid off in the original game for all your switching is to flip an infinite number of tails.

your money (if i understand this correctly) goes -1, 2, -2, 4, -3, 8, -4, 16, etc. you don't get a stream of $1s, you get a number that oscillates between -N and 2^N. i don't see how this proves anything about the OP game.

wtf? if you get 81, and then switch and get 243, you've just gotten paid off from switching.

[QUOTE=DarkMagus;33497109]your money (if i understand this correctly) goes -1, 2, -2, 4, -3, 8, -4, 16, etc. you don't get a stream of $1s, you get a number that oscillates between -N and 2^N. i don't see how this proves anything about the OP game.



you pay a dollar out of pocket each time for the right to keep doubling your lottery investment only

so first 1$ out of pocket as initial investment becomes $2 in the lottery. Now you have to pay another $1 out of pocket for the right to double the $2 into $4. then again another $1 out of pocket for the right to turn $4 into 8$. etc. so you keep paying $1 but when you cash out you get whatever huge sum you have invested. but if you never cash out you just keep paying $1 out of pocket forever with no actual return. point is. pure EV of each transaction tells you to keep investing. so you can't always follow pure EV all the time.



under the always switch strategy
we switch on 1 to obtain 3
but as often as we turn 1 into 3 we switch 3 back to 1 in the hopes of getting 9
and as often as we turn 3 into 9 we switch 9 back to 3 in the hopes of getting 27
and as often as we turn 9 into 27 we switch 27 back to 9 in the hopes of getting 81
and as often as we turn 27 into 81 we switch 81 back to 27 in the hopes of getting 243
ad infinitum

so?

why not?

your problem is you're not proving anything rigorously. you keep making up different examples, then making superficial conclusions about them, and coming to absurd results like you shouldn't take the highest EV option each time you have a decision.

in this (4th? 5th?) example game you've made up, you should obviously keep reinvesting until you run out of money to pay the $1 fees. then the game is over and you win your money. i don't see how this proves anything.

No one rigorously proved that you should always take the highest EV choice.

I wasn't trying to rigorously prove anything. i was basically told that if i accept the EV calculations then i must follow their advice. i was giving examples as to why i don't religously assume that one one must follow EV advice. consider them more like thought experiments.

The EV advice is leading to some seemingly absurd results, such as always-switch but only after you've looked.

and what if you could invest forever? then you have a stream of -1$ and no payback following EV advice. or if the cost to play was free after the first $1 then you'd keep reinvesting your winnings and never receive money.

no, the EV advice says always switch unless you receive the maximum possible value. if there is no maximum possible value then the EV is infinity, so yes it will give absurd results, but who cares because it's a game that is impossible to actually occur.

if you could invest forever (again, a game impossible in reality) then that means you start with infinite money. so who really cares what you do with it.

then keep playing until you have so much money that you could never possibly spend it all and cash out. lol. what is the point of all these impossible games?

Imagine you are comparing any two strategies that are identical for all numbers except 3^t. (t integer >= 0)

To compare the relative EV between them we only have to consider cases when you receive a slip with 3^t on it.

The strategy that switches with 3^t will do better than the strategy that stays with 3^t.
(Just do the EV calc of switching with this slip knowing the distribution.)

(Do you dispute this? The maths is indisputable even if the implications are uncomfortable.)

So when you see 3^t on the slip you should switch.
(Do you dispute this? It follows from the above assuming we want to maximise EV for trials when we receive this number.)

In simulations you have to give both strategies the same slips from the same pairs when comparing strategies.
(And not play both slips as this changes the distribution of slips and changes the original game.)

So switch WHEN you see 3^t on a slip. (For any integer t.)

If you don't look at the slip then it doesn't matter whether you switch or not.

It doesn't make sense but that is, I think, the solution.

????

In the absence of any external factors such as limited bankroll, utility of money etc, etc, why would you not take the most EV choice?

What is the purpose of this slip game if not to maximise your EV?

If you were given a gamble in this game with a 1,000,000,000-1 shot with 1,000,000,001-1 pay-off then to maximise EV you should take it.

The correct strategy would be to gamble.

(If you dispute this then there is no hope of convincing you.)

I suspect I was thinking about this wrong way. I was thinking that you are constructing a strategy for all N by attempting to maximize the EV for any given N. I'm not sure that is a valid thing to do. That would lead to a strategy of AlwaysSwitch and as far as I can tell AlwaysSwtich == AlwaysStay in a certain sense.

But that is not even what you are doing. That is not the problem at hand. You don't need a strategy for every N. You only need a strategy for a single N. The one you see.