All this really shows is the fact that divergent series can seem to give different answers depending on how you group the components. Consider the two series: 1+2+3+... and -1-2-3-.... If we add them together we get

1 - 1 + 2 - 2 + 3 - 3 + ...
= (1-1) + (2-2) + (3-3) + ...
= 0 + 0 + 0 + ...
= 0

right? But what happens if we group the numbers in a different way:

1 - 1 + 2 - 2 + 3 - 3 + ...
= 1 + (-1 + 2) + (-2 + 3) + ...
= 1 + 1 + 1 + ...
= infinity

Or:

1 - 1 + 2 - 2 + 3 - 3 + ...
= -1 + (1 - 2) + (2 - 3) + ...
= -1 - 1 - 1 - ...
= -infinity

There you go, three different answers. This is what is happening in your lottery ticket examples. It's not really a paradox, it's that manipulating divergent series in this way is not well-defined mathematically and can produce non-sensical results. There might be a name for this effect but I can't think of it.

Only if you're using the word "better" to mean something other than "has higher EV". Obviously the number of times you play a game doesn't affect its EV.

Yes exactly. I said the same thing before in this thread:

"...it is normal to get conflicting results when we attempt to group the terms in a divergent infinite sum in different ways..."

"when you take a divergent infinite sum and make different groupings out it. You get conflicting results.
+1 + -1 + 2 + -2 + 3 + -3 +... diverges
(a) (1 + -1) + (2 + -2) + (3 + -3) +... = 0 + 0 + 0...
(b) 1 + (-1 + 2) + (-2 +3) + (-3 + 4) +... = 1 + 1 + 1...
(c) -1 + (1 + -2) + (2 + -3) + (3 + -4) +... = -1 + -1 + -1...
"
So if we are going to play the game and compare AlwaysStay with AlwaysSwitch using EV, and we draw a tree of all possible outcomes with the first branch being how many tails are flipped and the next branch being which slip is drawn at random, we can calculate the EV of each node in the tree. When we go to sum them we get a divergent series. If we group the terms by number of tails seen we get two branches below that which cancel, so we get (0 + 0 + 0...) and the advice seems to be "It doesn't matter". If we group the terms by number seen N we get a sum of (2 + 2/9(3) + 2/9(9) + 2/9(27)...) and advice seems to be "switch". If we group the terms by number not seen, the sum is (-2 - 2/9(3) - 2/9(9) - 2/9(27)...) and the advice seems to be "stay".

But, yes, there is no paradox, once you see what's happening. The conflicting advice is due to different groupings of the diverging series.

So that takes care of any apparent paradox between AlwaysSwitch and AlwaysStay. It also implies that a person repeatedly playing LookAndAlwaysSwitch is not better of than someone playing AlwaysStay, even though the EV's at each N seem to favor LookAndAlwaysSwitch.

But still I am uneasy about the one case:
1. Person A decides to play the game 1 time only and wants to AlwaysStay
2. Person B decides to play the game 1 time only and wants to LookAndSwitch

When person B looks and sees N his EV for switching is +2/9N. He is not in an infinite EV situation any more. So it seems ironclad that he is better off switching and he is making a better choice than person A would make, no matter what N they see.

Yet prior to the game, if I am offered the choice to invest in person A or person B and receive corresponding winnings, there is clearly no reason for me to prefer one vs the other. I am back in the infinite EV situation.

I haven't really resolved that conflict in my mind yet. So for me that one case is still a paradox.

I'm not saying there really is a paradox. Ultimately I think any paradox like this is due to false reasoning or incorrect intuition. Once you discover the flaw in one of the arguments, or accept that your intuition is simply wrong, the paradox vanishes.

I am. But my usage is rational. If A has advantages over B and B has no advantages over A than we normally would say A is better. EV works better in finite situations because it always gives an answer (better, worse, equal) even when A and B both have some advantages over each other where as my usage only works in this one situation where A has advantages over B and B has no offsetting advantages. But if you take my usage of "better" and apply it to finite EV situations, it will never disagree with an EV calculation by saying "better" when EV says "worse" or "equal".

Right, but EV can't be used for comparison when EV's are infinite, yet in some situations a rational person would prefer A to B even though both EV's are infinite... like if they can only play 1 time and A is likely to make them rich and B is not.

I guess I missed that post.

The best explanation imo is because the optimal strategy isn't "Always switch", it's "Always switch except on the max value". Before we look, there's a chance we see the max value. Since we're letting the max value go to infinity, the chance we see it is infinitessimally small. Or I guess in your terms, you would keep flipping the coin forever and never reach the end of the game. That almost surely won't happen, but you can't say it's mathematically guaranteed not to happen. So it still remains a term in the EV equation of the overall game. Once a heads is flipped and we look at the paper, we know we haven't hit the max. So we switch.

I think what you're getting at is [url\http://en.wikipedia.org/wiki/Expected_utility]expected utility[/url]. If you're broke, winning $1 million is basically just as good as winning $2 million. They are both life-altering amounts of money. You might assign $1m an EU of 100 (on some arbitrary scale) and $2m an EU of 110 or something.

Unlike EV, EU does change depending on how many times you get to play the game. Once you win the first $1m and are no longer broke, winning another $1m or $2m is a nice bonus but no longer life-altering. So now the EU of $2m will be much closer to twice the EU of $1m. This is the same reason you don't play above your roll in poker even if you're +EV.

I don't really agree. I don't see that this game equates to the examples in that wiki article. I think if you consider the set of all possible sequences of coin flips including sequences of infinite length, that set does not contain any sequence consisting of an infinite number of tails followed by a heads followed by whatever. Every coin flip in some infinite sequence of coin flips is preceded by a finite number of coin flips. So how can you have an infinite number of tails followed by a heads?

I am not trying to include expected utility. If I include utility in what I am saying then I am saying that two choices A and B, with identical probability distributions for utility are equally desirable, even if the EU is infinite. Further, I am saying that if A and B have identical probability distributions for utility yielding infinite EU except that A beats B in one part of the distribution, then A is more desirable.

My reason for saying that (using the 99% +/- 100m example with only 1 chance to play) is that in most possible worlds I will be much happier with ticket A than B while in the rest of the possible future worlds A & B are equally likely to make me euphoric. With ticket A I am guaranteed to be very happy. With ticket B, I am probably miserable and possibly happy, but I am never a weighted average of the two.

EV doesn't change the more times we play, but EV does become more realistic the more times we play. EV equates [100% +$1] with [50% -$100,000 and 50% +$100,002] but in reality, these are not at all the same, unless we play lots of times.

I wish I could comprehend most of this.

I don't know what this has to do with anything. The point is the EV is undefined before the game. It isn't once you've looked at the piece of paper. It's not a paradox, it's the result of the game being non-sensical to begin with. I think you're overthinking this.

Err... this is exactly what EU is for. If you are broke you decide how much winning $100m is worth to you compared to how much being in debt $100m is. You come up with some arbitrary scale (utility) to weigh the two and decide if the EU of that bet is better than breakeven.

It's not because of how many games you play, it's because of your bankroll. Those two bets are only different if your bankroll is small and the EU of the outcomes are skewed. If you have a huge bankroll then you are basically indifferent between either bet.

The game as defined is going to be +EV no matter what. You pay me whats on my slip then the only thing in the air is just what magnitude +EV I end up. And that will be a matter of luck. Luck in how often I can pick the easy switches of case slip = 1 and luck in the cases that the first slip picked is the smaller number.

If the problem was that player A pays player B whats on his slip and player B pays player A then your payouts are different than shown but will still come down to the luck of the draw. The advantage going to the player who picks the smaller number in the cases where the numbers are very large and then switched and then more frequently the player who draws the 1 and has the choice to switch.

I was telling why I don't think it's correct to apply the wiki article about "almost surely" to this game and to view the game as having a maxiumum payout of +INF.

I understand that EV goes from undefined to finite. That still doesn't resolve the whole paradox for me.

Until the paradox is resolved for me, I can either continue thinking or give up. Paradox is really relative to the person. For someone who totally gets it (and hasn't made a mistake) there is no paradox.

I know what it is. My point is about comparing options A & B with infinite expectation, based on A having some advantage while B has no advantage. This applies to both $ and utility. It does not require conversion of $ to utility.

They have different characteristics. Std Dev of the sample mean is much different for 1 trial but becomes less with more trials. But yes, whether you care about that difference depends on bankroll.

Love this thread, but now very confused.

Anyway, I'm playing a one-off game with my brother and I've got a 9. Do I switch?

Ok then, what happens when you play the game and get infinite tails?

I suppose.

Sure, but I would argue that EU can't be infinite for the case of money. The more money you win, the less each win is worth to you, and at some point you could win so much that any more would make virtually no difference to your life.

If you think your brother has at least $27 in his bankroll and is being honest about the game conditions, then yes

As written the game enters an infinite loop and you never stop flipping. If that seems like a "bug" in the rules of game I think it can be rewritten as "Draw at random, a natural number T with probability 2^(T+1) and write on two slips of paper the numbers 3^T and 3^(T+1) etc."

I think if it converges you will lose the property of switching at every N.

Make that "Draw at random, a non-negative integer T with probability 1/2^(T+1) and write on two slips of paper the numbers 3^T and 3^(T+1) etc."

Okay, but you still have an undefined EV before looking, and a defined EV after looking. I really just don't see why you think it is a paradox that when you try to analyze a nonsensical game, you can get nonsensical results.

Probably. But this is getting a bit off topic... I don't think that really matters in terms of the paradox.

So, LookandSwitch (=AlwaysSwitch) is the best strategy for every possible game except the one where we get stuck in an infinite series of Tails and the game doesn't end.

That makes me think intuitively that AlwaysSwitch is a better strategy and hence seems paradoxical. But the Maths says 'no' (or does it say 'no comment'?).