is the EV for this game really infinity?
 

just did this in econ class but i didn't think it was true....
i thought i might have missed something so 2p2, analyze this game with me

basically we'll flip coins:

If you get head on the first flip, i will give you 2 dollars

if you get tails on the first flip and get head on the 2nd flip, i will give you 4 dollars

if you get tails on the first 2 flips and get head on the 3rd flip i will give you 8 dollars

if you get tails on the first N-1 flips and get head on the nth flip, i will give you 2^(n) dollars

what is the EV for this game?


the answer i got in class was infinity

because the EV is (1/2)x2 + 1/4 (4)+ 1/8 (8) +.... + 1/(2^n) x 2^n

hence the sum of all the EVs = infinty

there seems to be a flaw in the logic here and i can't point it out and it's bugging the crap out of me

solve this for me 2p2

Re: is the EV for this game really infinity?
 

ahhh i just got it after i posted it

the EV for the game is not the sum of the EVs, but it's just 1 of the EVs, hence the EV of the game is 1 since you only get 1 pay out

Re: is the EV for this game really infinity?
 

LDO?

econ prof here i come

Re: is the EV for this game really infinity?
 

This is the well-known Petersburg Paradox.

Re: is the EV for this game really infinity?
 

I will pay you $10 million to play this game

Re: is the EV for this game really infinity?
 

Do I ever loose money?

Re: is the EV for this game really infinity?
 

[ QUOTE ]
I will pay you $10 million to play this game

[/ QUOTE ]


you realize there is only 1 pay out right?


i will play this game with you

when will you pay me?

Re: is the EV for this game really infinity?
 

[ QUOTE ]
ahhh i just got it after i posted it

the EV for the game is not the sum of the EVs, but it's just 1 of the EVs, hence the EV of the game is 1 since you only get 1 pay out

[/ QUOTE ]
No, your original calculation was technically correct. Read the linked wikipedia article for proposed explanations on why most rational people would not pay a lot of money for this lottery. For me, the most compelling was even if it was backed by Bill Gates' entire wealth, it wouldn't be worth more than $18 dollars.

Re: is the EV for this game really infinity?
 

Interesting proposition. Lets say that you'd have to pitch in 3$ to play. Would this mean negative profit?

According to the EV=1 statement, it would.

However, you always win at least 2$.
So, you always win 2 or 4 or 8 or 16... and so on. So the EV has to be more than 2.

I'd say that the EV is a bit more than 2 with high positive variance. Just by thinking about cases how it could end up.

Think about it this way.
Would you play it if your bet would be 3$?

Also, lets assume you play it 100 times. Assuming perfect distribution you would end up with:
50 times you win 2$
25 times you win 4 and so on.
If we stop the flips at 5 tails and one head, which you'll get once you'd make ~600$. Divided by 100 games that makes 6$ per game.

Repeat that, but now 10 000 times. Assuming you never get to more than 9 tails and one head (around 1/10000) you'd make ~10 $ per game, thus you'd make about 100 grand.

I am not good at this kind of math (sums over ranges) but I think you could simulate the EV with monte carlo simulation for several amounts of games. Like once for 10, 100, 1000, 10000 and so on, and get some kind of result.

Re: is the EV for this game really infinity?
 

[ QUOTE ]
Interesting proposition. Lets say that you'd have to pitch in 3$ to play. Would this mean negative profit?

According to the EV=1 statement, it would.

However, you always win at least 2$.
So, you always win 2 or 4 or 8 or 16... and so on. So the EV has to be more than 2.

I'd say that the EV is a bit more than 2 with high positive variance. Just by thinking about cases how it could end up.

Think about it this way.
Would you play it if your bet would be 3$?

Also, lets assume you play it 100 times. Assuming perfect distribution you would end up with:
50 times you win 2$
25 times you win 4 and so on.
If we stop the flips at 5 tails and one head, which you'll get once you'd make ~600$. Divided by 100 games that makes 6$ per game.

Repeat that, but now 10 000 times. Assuming you never get to more than 9 tails and one head (around 1/10000) you'd make ~10 $ per game, thus you'd make about 100 grand.

I am not good at this kind of math (sums over ranges) but I think you could simulate the EV with monte carlo simulation for several amounts of games. Like once for 10, 100, 1000, 10000 and so on, and get some kind of result.

[/ QUOTE ]

didn't really read your post becuase your avatar gave me night mares

i think the EV is infinity but the utility of this is less and it's the paradox

i hope this enlightened someone cause it sure made my head spin

Re: is the EV for this game really infinity?
 

If you could have multiple iterations but it was relatively expensive, I assume this would some truly sick variance.

Re: is the EV for this game really infinity?
 

You can play this +EV game for $20 at this website. Repeatedly press the "play a single game" key and watch your positive EV losses mount!
http://www.mathematik.com/Petersburg/Petersburg.html

Re: is the EV for this game really infinity?
 

ok, I got curious so I ran my own simulations on this and got some plots. I didn't like the plot shown on the wiki because that's just a running average of your winnings over however many games you've played, not a running total win/loss. And it doesn't even take into account any fee that might be paid to play such a game anyways.

So here's some results, under 20,000 reps with various fees. This shows a running total of your earnings while playing this game repeatedly.
http://i16.photobucket.com/albums/b3...petersburg.jpg

soooo... ok you can pay up to 10 bucks a game and probably be pretty happy with the outcome over time, but... it really doesn't look good when you're playing for 20 bucks a game. But, this is ONLY 20,000 games! (hah) Let's see what happens when I jack it up!!

http://i16.photobucket.com/albums/b3...tersburg20.jpg

err... still doesn't look so good.

I will note, though, that while playing around with it, under some trials of 20,000 reps the $20 fee line did wind up slightly above $0 every now and then.

I'll leave you to draw your own conclusions.

Also here's my R code for those who are proficient in it and would like to see what I did, maybe let me know if anything looks fishy. I know it's a bit clunky too; I need to use less loops. Also note that the OP says the pot starts at $2 whereas the wiki version starts the pot at $1; I didn't notice this until now and just followed the wiki, so I guess things will differ by a factor of 2 somewhere, but the idea is obviously the same.

http://i16.photobucket.com/albums/b3.../pburgcode.jpg

Re: is the EV for this game really infinity?
 

[ QUOTE ]
I didn't like the plot shown on the wiki because that's just a running average of your winnings over however many games you've played, not a running total win/loss. And it doesn't even take into account any fee that might be paid to play such a game anyways.

[/ QUOTE ]
To factor out the effect of the fee, subtract this from the average win.

[ QUOTE ]

I'll leave you to draw your own conclusions.


[/ QUOTE ]
Your graph is misleading.

It's not hard to check that the median, and any (nondegenerate) percentile of the average win will grow as the logarithm of the number of trials.

Re: is the EV for this game really infinity?
 

[ QUOTE ]
[ QUOTE ]
I didn't like the plot shown on the wiki because that's just a running average of your winnings over however many games you've played, not a running total win/loss. And it doesn't even take into account any fee that might be paid to play such a game anyways.

[/ QUOTE ]
To factor out the effect of the fee, subtract this from the average win.

[/ QUOTE ]
While similar, and almost effectively the same thing, this still doesn't answer precisely the question that I was interested in. This is still an average win over how many games had been played up to that point. I wanted to know my net win/loss at each point. Yeah it's not really that different, but when I leave a casino, I tend to think in terms of $$ in my pocket.

[ QUOTE ]
[ QUOTE ]

I'll leave you to draw your own conclusions.


[/ QUOTE ]
Your graph is misleading.

It's not hard to check that the median, and any (nondegenerate) percentile of the average win will grow as the logarithm of the number of trials.

[/ QUOTE ]
Is my code wrong? If not, then my graph is exactly what I've claimed it to be: one realization of what will happen over 5 million games. This is why I personally concluded nothing from it, other than that you might not be happy after playing this game 5 million times. I actually expected things to look a little better than that over that many samples, but was wrong.

I didn't intend for the statement "I'll leave you to draw your own conclusions" to imply that one should conclude that a $20 fee will make you a long run loser, though I do see how it might look like that.

Re: is the EV for this game really infinity?
 

People keep mentioning this game's variance. Does this game even have a finite variance?

I would assume not.

Re: is the EV for this game really infinity?
 

Since the expected value is infinite, the variance has to be infinite.

Re: is the EV for this game really infinity?
 

peterchi: Interesting graphs, thanks for posting them.

The problem with the $20 fee version: After 20,000 games, your winnings will probably be around $150,000 (barring an extremely unlikely run of good or bad luck). If you're playing for $2 you'll be up $110k, but for $20 you'll be down $250k.

As your graph shows, the $2 and $5 games are profitable quickly, and the $10 game takes several thousand plays before you come out ahead. As you play for more money, it takes an exponentially longer amount of time before (with normal luck) you'll make a profit.

To make a profit in the first 20,000 plays, you'll have to hit a string of 17 tails in a row. The odds of this are 1 in 131,072, so you've only got a 1 in 6.5 chance of being profitable by then.

Hmm, now you're going to make me find out how long on average before a $20 game does become profitable. Seems to me like it should have happened in the simulation you ran... hard to say for sure though.

Re: is the EV for this game really infinity?
 

OK, I realized two things:

1) In my above post it should be 16 tails in a row (not 17), with 1 in 65k odds, so a 1 in 3.25 chance to have a profit

2) This is beyond my ability to calculate, so I quit. Here are some facts though:

After 5 million games, you've paid $100 million to play.
A string of 25 tails is worth $67 million. I assume you'll at least need that to happen once in order to make a profit.

The odds of 25 tails in a row is over 1 in 33 million, so you're even LESS likely to have made a profit at this point. So it looks like there are now two possibilities here:

1) I made another mistake

2) $20 is so far above the usual amount won, that the only way you can ever profit is to play an infinite number of games. Because no matter how much money you've lost, the odds of hitting some virtually impossible jackpot will never reach 0.

Re: is the EV for this game really infinity?
 

[ QUOTE ]
The problem with the $20 fee version: After 20,000 games, your winnings will probably be around $150,000 (barring an extremely unlikely run of good or bad luck). If you're playing for $2 you'll be up $110k, but for $20 you'll be down $250k.

[/ QUOTE ] I don't understand the basis of these statments. Are you using some sort of confidence interval? Without a mean it really makes no sense to me.

And since these are all independent events, I don't see any way you can expect to go down at first then eventually up. I understand how the lower bound of a confidence interval could begin negtive for some confidence level then become positive as the number of trials increases (given mean > 0), but the notion of an upper bound being below zero for a number of trials then the lower bound being above zero for a number of trails seems impossible, given a starting value of 0 and independent trials.

I'll wait for someone better at probability than I to weigh in.

Re: is the EV for this game really infinity?
 

Actually I just used the graph at http://en.wikipedia.org/wiki/St._Petersburg_paradox for an estimate for 20k hands. I also found another graph on a second site with similar results, so I assumed it's fairly accurate. (And now I see that the game being used on there gives $1 for an immediate loss, while the game the OP used gives $2 for an immediate loss, so all my numbers are completely wrong. sorry.)

As for losing at first, but then going up later, that's because of the infrequency of the large wins.

If you pay $6 to play the version that starts at $1, chances are you will be losing for quite a while. Your first handful of games will probably be a bunch of 1s and 2s with an occasional 4 or 8. Only after a few thousand games, when you will have come across some large "jackpot" wins, will you become profitable in the long run.

Re: is the EV for this game really infinity?
 

OK, here's how to get a realistic EV for a certain number of games. This method assumes average luck.

For 2,048 games with a $2 minimum prize, you will typically win the sum of:

1,024 wins for $2
512 * 4
256 * 8
128 * 16
64 * 32
32 * 64
16 * 128
8 * 256
4 * 512
2 * 1024
1 * 2048
(there is one game remaining, which is your "luckiest" game of the session, which is technically worth infinity. For the sake of convenience I'm calling it a win of 4096... yes I know this isn't mathematically correct)

Note that all of the above equations equal 2048. 2048 = 2^11. Adding in that final game, your total winnings will be 2048 * 13... meaning that your average per game will be $13... so all you have to do is find 2^X for your particular number of plays, then add 2.

# of plays ------ EV per game
8 --------------- 5
256 ------------- 10
1,024 ----------- 12
8,192 ----------- 15
32,768 ---------- 17
262,144 --------- 20
8,388,608 ------- 25
268,435,456 ----- 30

These numbers match up well with the graphs I've seen, so I'm pretty sure these are reasonably accurate. The EV rises quickly early on, but then slows down considerably. It takes only 8 games to reach an EV of 5, but then each increase of $5 EV requires 32 times as many games as you've played to that point.

Here's a quick chart... I don't have Excel installed so I used the retarded Google version that doesn't let you choose which info goes on which axis.
http://img456.imageshack.us/img456/229/failchartpm9.jpg

BTW I don't know why peterchi's results for $20 games didn't become profitable after 5 million games, it makes no sense for that to happen. It should fall sharply, slow down and hit a low point at ~110k games, then begin an upswing and break even at 262k games.

Yes, I was this bored... once I got started I just wanted to "beat the game".

Re: is the EV for this game really infinity?
 

The amount of completely incorrect math/probability theory in this thread is astonishing.

Re: is the EV for this game really infinity?
 

[ QUOTE ]
The amount of completely incorrect math/probability theory in this thread is astonishing.

[/ QUOTE ]

Other than assuming I could substitute a small number for something that is technically equal to infinity, was there anything in my last post?

Re: is the EV for this game really infinity?
 

I typed out a detailed description of how to estimate the probability that you win less than $100 million in 5 million trials, but apparently I took too long so the forum would not accept my post. Usually, I save the posts to the clipboard before trying to submit, but this time it didn't work. This time limit is annoying. What benefits are there that are supposed to outweigh the many lost posts?

Anyway, the answer was 1.43%, within about a factor of 1.02. The probability that you win more than $100 million is about 98.57%.

Re: is the EV for this game really infinity?
 

[ QUOTE ]
I typed out a detailed description of how to estimate the probability that you win less than $100 million in 5 million trials, but apparently I took too long so the forum would not accept my post. Usually, I save the posts to the clipboard before trying to submit, but this time it didn't work. This time limit is annoying.


[/ QUOTE ]

Agreed. For longer posts, I use notepad then copy the text into the form. Unfortunately, this screws up the formatting, so sometimes I just copy the text from the form to the Windows' clipboard before attempting to submit it. If the submission fails for some reason, I can paste it into a new form.

Re: is the EV for this game really infinity?
 

Don't try calculating the EV right off the bat. Calculate the number of expected coin flips the game will last:

1/2 of the time it lasts 1 flip
1/4 of the time it lasts 2 flips
1/8 of the time it lasts 3 flips
etc.

So, the expected number of flips is (1*1/2)+(2*1/4)+(3*1/8)+ . . .

That number equals 2.

So the excpected length of the game is 2 flips, and your EV is then 4.

Re: is the EV for this game really infinity?
 

[ QUOTE ]
I typed out a detailed description of how to estimate the probability that you win less than $100 million in 5 million trials, but apparently I took too long so the forum would not accept my post. Usually, I save the posts to the clipboard before trying to submit, but this time it didn't work. This time limit is annoying. What benefits are there that are supposed to outweigh the many lost posts?

Anyway, the answer was 1.43%, within about a factor of 1.02. The probability that you win more than $100 million is about 98.57%.

[/ QUOTE ]
Next time, just hit the back button on your browser, copy your post, re-reply and paste.

Re: is the EV for this game really infinity?
 

[ QUOTE ]
[ QUOTE ]
I typed out a detailed description of how to estimate the probability that you win less than $100 million in 5 million trials, but apparently I took too long so the forum would not accept my post. Usually, I save the posts to the clipboard before trying to submit, but this time it didn't work. This time limit is annoying. What benefits are there that are supposed to outweigh the many lost posts?

Anyway, the answer was 1.43%, within about a factor of 1.02. The probability that you win more than $100 million is about 98.57%.

[/ QUOTE ]
Next time, just hit the back button on your browser, copy your post, re-reply and paste.

[/ QUOTE ]
That was the first thing I tried, of course, and it didn't work. Browsers do not always save the information in textboxes.

Re: is the EV for this game really infinity?
 

[ QUOTE ]

So, the expected number of flips is [2]

So the excpected length of the game is 2 flips, and your EV is then 4.

[/ QUOTE ]
Non sequitur, and your conclusion is wrong. The correct answer has already been given along with justifications and links to other discussions.

Re: is the EV for this game really infinity?
 

Pzhon , do you use internet explorer ?

I've used internet explorer in the past but some of my posts would get timed out so I switched to Firefox . A simple page refresh and your post is good to go .

Re: is the EV for this game really infinity?
 

[ QUOTE ]
[ QUOTE ]

So, the expected number of flips is [2]

So the excpected length of the game is 2 flips, and your EV is then 4.

[/ QUOTE ]
Non sequitur, and your conclusion is wrong. The correct answer has already been given along with justifications and links to other discussions.

[/ QUOTE ]

So my conclusion is wrong just because you don't agree? I stand by my calculations, the expected length of the game is two flips.

Re: is the EV for this game really infinity?
 

Your solution does not incorporate any information concerning what the payoff is when the game lasts 1 flip or if the game lasts 3 or more flips as if this had no impact.

You can't just separate out the payouts, calculate the average length of a game and then later pick out the payout that corresponds to the average length of a game and ignore the rest. This isn't kosher mathematically and intuitively, the payout scheme ought to have an impact. Yes?

So your method of solution can't possibly be correct unless you were to show that indeed the longer or shorter games have no impact as in they cancel each other out or some such.

Not only do you not do this, I don't think you can do this, so your solution is incorrect for at least this reason.

Keep trying.

Re: is the EV for this game really infinity?
 

Perhaps we're answering two different questions. I am answering the question of what the expected payout is if you play the game one time.

That could be a very different answer from what the expected average payout is if you play the game many times.

If you are just going to play the game one time, you should expect the game to last two rounds and expect to win $4.

Re: is the EV for this game really infinity?
 

[ QUOTE ]
Perhaps we're answering two different questions. I am answering the question of what the expected payout is if you play the game one time.

That could be a very different answer from what the expected average payout is if you play the game many times.


[/ QUOTE ]We are answering the exact same question.

The expectation is infinite.

Re: is the EV for this game really infinity?
 

[ QUOTE ]
ahhh i just got it after i posted it

the EV for the game is not the sum of the EVs, but it's just 1 of the EVs, hence the EV of the game is 1 since you only get 1 pay out

[/ QUOTE ]

I believe someone already said this, but this is wrong. The EV of the game is infinity, but the reason we are not willing to play this game for millions of dollars is utility.
Think about the following scenario...
You have a choice between these two cases:
You can have a 100% shot of winning 100 million,
or a 10% shot of winning 2 billion.
Which do you choose? Which one gives you the higher EV?

Re: is the EV for this game really infinity?
 

There are games in game theory where how many times they are played makes a difference, but this is due to the exercise of memory by one or both participants. Past decisions have an impact on future decisions. Here however, the game does not feature any decision making, so whether the game is repeated or played singly makes no difference. The games are fully independent of each other.

Here is a link to the game theory game that is most "famous" for the effect you mention in case you are interested:

http://en.wikipedia.org/wiki/Prisone...er.27s_dilemma

In the game in question here, the relevant infinite series boils down to EV = 1 + 1 + 1 + 1 + ...

This sum of course diverges. However, there is something called Ramanujan Summation where a different notion of summation of such series is taken seriously. Apparently, this has its uses.

Within that framework, which is not applicable in our case, the sum is unexpectedly -1/2 of all things. Here are some links concerning this:

http://en.wikipedia.org/wiki/1_%2B_1..._%C2%B7_%C2%B7

http://en.wikipedia.org/wiki/Ramanujan_summation

Crazy stuff. I don't pretend to understand it.

Re: is the EV for this game really infinity?
 

[ QUOTE ]


If you are just going to play the game one time, you should expect the game to last two rounds and expect to win $4.

[/ QUOTE ]
I think your confusion may be because are using a standard English definition of the word 'expect,' and not the precise probability theory definition of the word 'expectation.'

To illustrate: If a friend of mine offered to roll a single dice and pay me $1 if it came up 1-5, but $100 if it came up 6, I would "expect" to win $1, but my "expectation" would be $17.50

Re: is the EV for this game really infinity?
 

cabiness42: Why do you think the game will last for 2 flips? There is only a 25% chance of that result.

If you go by the most common result, that will be the 50% chance of the game lasting 1 flip, with a payout of $2. Of course, this number means nothing... suppose I offered a game with 3 equally likely outcomes: you win $1, you win $1, you win $1,000,000. I'm sure you wouldn't say this game is worth only $1 to play.

The only other way to do it (which happens to be the correct way) is to add up all the possible payouts (divided by their chances of happening, of course). The result is

(0.5 * 2) + (.25 * 4) + (.125 * 8) + ...

It never ends... it's an endless string of 2+2+2+2+2+2+2+2+2, so the result is infinity.


It shouldn't be surprising that you can't get a mathematically correct answer by going "well, half the time it's one flip, but sometimes it goes for 3 or more, so it'll probably be somewhere around 2".

That works as well as going "well, AJ doesn't win as much as AA, but it still wins sometimes, so I guess it's around 1 in 3 to beat AA".

Re: is the EV for this game really infinity?
 

[ QUOTE ]
cabiness42: Why do you think the game will last for 2 flips? There is only a 25% chance of that result.

If you go by the most common result, that will be the 50% chance of the game lasting 1 flip, with a payout of $2. Of course, this number means nothing... suppose I offered a game with 3 equally likely outcomes: you win $1, you win $1, you win $1,000,000. I'm sure you wouldn't say this game is worth only $1 to play.

The only other way to do it (which happens to be the correct way) is to add up all the possible payouts (divided by their chances of happening, of course). The result is

(0.5 * 2) + (.25 * 4) + (.125 * 8) + ...

It never ends... it's an endless string of 2+2+2+2+2+2+2+2+2, so the result is infinity.


It shouldn't be surprising that you can't get a mathematically correct answer by going "well, half the time it's one flip, but sometimes it goes for 3 or more, so it'll probably be somewhere around 2".

That works as well as going "well, AJ doesn't win as much as AA, but it still wins sometimes, so I guess it's around 1 in 3 to beat AA".

[/ QUOTE ]

Well, if you read my post, I didn't just pull 2 out of the air. I calculated an expected number of flips the game would last based on the probabilities of each individual number of flips. The result was a series that converged to 2. That has absolutely nothing to do with your example of AA vs. AJ.