is the EV for this game really infinity?

[jay_shark]

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 .

[cabiness42]

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So, the expected number of flips is [2]

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

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

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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.

[R Gibert]

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.

[cabiness42]

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.

[T50_Omaha8]

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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.

[iloveny161]

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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

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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?

[R Gibert]

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.

[WhiteWolf]

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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.

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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

[pococurante]

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".

[cabiness42]

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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".

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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.