Probability: Take a Coin

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Rhombic
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Probability: Take a Coin

Postby Rhombic » Sat Jan 24, 2015 7:23 pm UTC

There is this game where a player gets a coin from a small barrel. The player then looks at the coin and depending on whether it is fake or not, he'll have to pay some money or receive some money. The player can tell whether the coin is a fake one or not with a maximum accuracy. There is no room for doubt. The game is planned to proceed as follows:

If the coin was counterfeit: The player pays 1€ to the organiser and returns the coin to the barrel.

If the coin was genuine: The player receives 6€ and discards the coin. The player grabs another coin and discards it (without paying or receiving money, regardless of whether the coin was fake or not).

There are 24000 coins in the barrel.
Knowing that the expected value is -0.5€ (per turn, over the course of the game), how many counterfeit coins are there in the barrel?

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Dopefish
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Re: Probability: Take a Coin

Postby Dopefish » Sat Jan 24, 2015 9:01 pm UTC

As long as there is at least two fake coins in the barrel, isn't it possible the game will go forever if the game gets to the point where there are no genuine coins left? Or is there an alternative condition for the game ending other than all the coins being removed?

FancyHat
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Re: Probability: Take a Coin

Postby FancyHat » Sat Jan 24, 2015 10:13 pm UTC

Dopefish has a good point.

My answer, which I haven't checked, is that there is just one counterfeit coin in the barrel at the start. That's assuming there is an answer where the expected value is -0.5€. My reasoning is in the spoiler.

Spoiler:
If just two of the 24 000 coins are counterfeit, then there is a non-zero probability of removing all 23 998 genuine coins first, and then the game is endless. That would make the expected value -1€, wouldn't it? Having more counterfeit coins doesn't help, since there's still a non-zero probability of ending up with only counterfeit coins remaining. So, there have to be less than two counterfeit coins, otherwise the expected value is -1€, because of the non-zero probability of ending up with infinitely many turns where a counterfeit coin is inevitable.

Have I got that right?

But if there are no counterfeit coins, then the expected value has to be 6€.

So, that leaves the possibility of just one counterfeit coin. There is the possibility of ending up endlessly drawing that one counterfeit coin, and never drawing more than 11 999 genuine coins (so that no more than 23 998 coins get removed) while the counterfeit coin remains. The probability of ending up with just two coins left, including the counterfeit coin, is non-zero, but you'd almost surely end up drawing the genuine coin eventually.

So, with just one counterfeit coin and 23 999 genuine coins, is the expected value -0.5€? If I've got this right, it has to be, otherwise there's no solution to this problem.

(I'm assuming there's no 1+2+3+... = -1/12 stuff going on for this.)
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jaap
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Re: Probability: Take a Coin

Postby jaap » Sun Jan 25, 2015 9:12 am UTC

FancyHat wrote:
Spoiler:
If just two of the 24 000 coins are counterfeit, then there is a non-zero probability of removing all 23 998 genuine coins first, and then the game is endless. That would make the expected value -1€, wouldn't it?

It wouldn't. Once you have reached this state, all further turns each have that expected value, and the average value of the turns in this particular run of the game would be -1. However, that does not make the expected value of the turns at the start of the game as a whole equal to that value, as it is not guaranteed to end up in that state.

FancyHat
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Re: Probability: Take a Coin

Postby FancyHat » Sun Jan 25, 2015 11:22 am UTC

jaap wrote:It wouldn't. Once you have reached this state, all further turns each have that expected value, and the average value of the turns in this particular run of the game would be -1. However, that does not make the expected value of the turns at the start of the game as a whole equal to that value, as it is not guaranteed to end up in that state.

But if you end up in that state, you've got infinitely many turns, whereas if you avoid ending up in that state, there will only be finitely many turns (though there's no upper limit to what that finite number can be). And the longer a terminating game runs for, the closer the mean value per turn of that game gets to -1€. Doesn't that mean the expected value per turn for the game generally has to be -1€? I'm afraid I'm failing to see where I've gone wrong :?
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Rhombic
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Re: Probability: Take a Coin

Postby Rhombic » Sun Jan 25, 2015 12:11 pm UTC

The game stops when there are only fake coins left, for obvious reasons.

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jaap
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Re: Probability: Take a Coin

Postby jaap » Sun Jan 25, 2015 6:53 pm UTC

FancyHat wrote:
jaap wrote:It wouldn't. Once you have reached this state, all further turns each have that expected value, and the average value of the turns in this particular run of the game would be -1. However, that does not make the expected value of the turns at the start of the game as a whole equal to that value, as it is not guaranteed to end up in that state.

But if you end up in that state, you've got infinitely many turns, whereas if you avoid ending up in that state, there will only be finitely many turns (though there's no upper limit to what that finite number can be). And the longer a terminating game runs for, the closer the mean value per turn of that game gets to -1€. Doesn't that mean the expected value per turn for the game generally has to be -1€? I'm afraid I'm failing to see where I've gone wrong :?


This side discussion is now somewhat moot since there are no infinitely long games:
Rhombic wrote:The game stops when there are only fake coins left, for obvious reasons.


It still illustrates an ambiguity in the question. The "expected value of a turn of the game" can be interpreted in two ways:

1. (FancyHat's interpretation) If the game were played many times, what would be the expected value of the turns in all those games combined?

2. (My interpretation) The game is played once, and its average value per turn is calculated, call this v. What is the expected value of v?

In the first case, longer games will be more dominant, as they provide more turns. All turns are on an equal footing (weighted only by the probability of the game they are in actually occurring).
In the second case, it is the games that are on an equal footing (again, weighted only by the probability of that game actually occurring). The turns in a long game are not worth as much, compared to the turns in a short game, because they don't contribute as much to the average value per turn of the particular game they are in.


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