re: Statistics Spinoff Thread: How much would you pay to play this "game of chance"

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True, but all the casino has to do is make enough to win in the long run. If as you say, the average payout of 100 players is 600$. Lets say the casino sets the game at 8$ to play. You flip at least twice and you make your money back. 3 in a row is a winner. Casino makes an average of 200$ off of every 100 players and the chance of someone winning 128$ is < 1 in 100. My numbers aren't as clear as yours but does that make sense? I may be way off...

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This is correct. The casino obviously isn't going to set the price at the break even point. If this game were truly in a casino I think you would see a price of $8-10 per game.

game might work in casino set it up with a glass dome like video craps it would get action for $9-$10

The complete lack of understanding on the first page is hilarious.

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I would guess that over say 100 games, the mean amount won would be around $8, so that's how much you should pay. (Maybe $10)

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That is one way to view it, but if you look at it from an expected value standpoint, the EV is infinite as far as I can tell - .5*2+.25*4+.125*8... (.5^x)*(2^x) = 1+1+1+...1. To put it in perspective, the EV of the Powerball is usually around -$1.70ish or so when it originally starts.

As far as I can tell, the game should not be offered as there is no "cost" that the game could charge that would deter you from playing. In other words, you should pay whatever the game is asking because in the very, very long run you'll come out ahead.

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Operator?

Let's say the price is $100. You have a 50/50 shot at collecting $2 on the first flip and the game ending.

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The expected payout for every possible sequence is $1, and there are infinite possible sequences. The odds get twice as bad as the payout gets twice as good. This goes on forever, or $(infinity symbol).

Graphically...

T = .5 * $2 = $1
HT = .5 * .5 * $4 = $1
HHT = .5 * .5 * .5 * $8 = $1
HHHT = .5 * .5 * .5 * .5 * $16 = $1
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Add them all up to get the total expected payout, which is $1+$1+$1...(forever), or infinite dollars.

Practically, probability reaches zero at about 6 or 7 straight heads, but it never truly gets there unless you force a limit.

I'd probably pay $5, but would argue the game would never be offered because of the potential for infinite losses.

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As far as I can tell, the game should not be offered as there is no "cost" that the game could charge that would deter you from playing. In other words, you should pay whatever the game is asking because in the very, very long run you'll come out ahead.

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This doesn't make sense. You'll need to show more of your work.

Yeah realistically I would pay 5 bucks to play that game

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Yeah realistically I would pay 5 bucks to play that game

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Realistically, no casino would allow it with an average 6$ per game payout.

I'd charge $10 to play the game.

In order for the house to lose money, there would have to be three consecutive heads. (12.5% chance).

The house has an 87.5% chance of making money.

A 50% chance of making $8
A 25% chance of making $6
A 12.5% chance of making $2

There is a 1 in 8 chance of the house losing money, and while the pot could theoretically grow infinitely, there is only a 6.25% chance of losing $6, a 3.125% chance of losing $22, and a 1.6125% chance of losing $54.

So if statistics held, and 8 players played, putting in $80, the house is going to make $8 off of four them ($32), $6 off of two of them ($12), $2 off of one of them ($2) and then the eighth player would win with the house up $46. It would take two more consecutive tails (3.125% chance) before the house would lose money.