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re: Statistics Spinoff Thread: How much would you pay to play this "game of chance"
Posted on 1/15/16 at 2:50 pm to PearlJam
Posted on 1/15/16 at 2:50 pm to PearlJam
quote:
I get this part, but if the expected payout is $1, why is the answer an infinite amount of money?
Because you ADD the value of each of the possible results.
0
Posted on 1/15/16 at 2:51 pm to GenesChin
Posted on 1/15/16 at 2:56 pm to PearlJam
quote:
I get this part, but if the expected payout is $1, why is the answer an infinite amount of money?
Also, why is the expected payout $1 on the first line of your graph when it is know to be $2?
The formula seems to be missing a component.
There is a .5 chance you win $2, so the expected payout there is $1 ($2 *.5).
Each line on my chart is a potential outcome. You have to add up all the potential outcomes to get a total expected payout. Each potential outcome has the same expected payout ($1), because the payout keeps doubling while the probability keeps halving. So, the total expected payout is $1+$1+$1... forever.
To make the game real, you have to cap the payout.
T = 1/2 * $2/1 = $1
HT = 1/4 * $4/1 = $1
HHT = 1/8 * $8/1 = $1
HHHT = 1/16 * $16/1 = $1
HHHHT = 1/32 * $32/1 = $1
.
.
. (forever)
(sigma) = $1 + $1 + $1 + $1 + $1 + ... (forever) = $(infinity)
Posted on 1/15/16 at 2:58 pm to slackster
St. Petersburg paradox. I see the math, but still don't understand the probabilities. I'll look at it later when I have more time.
Posted on 1/15/16 at 3:01 pm to Speedy G
quote:That isn't exactly accurate. There is a 100% chance you win at least $2. There is a .5 chance you win at least $4. If the expected value mathematics can't resolve that point, how can I trust them?
There is a .5 chance you win $2, so the expected payout there is $1 ($2 *.5).
Posted on 1/15/16 at 3:06 pm to anc
quote:
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.
Fair enough, but that only works if you're capping it to 8 players. If you let 1000 people play, statistically 125 people are expected to win $16 or more, one of which would be expected to win $1,024 and another would be expected to win $2,048 or more. If he wins $2,048, you're going to have lost ~$2,060. If he wins "or more", it could get out of hand quickly.
Posted on 1/15/16 at 3:12 pm to PearlJam
quote:
That isn't exactly accurate. There is a 100% chance you win at least $2. There is a .5 chance you win at least $4. If the expected value mathematics can't resolve that point, how can I trust them?
You're looking at this the wrong way.
You don't actually "win" anything until you flip a tails, so use that to define your success - when will you flip a tails and what is the corresponding payout.
There is a 50% chance you flip a tails on the first flip and a $2 payout for that flip - .5*$2=$1 towards EV of the game. There is a 25% chance that you'll flip get to the 2nd flip and then flip a tails a tails - .25*$4=$1 toward EV. So on and so forth until infinity.
Posted on 1/15/16 at 3:17 pm to slackster
quote:So say I change the game and if you flip tails on the first flip you get $0. Everything else remains the same. What is the EV?
There is a 50% chance you flip a tails on the first flip and a $2 payout for that flip - .5*$2=$1
Posted on 1/15/16 at 3:18 pm to GenesChin
This is not statistics. This is probability. Some of you are just dumb as a box of rocks.
Posted on 1/15/16 at 3:29 pm to PearlJam
quote:
So say I change the game and if you flip tails on the first flip you get $0. Everything else remains the same. What is the EV?
Still infinity for the entire game.
Posted on 1/15/16 at 3:30 pm to slackster
quote:What about the first flip?
Still infinity for the entire game.
Posted on 1/15/16 at 3:31 pm to sullivanct19a
quote:
This is not statistics. This is probability. Some of you are just dumb as a box of rocks.
Statistics and probability are closely related - enough so that bitching about it is petty.
Posted on 1/15/16 at 3:32 pm to PearlJam
quote:
What about the first flip?
$0 assuming no cost of the game.
If the game has a cost it would be .5*(0-x) where 'x' is the cost of the game.
Posted on 1/15/16 at 3:34 pm to slackster
quote:Why 0? I have a .5 chance of winning $2.
$0 assuming no cost of the game.
Posted on 1/15/16 at 3:36 pm to PearlJam
quote:
So say I change the game and if you flip tails on the first flip you get $0. Everything else remains the same. What is the EV?
Still infinite, but grows more slowly: $0 + $.50 + $.50 + $.50 +...
Still, nothing to prevent the game from going on forever.
Mathematically, the problem is that the payouts and probabilities and growing/shrinking in proportion to one another, so there is no convergence. With no convergence, there is not limit, and thus no real answer.
Posted on 1/15/16 at 3:37 pm to PearlJam
quote:
What about the first flip?
You cannot calculate an expected value for the first flip, b/c it can lead to infinite subsequent flips.
Theoretically, you can just keep flipping heads until the world ends and never get paid.
Posted on 1/15/16 at 3:45 pm to Speedy G
quote:Got you. I understand it mathematically. Interesting paradox. Pragmatically, when math is in the form of human behavior (games of chance) it is difficult for the mind to imagine infinite flips and an infinite bankroll so as to allow for enough opportunities to guarantee success.
Mathematically, the problem is that the payouts and probabilities and growing/shrinking in proportion to one another, so there is no convergence. With no convergence, there is not limit, and thus no real answer.
Posted on 1/15/16 at 3:50 pm to PearlJam
quote:
Why 0? I have a .5 chance of winning $2.
Not on the first flip, if the pot starts at $0.
You can win $2 on the second flip. That outcome has a .25 chance (.5 * .5), but there are infinite other possibilities, with increasing payouts and decreasing probabilities.
The easiest way to look at probability problems is to consider every possible outcome. Assign each outcome a probability. The sum of those probabilities must equal 1. Then, assign each outcome a value (in games of chance like this, some payout times the probability). Then sum up all the values to get the total expected value. The problem here is that there are infinite possible outcomes and no convergence (they just keep going forever in a straight line).
Posted on 1/15/16 at 3:57 pm to Speedy G
quote:If I flip heads on the first flip I win $2. My options are winning 0 or winning $2. So I have .5 chance of winning $2.
Not on the first flip, if the pot starts at $0.
quote:Perfect. With the ops question on the first flip I have 2 outcomes- Get $2 or get more than $2 - the probability that I get at least $2 is 100% so my expected value is at least $2, right?
consider every possible outcome. Assign each outcome a probability.
This post was edited on 1/15/16 at 4:06 pm
Posted on 1/15/16 at 4:05 pm to PearlJam
quote:
Why 0? I have a .5 chance of winning $2.
Huh? Your question:
quote:
So say I change the game and if you flip tails on the first flip you get $0.
You have a 50% chance of getting $0 on the first flip now.
You also have a 50% chance of getting $4 or more assuming you flip heads first, but probability/EV doesn't work that way. You multiply the unique outcomes by the probability of that outcome - 25% of $4, 12.5% of $8, ...
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