Expected Value
The Wizard's been discussing The Lion's Share slot machine at the MGM Grand and how it may now have a positive expected value given value of the progressive jackpot currently. If one were to play the Lion's Share machine until it hit the jackpot, if the Wizard is correct about it being +EV, then on average you would win more money than you put in to win that jackpot. However it's possible that you never win on any spin on the Lion's Share prior to hitting the jackpot AND it took more money to hit the jackpot than the jackpot is worth. Similarly you could win infrequently enough during play such that the amount of money you spent chasing the jackpot exceeds the total amount of money you won. Are these two things correct?
So what's at play here? I presume it's variance. Is it more or less likely that if you were to play the Lion's Share until its jackpot that you would come out ahead? The variance on the machine would presumably determine this and expected value would have less bearing here. Is that correct?
How many hands of blackjack would one have to play before seeing your winnings/losses converge to the expected value? How many pass line bets with odds in craps? How many hands of video poker?
The Wizard's signature is that it's not whether you win or lose, but if you had a good bet. I agree to a point. Not to rub salt in the wound, but the Wizard made good props bets on the previous Super Bowls -- no safety etc. -- and still lost. His losses are not an outrageously unlikely outcome and the Wizard shouldn't feel bad about losing, but it's still tough to swallow. I believe Buzzard commented in the 'Safety' thread that if you bet $100 on a safety in the last 20 Super Bowls you'd only be down $580. Not as big a loss as I would have thought.
So I guess my overall question is given that I'm not going to play forever shouldn't I worry less about expected value and more about the variance of the games I play?
--helpmespock
Quote: helpmespockI've been thinking about expected value and what it means. I believe the mathematical definition of expected value is if you were to play something infinitely then the expected value is what your average winnings would converge to. That means for a +EV game you'll make money, and in a -EV game you'll lose money if you were to play forever. However none of us play forever. Casinos do, but players don't.
The Wizard's been discussing The Lion's Share slot machine at the MGM Grand and how it may now have a positive expected value given value of the progressive jackpot currently. If one were to play the Lion's Share machine until it hit the jackpot, if the Wizard is correct about it being +EV, then on average you would win more money than you put in to win that jackpot. However it's possible that you never win on any spin on the Lion's Share prior to hitting the jackpot AND it took more money to hit the jackpot than the jackpot is worth. Similarly you could win infrequently enough during play such that the amount of money you spent chasing the jackpot exceeds the total amount of money you won. Are these two things correct?
So what's at play here? I presume it's variance. Is it more or less likely that if you were to play the Lion's Share until its jackpot that you would come out ahead? The variance on the machine would presumably determine this and expected value would have less bearing here. Is that correct?
How many hands of blackjack would one have to play before seeing your winnings/losses converge to the expected value? How many pass line bets with odds in craps? How many hands of video poker?
The Wizard's signature is that it's not whether you win or lose, but if you had a good bet. I agree to a point. Not to rub salt in the wound, but the Wizard made good props bets on the previous Super Bowls -- no safety etc. -- and still lost. His losses are not an outrageously unlikely outcome and the Wizard shouldn't feel bad about losing, but it's still tough to swallow. I believe Buzzard commented in the 'Safety' thread that if you bet $100 on a safety in the last 20 Super Bowls you'd only be down $580. Not as big a loss as I would have thought.
So I guess my overall question is given that I'm not going to play forever shouldn't I worry less about expected value and more about the variance of the games I play?
--helpmespock
You should worry about both. Apply the Kelly Criterion to determine if you can afford to make a particular bet.
Or in the words of Henry Gordon, East Coast bookie and gambler, subpenaed by the Kefauver Committee, arrested several times, never spent a day in jail, his motto " I DON'T WANT NOTHING TO DO WITH AN EVEN BET "
Quote: Buzzard" I DON'T WANT NOTHING TO DO WITH AN EVEN BET "[/q
I almost thought you were going to say EVEN BOB :)“There is something about the outside of a horse that is good for the inside of a man.” - Winston Churchill
Quote: Buzzard" A good bet " is one in which you have the best of it. PERIOD....
No, this is not true. Consistently overbet your bankroll by too much and you will go broke, even if you had the edge on every bet.
Betting too much, even with an edge, is a bad bet.
If a game's expected value is $1.00, you can expect to make $1 each and every time you play. Doesn't matter if you play once or a million times.
You are right in concluding that expected value is not the only thing to consider when gambling. You should also be familiar with the concepts of risk of ruin, variance, and the utility of money.
For example, it would be silly to risk $1 for a 1 in trillion chance of $ 2 trillion -- no matter how many times you can play. Your risk of ruin is basically 100%, your variance is off the charts, and you're risking money that has utility to you now for money that wouldn't have any utility if you won. Winning $2 trillion is basically the same thing as winning $500 billion.
https://www.facebook.com/pages/Your-so-fucked-up-not-even-bob-can-fix-you/275195652497240
Quote: sodawaterIf a game's expected value is $1.00, you can expect to make $1 each and every time you play. Doesn't matter if you play once or a million times.
I'm not sure this is true. If I play $0.25 9/6 JoB with perfect strategy my expected value is 99.54%. If I only play the game for one hand I will not win $1.24425.
--helpmespock
Quote: helpmespockI'm not sure this is true. If I play $0.25 9/6 JoB with perfect strategy my expected value is 99.54%. If I only play the game for one hand I will not win $1.24425.
--helpmespock
No, but $1.24 is what you can expect to "win." Or to put it more accurately, 1 cent is what you can expect to lose.
Expect is the key word.
Quote: sodawaterNo, but $1.24 is what you can expect to "win." Or to put it more accurately, 1 cent is what you can expect to lose.
Expect is the key word.
Well, not with the english definition of expect :)
But, yes, that is the expectation of the random variable. It's probably a poor choice of naming -- "expect" seems to imply some sort of median, but really, it's a mean that has no chance of actually happening.
Quote: AxiomOfChoiceWell, not with the english definition of expect :)
But, yes, that is the expectation of the random variable. It's probably a poor choice of naming -- "expect" seems to imply some sort of median, but really, it's a mean that has no chance of actually happening.
Sure, it's got zero chance of happening, but it's still what you can "expect."
Say you and I flip a coin for $10. You can't break even. You can either win or lose 10. But if you had to pick one number for what you "expect" to happen, you have to say $0. Nothing else fits. In that case it's the median, fine.
There's a scratch-off lottery ticket I like that costs $10. You can win anything from your $10 back to $1,000,000. You cannot win less than $10. The prize pool for that ticket is 75% of sales to winnings, and 25% of tickets win. Every time I buy that ticket, I expect to lose $2.50. What actually will happen is I will lose $10, or break even, or win something bigger. But there's only one number I can say I "expect" to lose, and that's $2.50. Every time I buy one of those tickets, I mentally deduct $2.50. Here it's the mean and not the median. And it still works.
So I would argue it holds true for both the statistical and common language definition of "expect."
Quote: AxiomOfChoiceWell, not with the english definition of expect :)
But, yes, that is the expectation of the random variable. It's probably a poor choice of naming -- "expect" seems to imply some sort of median, but really, it's a mean that has no chance of actually happening.
It has no chance chance of actually happening in the SHORT term. But given enough trials, it becomes almost a certainty in the LONG term.
Not true IMOQuote: anonimussKeep in mind the higher the edge the lower the variance.
Quote: AxelWolfNot true IMO
Definitely not true. Risk of ruin often decreases with a greater player edge, which is what I am thinking he meant. It's a more important thing to decrease the chances of anyway.
