Wounded Ronin
Dec 14 2007, 10:55 PM
My math-fu is weak, so I need help on this simple statistical question.
Let's assume that I'm playing Roulette at a casino.
Rules:
http://wizardofodds.com/rouletteThe point is that if I consistiently place a bid on Red, I've got a 47.37% probability of winning a 1 to 1 payout.
Let's say that I have $5,000 dollars. My goal is to make $1,000 in winnings. I use the following strategy: I bet $100 dollars on red. If I win, I bet $100 dollars on red again. If I lose, I double and bet again.
So, if I lost my first $100 dollar bet, my next bet would be $200. If I lost my $200 bet, my next bet would be $400. If I lost the $400 dollar bet the next one would be $800, and so forth. However, as soon as I win, I revert back to my next bet being the $100 bet.
What, under these rules, are my odds of increasing the amount of money I have to $6000 versus the odds of my losing all $5000? Suppose that my goal is more modest and I only want to have a total of $5500 in my pocket; what would be the odds of success versus the odds of losing eveything in that case?
Thanks very much for anyone who can answer this question.
Critias
Dec 15 2007, 06:54 AM
I'm no math guru, but I don't have to be. The simple fact is they don't build those big, fancy, expensive, buildings because the house loses money every day.
I would never, ever, walk into a casino with a plan to leave there with more cash than I walked in with. You go to be entertained, not to make any money.
Wounded Ronin
Dec 15 2007, 05:26 PM
QUOTE (Critias) |
I'm no math guru, but I don't have to be. The simple fact is they don't build those big, fancy, expensive, buildings because the house loses money every day.
I would never, ever, walk into a casino with a plan to leave there with more cash than I walked in with. You go to be entertained, not to make any money. |
The basic flaw with the "martingale" system, which is the one I'm describing, can be summarized as follows: you're usually going to win, but eventually you will lose many times in a row and lose really big. So if someone uses this strategy consistiently in casinos over the course of his life he is likely to end up losing more money than he wins.
The reason that the above strategy is different is as follows: the amount of money the player is trying to win is small compared to his bankroll. The bankroll, versus a small amount of desired winnings, allows for a large number of consecutive failures to be dealt with. Since the player is only trying to win a little bit of money and not continuing indefinitely the odds of a long string of consecutive losses are low.
So while a player using the strategy above has a probability of losing big, it seems more likely that he'd succeed in making a small amount of money if he stops playing as soon as he is reached his goal, and only gambles once or twice in his life instead of doing it regularly.
hyzmarca
Dec 15 2007, 07:14 PM
Your strategy fails after 5 permutations. 100+200+400+800+1600 = 3100. 1600*2 = 3200 5000-3100 = 1900. At this point you can no longer double your bet.
The probability of it failing can be represented as the number of (losing conditions/the number of conditions)^5, in the case of a 00 roulette wheel, it is the fraction (20/38)^5 = (20^5)/(38^5) = 3200000/79235168, which can be reduced to 100000/2476099.
Your probability of winning any set is thus (2476099-100000)/2476099 or 2376099/2476099, which is 95.96%
You have to win 10 sets so we must raise that fraction to the tenth power, which produces numbers too large for my calculator. Or we can raise the percent to the 10th power, which gives us 66.2%
More simply, using the probability you gave, we can make the formula (1-(1-0.4737)^5)^10 which gives us 66.2%
This is good, but it isn't very good. You're more likely to win than not, but the margin is far too small to call it a sure thing. The more sets you need to play in order to reach your intended goal, the lower your chances of winning are.
The larger your bankroll and the smaller your intended returns relative to initial bet, the better your chances are.
However, it is extremely important to remember that roulette probabilities are discrete per event, and are thus unaffected by past events. If you lose once then your probability of losing four more times is (20/38)^4. If you lose twice then the probability of you losing three more times is (20/38)^3 and so on. If you lose four times in a row, then your next spin has a 20/38 probability of losing.
Wounded Ronin
Dec 16 2007, 07:23 AM
Thanks very much for the good analysis. I cannot believe that all the tourists who go and gamble in casinos do not sit down and do the math first.
Crusher Bob
Dec 16 2007, 08:26 AM
Here's a slightly clearer version, we hope.
We wish to win $500, and have $7,500
so our betting patter would be
500
1000
2000
4000
Our chance of losing 4 times in a row if (.527)^4 = ~7.7%
Now we run a risk analysis on this:
~7.7% of the time, we lose $7,500
and the other ~92.3% of the time we make $500.
So our amortized loss is 7,500 * ~7.7% ~= 578.50 (used actual percentage which is 0.077133397)
and our amortized gain is 500 * ~92.3% ~= 461.43
which means that if we played an infinite number of 4 set games, we'd lose around $117.07 every set of 4 games.
Cthulhudreams
Dec 16 2007, 01:21 PM
I don't think the majority of people who go to Las Vegas and gamble plan on winning money, the show manship and artistry is at some point the draw card.
What I love is people who play slot machines - you don't even need to caculate the rate of return, in australia, *by law* it is actually printed on a specified label next to the money insert point.
Usually it's surprisingly high (slot machines at my local club return 90-something%), but they are counting on people just feeding their 90 something % back in, which they do.
Kagetenshi
Dec 16 2007, 06:29 PM
QUOTE (Cthulhudreams @ Dec 16 2007, 08:21 AM) |
Usually it's surprisingly high (slot machines at my local club return 90-something%), but they are counting on people just feeding their 90 something % back in, which they do. |
No, actually, they're just counting on someone feeding something back in. Naive blackjack (without counting cards, or with too many decks to count) gives the house an incredibly small advantage over a good player, but casinos don't play a few dozen games against a small handful of players a day—they play millions of games against thousands of players. It's the Law of Large Numbers in action.
Anyway, while it has some variance at the small-possible-repetitions end of the spectrum, IIRC the proper way to apply the bet-doubling strategy is to make your initial bet what you want to win. From there, you just make your bankroll large enough to ensure that you can wait until your number will most likely come up. There are, however, other things that make it impractical (like having a sufficiently large bankroll, or the possibility of a bet limit which would start requiring you to win multiple times).
It should be noted that there is at least one casino game at which you can, if you're good enough, expect to reliably make money and not get kicked out. That game is poker, and the reason is because no matter what happens, the house gets its cut—you're playing against the other players, and only against the house to the degree that you need to win enough to outweigh the loss of the rake. The same would hold true of any game where the payout comes from the other players rather than from the casino and where skill can gain at least enough advantage to outweigh the rake. On the other hand, you then have to play against the unknown skill of the other players instead of against a known set of odds for the house…
~J
DTFarstar
Dec 16 2007, 07:51 PM
It should be noted that even if they can do the math, they don't care because most of them think there is some amount of skill in every game, even slots. They think their "system" will work. Note the above large numbers notation, and the fact that there is free alcohol and if you start doing well you might attract a crowd to egg you on in some games and you end up with one cold blooded sober bastard that can actually stay on his game and being cold blooded and especially being sober are generally not things people go to Vegas for.
Human arrogance leads to human folly.
Chris
DTFarstar
Dec 16 2007, 07:53 PM
Craps is the same way as poker, bet well and you can get a decent return if you know what you are doing and stick to it and the house doesn't care because they are raking in money from everyone at the table that bet differently than you.
Chris
Kagetenshi
Dec 16 2007, 08:15 PM
Oh, yeah, Craps. Craps is not actually the same way, in that you do actually play against the house with fixed odds that aren't in your favour, but the house advantage on the basic pass/don't pass is quite slim (>2%, IIRC).
~J
PBTHHHHT
Dec 18 2007, 09:48 PM
QUOTE (Wounded Ronin) |
Thanks very much for the good analysis. I cannot believe that all the tourists who go and gamble in casinos do not sit down and do the math first. |
You're assuming most people can calculate the math. We're already an odd bunch here on the forums, one dedicated to an rpg... it's a different demographics compared to the rest of the population.
Cthulhudreams
Dec 18 2007, 11:25 PM
QUOTE (Kagetenshi @ Dec 16 2007, 01:29 PM) |
QUOTE (Cthulhudreams @ Dec 16 2007, 08:21 AM) | Usually it's surprisingly high (slot machines at my local club return 90-something%), but they are counting on people just feeding their 90 something % back in, which they do. |
No, actually, they're just counting on someone feeding something back in. Naive blackjack (without counting cards, or with too many decks to count) gives the house an incredibly small advantage over a good player, but casinos don't play a few dozen games against a small handful of players a day—they play millions of games against thousands of players. It's the Law of Large Numbers in action.
|
Yeah, I agree, many places the thinness of the house margin is actually an incentive to go there. Poker players definitely prefer a milder rake all things being equal.
FrankTrollman
Dec 19 2007, 01:27 AM
The original point remains. Given infinite bankrolls on each side and practically limitless numbers of plays, then there will always be a point at which people on both sides of the table will be ahead. As it happens, the House has an essentially limitless bankroll, and generally speaking they will let you plop down money as often as you want.
But you don't have a limitless bankroll. Sad, but true. As your bankroll becomes larger you chances of walking away with $1,000 dollars increases and the chance of you losing everything decreases. But the amount of money you'd lose if you lost everything keeps going up. And to that extent the House actually doesn't even care so long as you have a specific finite bank roll.
-Frank
nezumi
Dec 19 2007, 01:50 PM
In other words... If we had an infinite source of money... We could like... Get infinite money!! We'd be rich!
Wounded Ronin
Dec 19 2007, 10:51 PM
More to the point, if we had a large amount of money we could safely win a smaller amount, but there'd always be at least a tiny chance of losing huge.
hyzmarca
Dec 19 2007, 10:59 PM
Which is why the house always has betting limits, so that Bill Gates can't come in, put 50 billion on black, and bankrupt the casino.
They also screw the Martingale, since even if you have a sufficient bankroll you will eventually hit the betting limit.
The house counts on winning tiny amounts of money from a thousands of people a day, which adds up. They don't care about winning big. They don't really want to win big because a bankrupt player can't continue playing. And they don't really care if a player wins big, because it is rare enough that they'll make up for it and it encourages more people to gamble in the hopes that they too will win big.
Kagetenshi
Dec 19 2007, 11:10 PM
So here's a question many of you will recognize: there is a game you can play where you pay to enter, and then you start with a score of zero and start flipping a coin. Every time the coin comes up heads, you add one to your score and then flip again. When the coin comes up tails, the game is over and you win one dollar, doubled for every time you got heads (total $2^s, where s is your score).
How much should you be willing to pay to enter this lottery?
~J
Moon-Hawk
Dec 20 2007, 06:34 PM
QUOTE (Kagetenshi @ Dec 19 2007, 06:10 PM) |
So here's a question many of you will recognize: there is a game you can play where you pay to enter, and then you start with a score of zero and start flipping a coin. Every time the coin comes up heads, you add one to your score and then flip again. When the coin comes up tails, the game is over and you win one dollar, doubled for every time you got heads (total $2^s, where s is your score).
How much should you be willing to pay to enter this lottery?
~J |
Hmmm, my math doesn't seem to be working right, 'cause I keep figuring that you should pay anything to get into this game, but that seems wrong.
I need to recheck this.
edit: Is that the trick of it? The expected value is infinite?
Kagetenshi
Dec 20 2007, 06:41 PM
We have a winner
No, you're completely correct. Since the expected payoff of the game is infinite, the rational action based on expected payoff is to be willing to pay an arbitrarily large amount of money to enter.
Google for the St. Petersburg Lottery if you want to know more.
~J
Moon-Hawk
Dec 20 2007, 06:46 PM
Cool trick! Thanks Kagetenshi, I'm reading the wikipedia article about it now.
hyzmarca
Dec 20 2007, 06:58 PM
That doesn't take into account factor time available and time/flip, which limit the number of flips that can be made. Assuming that ending the game early (including by falling unconscious or dying) results in a forfeit and that you can't continue past the Heat Death of the Universe, this creates a both soft caps and a hard cap on potential winnings.
Wounded Ronin
Dec 20 2007, 07:03 PM
QUOTE (Kagetenshi) |
We have a winner
No, you're completely correct. Since the expected payoff of the game is infinite, the rational action based on expected payoff is to be willing to pay an arbitrarily large amount of money to enter.
Google for the St. Petersburg Lottery if you want to know more.
~J |
If the amount you pay makes no difference why is it not totally basic that you pay the least possible amount to get the goods, i.e. find the least-valued banknote in the entire world and pay that?
Kagetenshi
Dec 20 2007, 07:08 PM
QUOTE (Wounded Ronin @ Dec 20 2007, 02:03 PM) |
If the amount you pay makes no difference why is it not totally basic that you pay the least possible amount to get the goods, i.e. find the least-valued banknote in the entire world and pay that? |
The idea is to, if you come across someone offering entrance to this lottery at $
n, determine whether you should enter the lottery or not. You don't get to choose the price the person is offering
~J
Wounded Ronin
Dec 20 2007, 07:12 PM
QUOTE (Kagetenshi) |
QUOTE (Wounded Ronin @ Dec 20 2007, 02:03 PM) | If the amount you pay makes no difference why is it not totally basic that you pay the least possible amount to get the goods, i.e. find the least-valued banknote in the entire world and pay that? |
The idea is to, if you come across someone offering entrance to this lottery at $ n, determine whether you should enter the lottery or not. You don't get to choose the price the person is offering ~J |
Oh, I see. So even if he's asking for billyuns and billyuns of dollars you should pay that if you have the ability to do so.
hyzmarca
Dec 20 2007, 10:47 PM
It is really a case where the mathematically expected winnings do not match the most probable winnings.
In all likelihood, you'll win $2 or $0 $1 (75% chance). My calculator tells my that the Log2 of 1,000,000,000 is 29.8973528539868. The first value over 1,000,000,000 that you can win is 2^30 = 1,073,741,824 the probability of you reaching a score of 30 is 1/2^30 or 1/1073741824.
In other words, your probability of winning an amount of money or more (given that it is an integer power of 2) is equal 1/(that amount). Someone who would pay billions per play for this is a moron. This is a point where the math is wrong and common sense is right, which isn't often, and thus is often called a paradox.
It isn't a paradox so much as the fact that the mathematical formula used to determine expected value can't handle divergence, and thus a different formula is needed for determining the expected value of potentially divergent system.
However, given infinite possible itterations and the ability to quit when one is ahead, then you will always come out ahead, eventually. The problem is that you can't play infinite itterations unless you also have an infinite bankroll.
Thus, if you have infinite money then you'll have infinite money.
Kagetenshi
Dec 20 2007, 11:39 PM
You'll never win $0 according to the question as posed, as the payoff if you immediately come up tails is $1
~J
Moon-Hawk
Dec 21 2007, 05:07 PM
QUOTE (hyzmarca) |
In all likelihood, you'll win $2 or $0 (75% chance). My calculator tells my that the Log2 of 1,000,000,000 is 29.8973528539868. The first value over 1,000,000,000 that you can win is 2^30 = 1,073,741,824 the probability of you reaching a score of 30 is 1/2^30 or 1/1073741824.
In other words, your probability of winning an amount of money or more (given that it is an integer power of 2) is equal 1/(that amount). Someone who would pay billions per play for this is a moron. This is a point where the math is wrong and common sense is right, which isn't often, and thus is often called a paradox. |
I marvel at your ability to make that problem so complicated.
That's.......a slightly different approach than I took.
This is a "lo-fi" version of our main content. To view the full version with more information, formatting and images, please
click here.