Am I wrong in pointing out that when the house wins they win 1 chip and when the player wins they win 2 chips?
Obviously you didn’t read my post where I explained it clearly. The point is that you play until one or the other is reduced to 0. I have worked out that if the house starts with n chips and you with m, the house will wipe you out with a probability of n/(n+m) and you will wipe out the house with a probability of m/(n+m). I was a bit surprised that the answer was that simple, although the proof that that is the unique solution is a bit tedious.
On average you will win 50% of the bets, but once you are wiped out, you are finished. In the n = 2, m = 1 scenario you will lose the first bet and the game half the time.
I read it, I just disagree with your logic as applied to this problem. The problem as posed was “Even if the odds were 50%, the house would still win because in general people keep playing until they bust.”
I agree that for a given random person, if they play until the have $0 dollars, then relative to that person, the house will win. (Yeah, hard to argue with that.) But you can’t extrapolate from that and say, therefore, the house will always win in such a situation…
Consider this:
If there was 1 bet total: 50% win, 50% lose.
If there were 2 bets total: 50% win, 50% lose.
If there were n bets total: 50% win, 50% lose.
I don’t see how the house can possibly win by outlasting with deep pockets…
The house is finished if they’re wiped out too.
In your n=2 m=1 game, out of 3 tries, the house wins 2, the player wins 1, right?
When the house wins, they win 1 chip. When the player wins, he wins 2 chips.
How does the house gain an advantage?
This is exactly the reason that Martingale systems don’t work. Sure, you have a high probability of winning a small amount, but you lose big when you lose, and statistically offset all of your winnings.
I worked for a while at a video gaming technology company, so I have some information about the inner workings of video slot machines, though the devices the company manufactured were specifically created for Native American casinos, primarily in Oklahoma.
Those games all played Bingo, though the customers would be hard-pressed to identify it as such. When you pressed the “Play” button, the device contacted the game server, tucked away in a part of the casino that you would never know was there. The casino would generate a random Bingo card for the device on the floor and add the device to a game already in progress.
Some of the game machines had physical reels, but most of them had video reels; the company was phasing out the physical reels, because some of the players had figured out that if they pressed on the glass, it would bend inward and they could affect where the reels stopped. The machine would report no win, but the customer would call over someone and point to the reels and try to get paid.
Before the reels (mechanical or video) ever started spinning, the server would have already determined if the game machine had los or won; if it had won, the server would also determine how much that machine had won. The reels stopped in the appropriate positions for a win of the given magnitude.
Payout values were set by “pay tables” that were loaded into the game server when they were booted up. Each “win” type, from simply returning the amount of the bet, all the way up to the maximum jackpot the games were allowed to pay out, had a certain percentage of occurrence.
While the total of payout percentages was typically less than 100%, it need not be. We had several game machines in our conference room, each of “different” games, for demonstration purposes. Those games were connected to the same server, and the pay tables for those machines were set to pay out over 100% – it was simply impossible to lose on those machines. They had to be reset periodically to 100 “coins” (while the machines in the casinos had bill readers that accepted all current currencies from $1 to $100, the customer received “coins”, making it more difficult for the customer to know how much money they were spending with each play). This kept the visitors (prospective customers) from thinking that the games that company produces would pay out more than they took in, while still giving prospective customers a “winning” experience.
In terms of rate of return, casinos only need a small profit margin, since they provide hundreds or thousands of machines. If a machine has $100,000 played on it each day, and the casino has 1000 machines, a 98% return still nets the casino $2 million each day.
Obviously, returning less than 98% gives the casino a higher input of cash, but lower that too much, and you lose players. The machines help, however, by paying out periodically. If you lose $100 dollars in ten minutes, you will probably never play that machine again. The longer it takes for you to lose all of your money, the more likely you are to get more money to put into it.
Of course, there are different sets of machines for all comfort levels of spending, from $.01 all the way up to $100 (possibly even more!) per play. The casino gets a bigger payout per machine with higher stakes, but can make that up by providing more machines that play at lower stakes. And, of course, even the penny machines (which still only takes bills) pay more for a win, the mode “coins” you wager each play. On most games, you can only win the “super duper extra jackpot” if you wager the maximum amount per play.
Simply put, if a machine doesn’t pay out, ever, not only will no one play it, they may contact the gaming commission about that machine. Casinos do not want either of those situations, so it is in their interest to keep the games “honest”.
Where does that come from?
If n=2 and m=1, the house wipes the player out 2/(1+2) or 2/3rds of the time, and the player wipes out the house 1/3rd of the time.
However, since the house starts with n chips, 2 chips, when the player wipes out the house he wins 2 chips. The player starts with 1 chip, so the house can only win 1 chip from him.
The house has a 2/3rds chance of winning 1 chip and the player has a 1/3rd chance of winning 2 chips. Dead even odds, which is completely consistent with even odds betting.
You aren’t measuring odds, you’re measuring expected value.
ETA: To clarify, you’re not measuring the odds of the player losing all his money. The player is more likely to lose all his money than the house is.
Let’s put it this way. I have a million quarters. You have two quarters. We’re going to play a game where the winner is whoever ends up with all the quarters. They way you play is, you each flip one of your quarters. If they match, I get them both. If they don’t match, you get them both.
Do you expect to win this game?
Sure, you will lose the vast majority of the time, but you’re only losing 50 cents each time. When the house loses, it loses a million quarters. The overall EV should even out.
So, I suppose that’s your point. The house edge is 0. For any individual player it’s rather unlikely they will win, but, overall, it’s an even game and the house isn’t going to make any money on it.
Why would I expect to win a ridiculous game that you have engineered such that I’m virtually guaranteed to lose?
Real world gambling doesn’t involve gamblers who will bust after two losses. It doesn’t involve players who will attempt to play until the house busts. It doesn’t involve just one underfunded gambler vs. the house. Gambling is a pastime that involves hundreds or thousands of bets made over time. When you start talking that number of tries, the payout is going to closely match the odds. Even odds will mean even payout, the house isn’t going to take in more money than they pay out.
Of course you shouldn’t expect to win it. But this game is just illustrative of the point Hari Seldon is making, as I understand it.
I don’t think any of the above is actually relevant to the point under discussion!
People are talking past each other here.
Yes, I agree with all of those sentences. 
Hari Seldon is, as I understand it, pointing out that the odds of losing all your money is very high. This is apparently what he originally did not believe to be true. And it’s not at all incompatible with the idea that EV is zero overall–and Hari Seldon hasn’t said otherwise.
Pretty much like Elections then, even before machinery.
Plus Diebold now.
People buy lottery tickets which give them a very small chance of winning a very large prize all the time, even when the odds are stacked against them. I think you’d find a lot of people willing to risk 50c to win $500,000, at 1:1,000,000 odds.
jtur88 claimed that the house would “still win” even with even odds, because they have deep pockets and will outlast the players. Hari is agreeing with him.
If the EV is zero, the house isn’t winning. There is no further discussion needed, the house isn’t making money on the players if the EV is zero.
Yes, an individual player with a hilariously undersized bankroll is more likely to go bust as a result of a random walk. He also is more likely to win 5 or 10x what he walked in with, which the house cannot win off of him. The EV is still zero for him, even though his risk of bust is higher. The house is not going to “still win” unless their EV is greater than zero.
It’s true a player should have a bankroll at least two or three times the standard deviation, otherwise he can go bust and lose on an even odds game. A low bankroll is vulnerable in a fair game with high variance.
But even if an individual player busts and loses his bankroll, the even odds game will still not turn a profit for the house.