You’re stuck in the “Purgatory Casino” until you make a profit. You start with $1Quadrillion and you can only flat bet $2 a spin on “red” or “black” It’s a fair 37 slot wheel with a 0, if it lands in the 0 you get $1 back, so each spin there’s a ~48.65% of gaining $2, ~48.65% of losing $2 and ~2.7% of losing $1. What are the odds of never ever making a profit?
I’ve put it in GQ as it’s mainly a maths question, feel free to move it to The Game Room if you think it must be there.
What happens if you lose all the 10 quadrillion? Can you borrow more money? If so, I think the answer is eventually you will make a profit, guaranteed.
If not, I think the answer is 1 minus the probability of losing al the money. I’ll wait for someone smarter than me to work this out.
No, you will never reach a point where you have 0% chance of making it back up, but as you’re digging yourself deeper each time, the probabitlity will get less and less so that in calculations of the genearal probability it will become negligible.
For simplicity, let’s approximate your 18-18-1 proposition with a simple even money bet with win chance 36/73.
The chance that you will never turn a profit is 1/37.
More generally, the chance you will ever get B bets ahead on even-money bets won with probability p is
(p/(1-p))[sup]B[/sup]
(Once you convince yourself that the probability of getting 2 bets ahead is the square of the 1-bet-ahead probability, and so on; the form of the forumla is known and readily solved.)
So in that case, your problem is equivalent to starting with 0, being able to accumulate debt without limit, and whenever you have a positive amount of , you get out.
If I have understood correctly, then I propose the following answer, with almost no math.
If we go back to the 10 Quadrillion version, the only way to not get out is to lose everything. So the probability of getting out is 1 minus the probability of losing everything. Now consider what happens to the probability of losing everything as we increase the starting amount of . It gets smaller and smaller. In the limit as goes to infinity, the probability is zero. Therefore, the probability of getting out is 1 minus zero which equals 1. You will eventually get out.
A more interesting questions is, what is the average waiting time for people in purgatory casino?
This is quickly refuted with an argument frequently mentioned in these gambling GQ threads. In this case(*) all we need note is that net average vigorish is positive. With our stopping rule (stop after net win of $1 or loss of $500 trillion), vigorish would average NEGATIVE $1 per session, impossible.
Indeed the constancy of constant vigorish lets us calculate average life. You lose (500 trillion * 36/37 - 1/37 - **), or about $486 trillion on average; since the vigorish (1/73) multiplied by average number of tosses (T) will also equal average vigorish. we see that you get 35.5 quadrillion tosses on average. To spur further interest, I’ll leave it as an exercise how this average is distributed across the two cases:
T = 35.5 quadrillion = .973 * T[sub]$1 victory[/sub] + .027 * T[sub]$500 trillion defeat[/sub]
** – Oops – it looks like I reverted to a non-European variation in the numbers. Oh well, you get thie idea. 
[quote=“Mr_Shine, post:1, topic:736886”]
You’re stuck in the “Purgatory Casino” until you make a profit. You start with $1Quadrillion and you can only flat bet $2 a spin on “red” or “black” It’s a fair 37 slot wheel with a 0, if it lands in the 0 you get $1 back, so each spin there’s a ~48.65% of gaining $2, ~48.65% of losing $2 and ~2.7% of losing $1. What are the odds of never ever making a profit?
I’ve put it in GQ as it’s mainly a maths question, feel free to move it to The Game Room if you think it must be there.[/QUO
I worked in the casino industry for 10yrs. and the wheels were set up with 36 black/red numbers which would be 50/50% odds. However, the wheels also had 2 green numbers-0/00. This gives the House a slight advantage which, over time, will prevail. Betting red/black-odd/even numbers. These bets are placed on the “outside field”, which means you’re not playing “in the field”, ie., not on any of the 36 numbers.
With that being said, other than straight poker, the thee best games for a customer to play are roulette(r./b.-odd/even), Pai Gow and Baccarat, which gives a slight edge as compared to other “table games”.
Also, a huge Caveat emptor for what the casinos call “Carny games” like Spanish Black Jack, 3 Card Poker, Caribbean Stud, Let it Ride, etc.:smack:. These are all variations of regular poker. Some of these are peoples’ favorite games. Younger people usually don’t like Pai Gow because it moves slowly. Young people, especially guys, like quick popping action like a well dealt Blackjack game.
Also, another Caveat emptor for what the House calls a “bonus bet”. Usually, a small circle bet outside of the actual main bet. The House makes a killing on these bets.
All of the numbers for all games have been crunched ad nauseam. They know exactly what they’re doing and why they’re doing it.
No gambling game with a house PC has anything but a dead loss outcome for the long-term player.
That isn’t what he is asking though. He is asking what are the odds that you will NEVER be ahead (not even once) if you are forced to play the red versus black bet in Roulette with an incredible amount of money ($1 quadrillion dollars). I know one way to calculate it but it is brute force statistics and not something I am willing to do on a Saturday night.
The question is basically the same as asking about a coin flip but with a small bias against you no matter which way you bet. 48.65% of people will win on the first try and be allowed out of Purgatory. Of those that lose the first bet, 48.65% of the remaining people will win on the next play to get back to even money. Out of the people that fought their way back to even money, another 48.65% will win on their next try like they just started playing. That already accounts for a significant majority of people being ahead on just their first three plays. Repeat until you find out how many people burn through $1 quadrillion dollars without ever making it to a positive balance. I don’t know an elegant formula to calculate the exact result but I believe the final answer for players that will never be ahead at least once is incredibly small.
If I was forced to calculate this in real life to get out of Purgatory myself, I would write a computer program to figure out the odds of every possible scenario but it would probably take a very long time to return an exact result.
BTW, there is a big difference between infinite money and $1 quadrillion dollars. The answer for infinite betting money is all players will eventually hit an incredibly improbable streak that gives them positive money. That is not a guess either. That is just the way infinities and statistics work. $1 quadrillion is a whole lot of money but it isn’t truly infinite and there will be a very few players that blow the whole wad of cash without ever being in the red.
There’s a 48.65% chance you’ll be ahead after one spin …
There’s is a non-zero probability that one would lose 500 trillion minus 1 times in a row, and then win 500 trillion times in a row …
Somehow we have to add up these and all the probabilities in between.
I don’t think that’s actually true, and it’s certainly not a given. Yes, if you play long enough, then eventually you’ll get a streak of any given length… but it’s not a single given length of streak that you need. On average, the longer you play, the greater the streak you need to recover. And the length of the needed streak increases linearly with number of plays, while the odds against getting any given streak are exponential in the length of the streak. So the streak you need grows much faster than the streak you have.
Or, to put it formally: For any given bankroll b, there is a probability P(b) that you will profit before busting. And the limit of P(b) as b -> infinity is a number strictly less than 1.
You are probably much better at math than I ever hope to be but I think I have a good handle on statistics and infinities. I stand by my statement. If the money is truly infinite, the player will ALWAYS win 100% of the time.
You are talking about limited bankrolls even if they are extremely large. I agree that an extremely small percentage of people will lose this bet given a $1 quadrillion dollar bankroll but it will not happen with a truly infinite one.
I think your error is imposing a limit on infinity when it doesn’t have one by definition.
Quite the contrary, limits are the only way we can meaningfully discuss infinity.
I don’t have any reason to disbelieve you. Both statistics and infinite concepts can be quite counter-intuitive. Can you explain in laymen’s terms why an infinite sequence of bets would not result in a least one of them being positive? I am not seeing it.
In short, even a biased infinite string will contain a quadrillion reds or blacks in a row just by chance alone plus every other possible combination. What is preventing some players from ever going positive?
No Chronos is correct. There is a non-zero probability that you will never be ahead at any point. I don’t really have the time to do the given problem, but a similar one is straight-forward. Suppose the 0 loses as well, and just to make things simple make the bet $1 each time.
Then let Q(n) = the probability you’re never ahead when you’re currently n dollars in the hole. There is a p = 18/37 chance of winning and 1-p = 19/37 chance of losing the next bet so
Q(n) = p*Q(n-1) + (1-p)*Q(n+1)
This is a difference equation. It’s general solution is Q(n) = Ax[sup]n[/sup] which can be verified by substituting that into the equation giving
Ax[sup]n[/sup] = p*Ax[sup]n-1[/sup] + (1-p)*Ax[sup]n+1[/sup]
or after dividing through by Ax[sup]n-1[/sup]
x = p + (1-p)*x[sup]2[/sup]
The two solutions for x are 1 and p/(1-p). In this case 1 and 18/19 = 0.947368 = q. The whole solution is
Q(n) = A + Bq[sup]n[/sup]
For large n, Q(n) -> 0 so A must be 0. For n = -1, you’ve just won so Q(-1) = 1 and 1 = Bq[sup]-1[/sup] and B = q. Therefore
Q(n) = q[sup]n+1[/sup]
and Q(0) = q = 18/19. So there is a 1/19 chance of never being ahead of your staring point at 0.
For the original problem the game is not quite so bad as a green zero only loses half your stake rather than all.
Thanks to septimus, that answer looks good.
You will get a quadrillion redds in a row but only once every 2.05^1000000000000000
spins wich is such a huge number Google won’t even approximate it for me. By tat time you’ll be so far in the hole, you would need an even more staggering run of luck, A quadrillion spins would not be enough at that point.
No can’t be.
On the first throw there’s 48% chance. (W )
0% additional chance from the second. (LX can’t result in profit… X dont care)
12% additional chance from the third… (LWW)
0% from the forth (as LWLX can’t turn a profit, and the other cases were already accounted for).
3% from the fourth.
So there you go… the probabilities give you a chance short term wins (therefore not much money, relative to a constant bet you make) but guarantees a loss (of the sum of all bets) in the long run.
I don’t believe you guys. 
I wonder how many have me set to Ignore.
There’s the correct answer, right in #4, with a brief English clause guiding toward rigorous proof. OP accepts it; OldGuy knows the answer, though he failed to acknowledge being Ninja’ed and fails to complete the proof as my observatrion did.
Or, just read post #4.
Yes, it’s 36/37. See post #4.
I’d tell you what the error in your reasoning is … but am unable to discern your reasoning. You are aware that OP, in Purgatory, quits as soon as he’s ahead, right?
I certainly don’t have you set to ignore. This a good question right up there with the Monty Hall and the plane on a treadmill problem in that it is very unintuitive and will generate a lot of different answers based on the knowledge of the responder. I believe your elegant answer in an academic but not intuitive way. I felt the same way about the Monty Hall problem until I wrote a simple computer program to do it a million times and it worked just as they said and I understood it quite well after that. I might have to go through another simulation to understand this one. I have never been good at abstract math but I am good at programming computers to demonstrate the results of the underlying formulas.