h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

http://www.bbc.co.uk/h2g2/guide/A538616

I hope you find this helpful...

Hi, an interesting article but I'm unconvinced I'm afraid.

My understanding (having lost horribly once at blackjack - although I wasn't using your technique) was that if you matched the dealer, you lost, this thereby disadvantages you. Your rules say to never twist when you are at seventeen or more, this is the same as the dealer, hence you must beat the dealer and so you lose more often than them.
As in all casino games, as far as I can see, the casino wins more more money than you do. You don't see too many poor casino owners - this is because they bias the odds against you.

Also, what do you do if you get the opportunity to split?

I'm ready to be convinced so if my logic is wrong please tell me.

Thanks for your feedback orcus. I understand your point on losing if you and the dealer both get 17. Do you think i should add this, or should i keep the article simple? I have to tell you that i have never lost yet with this system, playing to $400, because even with your point taken into account, the chances of losing many times in a row are very slim. But i will change the article on that point, thanks. Also, did you think i used the first-person too much thereby straying from a 'just the facts' format?

Oops. Forgot to say that when i played the Venician in Vegas the dealer tied with me at 17. Also, if you decide to split thats up to you, but what a headache figuring out the math of how much you win and lose and how much to bet the next time to make up for your losses. I haven't ever strayed from the basics, so i dont double down.

I'd agree with your professor that this isn't a "secret way" to make money in a casino -- I've seen this method in pretty much every "blackjack" tip book I've ever seen.

Mikey

If you lost 4 times in a row then there's no better chance to win in the next game (ask any math prof, he'll instantly be talking about 'statistically independent processes' and the like). If you throw a die and get a '1', then the chances for another '1' in the next turn are still 1/6, and nothing else.

Oh, and the 1st person style shouldn't be difficult to eliminate!

No probs May, yeah I was going to also comment on the 1st person thing but as already said, that can easily be removed.
Just because I'm unconvinced, doesn't mean it shouldn't get into the guide. I actually quite liked the advice - walk away if you lose four times on the trot and walk way when you win $400 as this is certainly sensible advice when it comes to gambling - never get greedy. If I ever play Blackjack again (doubtful - twas a harrowing experience, I'll bear this system in mind )

Using the sequence X, 2X, 3X, 6X, with 50/50 odds

Half the time, you win first time (profit $10)
One quarter of the time, you win second time (profit $10)
One eighth of the time, you win third time (break even)
One sixteenth of the time, you win fourth time (break even)
one sixteenth of the time, you lose fourth time (lose $120 and go home)

For a single run of play, the average outcome is 10/2 + 10/4 +0/8 +0/16 - 120/16 = 0, and so in the long run, you break even.

Running a simulation 100000 times shows that on average, a 'game' (either winning $400, or leaving on the first loss) will involve 15 runs of play, and 28 deals of cards.


Blackjack *is* the only casino game where the odds are slightly in your favour (by something like 1.5%, I think), but that only applies if you're capable of remembering which cards have already been played. Very few people are good card-counters, and once the casinos have worked out who they are (which they will, if they keep winning), they're usually not welcomed back.

yeah - you can get 1.5% if you card-count, which is against the house rules in all cinemas and will get you chucked out...

Nope - this articles suffers from being factually inaccurate: it's a fairly old concept - keep playing double or quits and eventually you'll win and cancel out your losses - the problem is that eventually you'll either hit the house limit for a single bet - or, more likely, you'll run out of money. If you only have a finite amount of money to start with, you can't make a profit, on average, with this method.

However, it is *marginally* better than just betting $10 each time. The reason is that each round isn't independant - there are only four aces in the deck {well, casinos tend to use double or even quad decks, but the principle applies}. Picture cards and aces put the edge marginally against the house, while other cards put the edge marginally in favour of the house {which is how card-counters win}.

Anywho, if you've lost 4 times, then you're slightly less likely to lose again, because there's likely to be slightly more picture cards about to come up. But I doubt that this'll be enough to beat the house edge of winning on draws - it'd be a very small effect.

Oh - magic practitioners will no doubt see this as an entirely reasonable case of lesser magic - and they might even be right - but they don't write the textbooks, so we get to laugh at them...

i want to thank you who took the time to give me your helpful comments on this article. i think that based on your comments, it seems it is necesary to clarify the discrete math justification to this method. i will add the following explanation to the guide entry if you think all readers would also be interested.
to remind you after any win you dont double your bet, you always bet $10 after winning. the odds of losing 4 times in a row are 1/16 for a $10 table, 1/32 for a $5 table if you allow yourself to lose 5 times in a row maximum and then quit. losing less than 4 or 5 times in a row is not important since by this system, the next win will make up for all your losses. therefore, 15/16 or 31/32 (for 5 losses in a row) is the probability you will not lose 4 or 5 times in a row. for both cases you are ahead since 150-120 leaves you ahead and 310-120 leaves you ahead. also, im not surprised to hear that this system appears in many texts, since what works usually does not stay secret (i'll gladly change the article title to 'not so secret method' if you want).
i also took off the comment about the stats prof. i value your comments and enjoy reading your intelligent input (any other changes?). thanks again. mav

The original ($10, $20, $30, $60) isn't doubling the bet.
If you always *double* your bet on losing, the sequence for a run of play (when playing until you win, or lose the final bet) is :
$10, $20, $40, $80, and the results are

Half the time, you win first time (profit $10)
One quarter of the time, you win second time (profit $10)
One eighth of the time, you win third time (profit $10)
One sixteenth of the time, you win fourth time (profit $10)
one sixteenth of the time, you lose fourth time (lose $150)

For a single run, the average is 10/2 + 10/4 + 10/8 + 10/16 - 150/16, and the average over a number of runs is still break-even.

Starting at $5, doubling each time with a maximum run length of 5 gives a sequence of $5, $10, $20, $40, $80
with an average of 5/2 + 5/4 + 5/8 + 5/16 + 5/32 - 155/16, or break even again.

Whatever system you use, I don't think there is a way to beat straight 50/50 odds. If there *is* a way to get slightly better odds, you can win on average, but 50/50 is effectively a brick wall.

But the odds aren't 50-50. As I said earlier (certainly when I played the game anyway) you have to BEAT the dealer, if you draw with them you lose. Therefore the odds are worse than 50-50.
Anyway, i've had my say. I'll leave others to more constructive comments.

Let's put it like this: when you bet $X, you either win $X, or lose $X - and these two options are equally likely. Therefore your expected winnings from any single bet is zero. Therefore your expected winnings from any sequence of bets is zero.

It's not entirely clear how this can be made any more clear. If you are making money with this system you are doing so by your natural reserves of luck - not by the system.

I must back the mathematicians against the optimists. Sorry, gamblers.

I don't know about Blackjack specifically, but if your estimate of 50-50 is even roughly right:
You state that the chances of losing 5 times in a row are small (they're 1 in 32 if it's really evens).
To get to $400 (your cutoff) without means this 1 in 32 chance not happening for 30 games in a row. The odds of this happening are only 39%.

Basically, as Potholer demonstrated, you win small with this method until you lose big.

There are only 2 ways I believe in to win in a casino:

One is at roulette using a concealed computer, as documented in the excellent "Newtonian Casino" (by Thomas A. Bass), which needs technical know-how, practice, and a lot of nerve. It's also technically illegal.

The other is to be the House. That's the easy one.

GVI

making $100 turn into $400, 39% of the time doesnt sound too bad to me(you didnt mention that if you set your sights lower than $400 your chances increase)...i also wanted to suggest, to anyone interested, to download one of the many online casinos, and try the system for yourself without using real money. tell me how it you do, then try it in a real casino.

If that the odds of winning or losing any one hand are 50/50, and given that they are independent of any previous wins or losses (card-counters excluded), whatever system you use - quit at a profit of $400, $200,(or anything else), quit on the first (or second, or third...) full losing streak, on average you will break even in the long run.

However much you've won or lost at the table already, at the start of any individual 50/50 game where you bet $X, your odds of ending up with $X less are exactly the same as ending up with $X more, so on average, you must break even. However many games you play in succession, and however much you change the X for each game the overall average is still break-even.

The system given does seem to have potential from an intuitive point of view, but intuition often disagrees with mathematics. I think the reason the system is appealing is that in terms of runs of play (carrying increasing bets until winning or hitting your loss limit)you do win a run more often than you lose, but you win small and lose big.

If anyone is going to try this in a casino simulation, they should probably note down how many individual deals they win and lose, to check how many deals they win and lose - play-casinos may not always be unbiased.

Editorial Note: This conversation has been moved from 'The Writing Workshop' to 'The Flea Market'.

The author of the entry seems to have 'left the building'...

The problem with this method, put simply, is that when you draw with the dealer you do not win.

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The odds of you winning are less than 50%.

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In an average run of games, you will make a loss.



Hence this article is not a matter of fact, so there is no point in trying to get it into the Edited Guide.

Friendly tip

This thread was last posted to, four years ago, when I was trying to make the same point

Just adding my