Oh, I see, I thought you meant for Blackjack in general, not just for those few golden moments when the count was in your favor.
If it’s extreme examples you want, why not let the deck be a huge number of eights? Dealer is guaranteed to bust, and you can win as much as you want with repeated splitting! 
The Kelly criterion may not be as well known as it should be, though I think professional poker-players understand it at least at an intuitive level. It’s relevant not only to gambling, but banking, investment, insurance; is it applied there? (BTW, doesn’t Kelly Criterion post on the SDMB? 
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True (in a sense), but that doesn’t imply all systems are equally unefficacious. Since there are gambling theoreists in this thread and OP has been answered let me pose a small puzzle. (The answer is not as well known as it should be; prior discussion ended with L. N. and I calling each other morons!)
Puzzle: You have $1000, but want $2000 and have access to an American-style roulette game. (No, you don’t get to be the House. :D) Yes, your doubling chance is less than 50% whatever you do, but what should you do to maximize that chance?
That’s easy, if the requirement to be met is simply to double the money you have (and having $1999 counts just as much as a failure for the purpose of solving this puzzle as having no money at all would): Your strategy then is to put everything you have on one even money bet (like odd/even, red/black, or 1-18/19-36 - it doesn’t matter which bet you choose), then hope for the best. Your chances of soubling your money are 18/38 = 47.4 % in American Roulette (with double zeroes) and 18/37 = 48.6 % in European Roulette (with a single zero). There’s no way of increasing the chances beyond this value, since the more often you play, the more often the house edge will come into play to your disadvantage.
I just came back to check on this thread and i noticed that you have a typo.
The fraction should be 368467/1286390.
You fell right into his trap. You should have searched for the old thread.
http://boards.straightdope.com/sdmb/showthread.php?p=12049515
The solution presented there is:
That is not the correct solution to the puzzle, at least not in the format presented here. In that case, you have a 1/38 chance of doubling your money, and consequently a 37/38 chance of not doing so. It is true that the expected value of that bet is higher, since you suffer the house edge on only a $2 bet rather than a $70 bet on even money. However, the puzzle did not ask for minimizing your expected losses or maximizing your expected value in the game. What was asked for was a way to maximize the chances of walking out of the casino with twice the money I came in with. If this is the single goal to be met (and it was in the formulation of the puzzle presented here), then having no money at all is just as much a failure to meet the requirement as having $68 (losing your $2 bet on a single number), or having $69.89 (expected value of the game with a 2/38 house advantage and a bet of $2).
You only read the beginning of the discussion.
Posts 89, 92, and 94 hammer out all the details.
This is a misconception as well. Let me propose a game to illustrate this.
If I bet $1 on a single number in roulette. There is a 1/38 chance I can walk away with $36 which means I can expect to walk away with $36/38. This gives a house edge of 1-36/38 = 5.26% (approx.).
Say I have $20 and each spin I bet $1 on a single number until I win once or go broke.
How much money should I expect to walk away from the table with?
What is the house edge?
The house edge stays constant, of course, at 18/38 in American Roulette. You can’t change that, it’s hard-wired into the rules of the game. Your expected loss in the game does not stay constant, however, since the expected loss is the house edge multiplied by your stakes. Each time you play, you lose an expected 5.26 cents on each dollar you play. Remember I didn’t say that the more often you play, the higher the house edge will be - what I said was that the more often you play, the more often the house edge will come into play to your disadvantage.
Yes, they hammer out the details, but they still don’t support the point septimus was making. The point was: If you have $70 and your objective is to double that money, walking out with $140, then the strategy that maximizes the chances of achieving this is to keep betting on single numbers until you win. That point os not demonstrated in the thread you link to.
My expected value is actually (this would look better in Tex):
- 20 - 2 (1 - (37/38)^20) = 19.17 (approx.)
The house edge for this game is 1 - 19.17/20 = 4.15% (approx.)
- 20[(1/38)(20+35) + (1/38)(37/38)(19+35) + (1/38)(37/38)^2(18+35) + … + (1/38)(37/38)^19(1+35)] simplifies to 20 - 2 (1 - (37/38)^20).
This:
Should be: (1/38)(20+35) + (1/38)(37/38)(19+35) + (1/38)(37/38)^2(18+35) + … + (1/38)(37/38)^19(1+35)
Here you are assuming that you constantly bet $2 on single numbers. That’s not going to meet the requirement of walking out of the casino with $70.
If you bet $2 in your first bet and win, obviously you meet the requirement, walking out of the casino with the remaining $68 of your capital plus your winnings of $72, totalling $140. But if you lose (which you do with a probability of 37/38), it won’t be sufficient to bet $2 next time since even if you win, you’ll only have $66 in cash left plus $72 in winnings, giving you a total of only $138. To calculate how much you need to bet in your second bet, we need the following equation (x being the stake):
68 - x + 36*x = 140
So you would have to bet $2.06 in your second bet, leaving you with $65.94 in cash plus winnings of $74.16 if you win for a total of $140. If you lose, you will, again, have to increase your bet, since at each step you don’t just need to win $72 - you also need to make up for the losses you’ve incurred so far.
I apologize for the confusion. I was explaining how with the proper strategy you could decrease the house edge by using the game I introduced in post 28.
That is a different game than the puzzle proposed by septimus which he explained thoroughly and completely in the linked thread.
For both our games the house has a slightly smaller edge than it does for any single one play at roulette. The commonality is that both our strategies dictate walking away after winning a high payoff bet. I thought mine would be slightly clearer since I am betting exactly $1 each time.
No it doesn’t. The house edge is based on the rules of the game and is constant irrespective of betting behavior. You can change the amount of your bets and when you make them, but the house edge of 5.26 % of stakes played is nothing you can avoid or mitigate.
I laid out a roulette strategy in post 28 and explicitly calculated the house edge of 4.15% against this strategy in post 31 (with a slight correction in post 32).
Please point out my error.
So someone does agree with septimus. Yippee!!
The other thread seemed interminable. I was sure I’d finally made L.N. see the light (but thought it rude he didn’t say Aha! to give closure) and a few months later, he said my views on Social Security were probably as ignorant as my roulette “martingale”. :smack:
Since I’ve explained this all “thoroughly and completely” already (Thanks, Lance Turbo ) I’ll go away, but first offer a suggestion:
Presumably Schnitte thinks his intuition about roulette (18 reds, 18 blacks, 2 greens) would apply to a simpler setup (e.g. 4 reds, 4 blacks, 1 green – or even 4 reds, 4 blacks, 4 greens – with 1:1 payoff on red and 7:1 payoff on number). If you choose to verify the claim with tedious non-computer calculations you may find it easier to use a simpler setup.
Your formula is wrong to begin with, since to calculate the house edge you multiply the payouts (in terms of stakes) in the event of winning by the probability of winning (this is the expected value), then subtract this from 1. You’re not doing that, and that’s why your formula is wrong.
In a fair game, the payouts are the reciprocal of the probability of winning, giving you an expected value of 1 and a house edge of zero.
In American Roulette, if you bet a single number, you have a 1/38 chance of winning, and if you win you’ll get your stake back, plus 35 times your stake, giving you a payout of 36 times your stake in the event of winning. The expected value is thus
(1/38) * 36 = 36/38 = 94.74 %. The house edge is the remaining 2/38, or 5.26 %.
If you believe that by applying a particularly “clever” way of sizing or timing your stakes you can reduce or eliminate the house edge, then you fell for the fundamental fallacy on which the gambling industry is founded. The 5.26 % house edge is etched (pun intended) into the rules of the game, and you can’t change that no matter how you size or time your stakes. Of course you can influence the expected value (in dollars, not percent of stake) of the game by sizing your stakes - the less money you gamble, the less money will be subject to the 5.26 % expected loss. The per cent value, however, you cannot change.
Things are different in blackjack (if no reshuffle between hands occurs), since the cards that have been dealt already cannot be dealt again. Therefore, outcomes are not independent of each other; but they are in Roulette.
It’s the same for even money bets, btw: Winning probability 18/38, payout in case of winning 2 → expected value: 36/38
Or for the columns: Winning probability 12/38, payout in case of winning 3 → expected value: 36/38
This goes for every single bet in Roulette, since the payouts are calculated such that they would be fair if there were 36 numbers on the board - but in fact there are 38, reducing your winning probabilities by the coresponding margin.
That’s exactly what I’m doing, however I’m not labeling things precisely that way so I’ll do just that.
Let me reiterate the strategy first. My stake is $20 and I’m going to bet $1 each spin on a single number on a double zero roulette wheel until my number hits or I go broke.
There are 21 events to examine. Number hits first spin, number hits second spin, …, number hits 20th spin, and number never hits.
The probabilities of each of these events occurring are: (1/38), (37/38)(1/38), (37/38)^2 (1/38), …, (37/38)^19 (1/38), and (37/38)^20.
The payouts are: (20+35)/20, (19+35)/20, (18+35)/20, … (1+35)/20, and 0.
The division by 20 is so payout is given in terms of our original stake of $20.
So putting that all together using the formula you gave we get:
1-(1/38)(20+35)/20 + (37/38)(1/38)(19+35)/20 + … + (37/38)^19 (1/38)(1+35)/20 + (37/38)^20 (0) = 4.13 % (approx and note that I am correcting a rounding error from earlier)
So in calculating the house edge in precisely the way you specified we have beat the house edge we could achieve on any single spin.