Gandler
• Posts: 1797
Joined: Jan 27, 2014
March 24th, 2014 at 11:41:32 AM permalink
Ok so every table game has a positive house edge of what ever it may be. So somebody goes to the casino to play roulette which has a high house edge (2.63 on outside bets). And they have a set bankroll for their play for a the evening. Lets say 200 dollars.
I am certainly no mathematical expert, and I have been laughed at many times for suggesting this, but would the best betting system for any game with a house edge of any level be place one single bet, your whole bankroll in one bet? For this example if the player place 200 dollars on red. And if he won walked away and if he lost walked away.
My thought is the longer you play the more it benefits the casino, so placing your total gambling bankroll in one bet on every visit in my thoughts would always be the best system? Is there anyone who can validate or deny this? Obviously this only applies to positive HE games, if by some miracle there was a game where the player had an edge, I would assume the longer play time would benefit the player so the reverse would be true?

And also I randomly chose roulette as an example, I know a game with a lower house edge is always ideal, but the basics of the theory apply with any game with a positive HE regardless of how high it is?
Dalex64
• Posts: 1067
Joined: Feb 10, 2013
March 24th, 2014 at 11:50:06 AM permalink
That's the way to go if you want to maximize your chances either doubling your money or losing it all in a short a time as possible.

If, instead, you bet the table minimum, and play for some amount of time, you are more likely to be close to the middle of the bell curve of how much money you have left, and should lose money slowly.
geoff
• Posts: 368
Joined: Feb 19, 2014
March 24th, 2014 at 12:14:19 PM permalink
You can think of the house edge as how much you expect to lose for every \$100 wagered. To that end if bet \$200 in one shot or wager \$50 in 4 separate events you have the same theoretical loss. This all assumes you walk away after having that \$200 on the table. If you take any winnings and rebet them then you can lose more as an average.

For example if only one of your \$50 bets won and you decided to give it one last go to get back to even and you put \$100 on the table that money counts as an additional \$100 even though you didn't reach into the wallet.
chrisr
• Posts: 141
Joined: Dec 9, 2013
March 24th, 2014 at 2:10:44 PM permalink
I think you are spot on with your analysis. you would have a larger RSD (relative standard deviation) placing one bet then walking away rather than spreading your bank out over several bets identical bets.

If you get no utility out of playing the game and only want to gamble money.. large RSD is good. you can think of it as reverse insurance. you are paying money on average to increase risk instead of protect against it. So in that respect the larger the RSD the more reverse insurance you are getting for your buck.

Before you seek out gambling though one way to increase variance in your wallet is to get rid of some insurance policies. It's effectively the same thing but it's +EV (unless you are a bad driver in a good driver pool or something).

In conclusion, you shouldn't gamble if you have insurance.
mustangsally
• Posts: 2463
Joined: Mar 29, 2011
March 24th, 2014 at 3:27:59 PM permalink
Quote: Gandler

but would the best betting system for any game with a house edge of any level be place one single bet, your whole bankroll in one bet? For this example if the player place 200 dollars on red. And if he won walked away and if he lost walked away.

for the highest probability of hitting a win goal or a target bankroll, you are describing the concept of Bold Play.
Bet just enough to hit your target or everything, which ever is smaller

18/37 = 0.486486486 chance of doubling \$200. one spin and win!
if we tried to bet 1 number straight-up to take advantage of variance and the 35 to 1
I get only [1] 0.4699368

But with a 35 to 1 payout the ratio of the bet to the target becomes very important
as one can easily over-shoot the target and win more but lower the overall probability.
There is a formula to calculate this and get close but we will use just Excel

example shows this
\$2000 bet all on Red. still a 18/37 chance of winning. 0.486486486
Now bet each spin on your favorite number using this progression after each loss

58 <<< this is the 1st bet to make
59
61
63
65
66
68
70
72
74
76
79
81
83
85
88
90
93
96
98
101
104
107
110
lose all these spins and we are left with 53
keep going
53
60
62
64
66
67
69
71
73
75
78
80
82
84
87
89
92
95
97
100
103
106
109
of course one would have to be able to bet 110 straight-up on a number, so many tables may not allow this.

overall chance to double = [1] 0.4884156
higher than 0.486486486 for one bet on Red. and higher is better!
and because we have to make even \$ bets we will win more than 2000, looks to be about 2015 on average

this method of Bold Play on Roulette would also last more spins than just one (on average this example 5.63)
and get more comps (maybe)
and get way more thrill and excitement
IMO

win,win
fun,fun,fun
Sally
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