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  #1  
Old 11-18-2005, 01:24 AM
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Default Help with probability in Casino War.....no joke.

In no way am I trying to solicite any type of betting system when I write this just wondering if I could find some insight into this one. Anyways here it goes the system goes bet $1 if you win bet the same amount. If you lose, bet $2 if you lose bet $3. From $3 on you double the bet to 6, 12, 24, 48, 96, 192, 384. The basis of this system is that it would be very difficult to lose 10 hands in a row, and if you did it happens so rarely that you should be ahead. I didn't invent this system so don't flame. The person that invented this said it would work at Golden Palace because most casinos get the edge in casino war from when players tie and they go to "war". Sorry for making this so long, but I was just wanting some insight from people who can actually spell math. Thanks.
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  #2  
Old 11-18-2005, 01:45 AM
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Default Re: Help with probability in Casino War.....no joke.

You gave the answer in your post: "...it happens so rarely... ", but often enough I guess.
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  #3  
Old 11-18-2005, 01:54 AM
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Default Re: Help with probability in Casino War.....no joke.

Well that's the person said on the website. Not my personal experience.
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  #4  
Old 11-18-2005, 02:41 AM
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Default Re: Help with probability in Casino War.....no joke.

yes, this was thought of a long time ago, called nightingale theory i believe, only prob is one time out of say 1000 you will lose 10 coinflips in a row and lose your entire bankroll. Also most casinos have a stop to this system by enforcing a max bet. Say min bet is 5 max bet is 100 5,10,20,40,80- merely have to lose 5 coinflips in a row and your broke, while each time you only gain 5-basically if you do the math the odds of losing your whole bankroll are greater than that of the amount of money you gain winning over and over with the occasional loss. im sure someone could do the math for you with actual AC/vegas casino bet limts/max
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  #5  
Old 11-18-2005, 03:19 AM
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Default Re: Help with probability in Casino War.....no joke.

I tried looked up that theory, but got nothing. I tried the wizard of odds website, but couldn't find anything related. Mostly I was talking about 1 dollar bets going up to 384. The max on the site that is suggested on the website was 500. I tried this with play money and I guess got lucky and ran 400 up to 1200 in 2 1/2 hours. Yeah I need to get a life.
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  #6  
Old 11-18-2005, 03:23 AM
ThinkQuick ThinkQuick is offline
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Join Date: Jan 2004
Location: Alberta, Canada
Posts: 97
Default Re: Help with probability in Casino War.....no joke.

I don't know what it is.
Hopefully the problem is jsut that I had a long day, and not that I am tired of teaching people.

But this is getting ridiculous, I mean it.
Ok give me a few minutes, I will respond if nobody else has.

Meanwhile, while there are a number of arguments that can be made about why war is worse to try this on than anything else, I don't think you care about that, just about the 'system' itself.

Search "martingale", plain and simple. Note: Don't let anyone tell you it has to do with table maximums.
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  #7  
Old 11-18-2005, 03:30 AM
yellowjack yellowjack is offline
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Join Date: Nov 2004
Location: Vancouver, Canada
Posts: 263
Default Re: Help with probability in Casino War.....no joke.

google "martingale"

Below is a brief calculation I did for playing blackjack where you have less than a 50% chance of winning:

When you play for $1, it's -EV
When you play for $2, it's -EV

Even though you're doubling your bet each time, it's still -EV

I don't know any strategy for BJ, but I do know that surrendering, doubling down, etc. are only moderately +EV in comparison to the overall -EV in playing. To simplify this, I'll put the probability of winning a hand at 40%, even though I think it is a bit generous (esp. online)

Suppose you're playing a max of 4 hands, where you go from $1 to $2 to $4 to $8 if you lose, and you're stopping at $8 (we'll see later that the max amount doesn't matter, it's still -EV). Let's say you have a 40% chance of winning a hand. If you win at $1 you just play for $1 again.

Note that the win is always just +$1 since when you're playing for say $4, you have already lost $3. So your net gain is +$1.

EV = 0.40*(+$1) + 0.6 * [0.4*(+$1) + 0.6 * [0.4*(+$1) + 0.6 * [0.4*($1) + 0.6(-$15)]]]

EV = $0.40 + 0.6 * [$0.40 + 0.6 * [$0.40 + 0.6 * [$0.40 + (-$9)]]]

EV = -$1.0736 (OP should check this)

So in words, even though the chances of losing 4 bets in a row is small, when you do lose 4 bets in a row it costs a lot. It's worth noting that the more bets you're prepared to make, the worse your EV is.

Take for example just making $1 and $2 bets.

EV = 0.4(+$1) + 0.6[(0.4)(+$1) + 0.6(-$3)]
EV = -$0.44
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  #8  
Old 11-18-2005, 04:15 AM
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Default Re: Help with probability in Casino War.....no joke.

Wow.... I really need to check with you guys first before I go and try something know nothing about. In short I blew half my poker bankroll on this, and I feel like a complete moron as I should. I appreciate the help, and I'll be checking with you guys before I go and try to make stupid bets.
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  #9  
Old 11-18-2005, 04:34 AM
ThinkQuick ThinkQuick is offline
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Join Date: Jan 2004
Location: Alberta, Canada
Posts: 97
Default Re: Help with probability in Casino War.....no joke.

[ QUOTE ]
Wow.... I really need to check with you guys first before I go and try something know nothing about. In short I blew half my poker bankroll on this, and I feel like a complete moron as I should. I appreciate the help, and I'll be checking with you guys before I go and try to make stupid bets.

[/ QUOTE ]

Willy: I really appreciate that you truly listened to the posters here. I'm sorry that you had to lose that $400 + to learn this lesson though.
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  #10  
Old 11-18-2005, 04:39 AM
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Default Re: Help with probability in Casino War.....no joke.

Yeah I'm just glad I was conscious enough not lose my entire bankroll on the game. Oh well lesson learned and I will stick to poker except for the occasional trips to the local casinos in St. Louis for some five dollar blackjack.
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