Ninfb

All about Video Poker with casino gambling expert Michael 'Wizard of Odds' Shackleford. Author of the American Casino Guide, interviews gaming expert Michael 'Wizard of Odds' Shackleford about.

1. Wizard Of Odds Casino Wars
2. Casino War Online
3. War Card Game Vegas

Thread Rating:

Used a Fibonacci progression, biggest bet 13 units a few times, love the zero HE aspect, until there is a Tie.

I don't think you need to worry about any sort of 'heat' while playing this game. I doubt surveillance much cares about it. Being able to vary your bets is great, but there really is no 'count' to the deck.

There is a myth out there that there isn't a strategy in 'Casino War' but you are right and wrong. The reason you are right is because there is luck in this game but the reason you are wrong is because if you pay attention to what cards are coming out then you will have greater success than people who are just putting their money in the circle hoping their card wins. I play this game religiously and I'm successful about 60 percent of the time which is a pretty good percentage. I came in the Casino with $20 and ended up leaving with $400 2.5 hours later. I've lost alot and won alot playing this game. When I lose, many times is because someone isn't paying attention to the cards and adding an extra hand that isn't needed. Of course the House has the advantage but if you play your cards correct than you can be successful. I came in the Casino multiple of times with less than $100 and ended up winning $500 plus. I've also seen players leave the table with 10 stacks ($10,000.00) or more. I've seen 18+ hand runs multiple times to answer you question.
Here is my strategy (Remember this isn't a 100% you're gonna win method);
1) Don't play against the Table!!! The House is gonna eat meaning you aren't gonna win every hand, I repeat you aren't gonna win every hand. The object of the
game is to give the Dealer the lowest card possible. It's the table against the Dealer not the player against other players and Dealer.
2) Pay attention to the card flow. If 3 hands (3 Player Hands plus 1 Dealer Hand) is winning majority of the time, you ride that out. Momma always said if it isn't
broke, don't fix it. Don't change the number of hands being played because you lost once. If the number of hands being played is losing majority of the time then
change the number of cards being played.
3) The best number of hands to play are 3 or less. The more hands being played the better chances of the house to win.
4) If possible, you want to play with the House money not yours.
5) Press (Stack your winnings each hand) when you are on a multi card run and play minimum (Or not play at all) when you think you might lose or situation is iffy.
6) Play your ties (Second or red circle). Playing your ties will allow you to get ahead of the house. This is a game of runs, some are losing and some are winning.
You're gonna have some bad nights and you're gonna have some 'On Fire Nights'. Placing a $5 chip on the tie will save you everytime because it will give you the
choice to surrender or go to war. The Tie is 10 to 1, second best odds in the Casino (Baccarat is the best 30-1). If you happen to be lucky and play the tie at the
right time and win, surrender surrender surrender. Ties come in two's and three's meaning they repeat their selves. After War, repeat your bet!!!
7) Most of all, have fun!!!!!! This is a game of good vibes. Any negative energy around, the game will follow.

Well, you're correct on #2. Could have been worse.
ZCore13djatcteliot

My strategy is pretty effective.

We are all very happy for you.beachbumbabs
Administrator

Casino War online pays 10 to 1 for a tie and the bet range is $1 to $250.
Are the real casino rules the same?
Also, how often would one hit a streak of 40, 45, 50, 55 or 60 hands without a tie?
I have no idea how to figure this out.
Thanks,
Russ

For various rules and paytables both in casinos and online, see this link.
As to streaks, I'm sure the info is out there, because that had to be part of the calculation to determine odds paid on war, but I don't have a link. It would depend on how many decks are used, too.ThatDonGuy

Also, how often would one hit a streak of 40, 45, 50, 55 or 60 hands without a tie?
I have no idea how to figure this out.

If you assume an 'infinite deck' - that is, one where the number of aces, 2s, 3s, and so on through kings is always the same after each card dealt - then it is relatively easy to figure out the probability of a streak: the probability that a hand will not be a tie is 12/13 (since there are 13 possible ranks for the dealer's card, only one of which will tie the player's card), so the probability of N non-ties in a row is (12/13)^N.
For N = 20, this is about 1/5; for N = 30, about 1/11; for N = 40, about 1/25.
However, streaks are irrelevant - the only relevant numbers in terms of ties is, if you bet 1 on the tie, you win 10 1/13 of the time, and lose 1 12/13 of the time. You will expect to lose 2 out of every 13 bets, or a house advantage of 15.4%.mustangsally

Thanks for the reply,
I am playing War on an online casino which says it uses six decks and reshuffles after every hand.
So, if I dealt the first card, then dealt every other card one at a time I would match 23 times and not match 288, which would be the same for every hand.
I don't know how to figure a streak.

p=288/311
so for the next 50 games
p^50 or about 1 in 47
but there will be more games than just 50 games played
so a calculator (or spreadsheet) can do this
let's use WolframAlpha!
how about 50 in a row over 100 games?
1 in 10
≈0.100808
https://www.wolframalpha.com/input/?i=streak+of+50+successes+in+100+trials+p%3D288%2F311
how about 50 in a row over 200 games?
about 1 in 4
≈0.243779
hope this helps out some

Wizard Of Odds Casino Wars

have fun
Sallymustangsally

Also, years ago I heard a 'Bet you can't lose' that said in a shuffled single deck if you dealt the cards one at a time there would be at least one pair, and usually two or three. They gave the math, but it was beyond me.

you mean 2 in a row with the same rank. ie (8,3,6,5,J,K,7,7)
lots of math (and a computer) for that one
(i agree with your 'beyond me')
so
1 - p(not getting 2 in a row)
4184920420968817245135211427730337964623328025600.
Dividing by 52!/(4!)^13 gives the probability: .045476
https://math.stackexchange.com/questions/310971/no-two-identical-ranks-together-in-a-standard-deck-of-cards
1 - .045476
would be the answer
most would need to run a simulation to get
an approximate answer, I would think
*****
I think just dealing 12 or 13 cards would be more fun
and faster too
the math is still difficult without computer help
I simulated it instead(for now)
shuffle and deal 12 cards

Casino War Online

probability of no 2 in a row same rank about
[1] 0.508

War Card Game Vegas

shuffle and deal 13 cards
probability of no 2 in a row same rank about
[1] 0.478
I can feel the 13 cards when doing this by hand
hard to tell with 12 cards
thanks for the thoughts!
Sally

*****
I think just dealing 12 or 13 cards would be more fun
and faster too
the math is still difficult without computer help
I simulated it instead(for now)
shuffle and deal 12 cards
probability of no 2 in a row same rank about
[1] 0.508

Sally,
As always, love your posts.
The [1] suggests to me you may be doing your simulation(s) in R, as that is often how it reports output. If so, can you please share your code (for the R neophytes, such as myself)?
If not in R, then how simulated? (Curious minds want to know learn.)
Many thanx.

The [1] suggests to me you may be doing your simulation(s) in R, as that is often how it reports output. If so, can you please share your code (for the R neophytes, such as myself)?
If not in R, then how simulated? (Curious minds want to know learn.)
Many thanx.

thank you.
I 1st did this in Excel, not with vba but using an add-in
I saw some code online and just changed it for a question
someone else asked me
so my original code is lost (or missing in a folder)
here is what I used

hope this helps out
btw,
C = 52 for a complete single deck
Sally