russ451
Joined: Dec 18, 2017
• Posts: 9
December 18th, 2017 at 2:20:03 PM permalink
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
It's only impossible until somebody does it.
beachbumbabs
Joined: May 21, 2013
• Posts: 14230
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December 18th, 2017 at 2:48:15 PM permalink
Quote: russ451

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.
If the House lost every hand, they wouldn't deal the game.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 4460
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December 18th, 2017 at 4:58:28 PM permalink
Quote: russ451

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%.
russ451
Joined: Dec 18, 2017
• Posts: 9
December 18th, 2017 at 10:49:31 PM permalink

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.

With a coin flip, for instance, for a two flip scenario there are four possibilities with one being two heads, so, the chance of two heads is 1 in 4, the chance of three Heads is 1 in 8.

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.

I don't think this would apply in this case as the cards are shuffled after each hand.

Another one is that in a room of 50 or so people at least two would share the same birthdate ( not necessarily the same year, just the date.).

In any case, was wondering what the likelihood of no pair for 50ish hands was.

Russ
It's only impossible until somebody does it.
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
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December 19th, 2017 at 8:48:48 AM permalink
Quote: russ451

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?
≈0.243779

hope this helps out some
have fun
Sally
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mustangsally
Joined: Mar 29, 2011
• Posts: 2463
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December 19th, 2017 at 12:26:43 PM permalink
Quote: russ451

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

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

shuffle and deal 12 cards
probability of no 2 in a row same rank about
[1] 0.508

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
Last edited by: mustangsally on Dec 19, 2017
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LuckyPhow
Joined: May 19, 2016
• Posts: 659
December 19th, 2017 at 7:32:25 PM permalink
Quote: mustangsally

*****
I think just dealing 12 or 13 cards would be more fun
and faster too

the math is still difficult without computer help

shuffle and deal 12 cards
probability of no 2 in a row same rank about
[1] 0.508

Sally,

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.
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
December 19th, 2017 at 8:15:29 PM permalink
Quote: LuckyPhow

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
so my original code is lost (or missing in a folder)

here is what I used
`> V = 4 # number of values per rank> R = 13 # number of ranks in deck> # V*R = # of total cards in deck> C = 13 # last card to check for a run of 2> F = 2 # 1st card to check for a run of 2> > check_deck <-function(deck){+   for(i in F:C){+     if(deck == deck[i-1]){+       return(0)+     }+   }+   return(1)+ }> > N = 10000000   # number of sims> count = 0 > for(i in 1:N){+   d = sample(rep(1:R,V))  +   count = count + check_deck(d)+ }> > p = count/N> p # probability of NO 2 run[1] 0.478026> 1-p # probability of at least one 2 run[1] 0.521974> > ############################> V = 4 # number of values per rank> R = 13 # number of ranks in deck> # V*R = # of total cards in deck> C = 12 # last card to check for a run of 2> F = 2 # 1st card to check for a run of 2> > check_deck <-function(deck){+   for(i in F:C){+     if(deck == deck[i-1]){+       return(0)+     }+   }+   return(1)+ }> > N = 10000000   # number of sims> count = 0 > for(i in 1:N){+   d = sample(rep(1:R,V))  +   count = count + check_deck(d)+ }> > p = count/N> p # probability of NO 2 run[1] 0.5081397> 1-p # probability of at least one 2 run[1] 0.4918603> `

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

Sally
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russ451
Joined: Dec 18, 2017
• Posts: 9
December 19th, 2017 at 8:26:36 PM permalink
With my "Martingale" betting progression I can go 55 hands as long as I win hand 55 or less.

Yesterday I was playing for free with 1 - 250 limits and reset stake after 10 wins.

I played 5 rounds.

Round 1 50 hands + \$72

2 118 hands +140

3 109 hands + \$132

Looking good, I am Sooo excited!

4 170 hands - \$2486 ( I think the loss is a little high, but still a loss. Total streak went to 64.)

5 89 hands - \$2496.50 Lost on fourth progression on hand 55, did not continue playing to see when streak ended, had to go do stuff.)

Soooo.... Heavy Sigh.....

Back to Roulette.

Thanks.
It's only impossible until somebody does it.