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

Quote:russ451Casino 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.

Quote:russ451Also, 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%.

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

p=288/311Quote:russ451Thanks 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.

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

have fun

Sally

you mean 2 in a row with the same rank. ie (8,3,6,5,J,K,7,7)Quote:russ451Also, 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.

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

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

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

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.

thank you.Quote:LuckyPhowThe [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.

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

> 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

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.