Need Help Calculating House Edge On Never Seen Before SIDE BET

Ok so i spent about an hour trying to work this one out so far and ive failed miserably. PLEASE HELP ME GREAT WIZARD!
At my local casino(Sacramento) they have a side bet game for Blackjack that I have never seen mentioned here or anywhere for that matter it goes as follows:

Name: "RED/RED Bonus"

Rules: If the dealer has anything but two reds in his first two cards, it is a automatic no pay/loss. I.E. one black one red is a loss. Two black cards is a loss. However get any two reds is a pay and if the dealer needs to hit to additional cards IF they are RED the payout increase. However the payout increase is stopped if the dealer hits a black card. The dealer stops taking cards when he or she hits to 17 and above or busts. So if the dealer has two red tens then the most red cards you can get that hand is two.

Heres some Examples R=red card B=Black Card
Dealers Takes a total of 6 Cards in all these Examples
1. RRRRRR This counts for 6 Red Cards
2. RRBRRR This Counts for 2 Red Cards
3. RBRRRR This Does not count for bonus at all
4. RRRRBB This counts for 4 Red Cards.

Bonus payouts are as follows:

# Of
Red Cards....................Payout

2 ................................2X1
3 ................................3x1
4.................................8x1
5.................................50x1
6.................................100x1
7+...............................250x1

This sounds like a fun side bet! Is the game dealt from a continuous shuffle machine? And how many decks are used?

I may have made some mistakes, but I wrote a program to tabulate the number of cards that the dealer will get, on average, with an infinite deck. I only manually verified the probability of the dealer receiving two cards. These are the results:

Card Count	Probability
0	0
1	0
2	0.349112426035502
3	0.439235320892124
4	0.172788067644697
5	0.0344121758287876
6	0.00411659131322629
7	0.000318525456147501
8	1.63350474083679E-05
9	5.46466155182669E-07
10	1.11926365891922E-08
11	1.22756869715455E-10
12	5.5798577143389E-13
Total	1

From here, you can calculate the probability of getting a specific number of red cards. For instance, in the case of 4 cards, the probabilities are as follows:

Red Card Count	Formula	Probability
0		0
1		0
2	(.5)^2*.5	0.125
3	(.5)^3*.5	0.0625
4	(.5)^4	0.0625

When I do all the calculations, I get a return of 0.702 if the pays are "for 1" and 0.952 if the pays are "to 1" (that is, you have your original bet returned).

I was just interested, because of my calculations, how on earth does a dealer get 12 cards:

It turns out to be: A A A A A A 6 A A A A XX (where the XX card can be anything).

If the dealer hits soft 17 then the most cards he/she can get is 13: A,A,A,A,A,A,A,5,A,A,A,A,X where X can be any card.

No its actually a 6 deck shoe game with about 5 deck penetration

Quote: CrystalMath

I may have made some mistakes, but I wrote a program to tabulate the number of cards that the dealer will get, on average, with an infinite deck. I only manually verified the probability of the dealer receiving two cards. These are the results:

Card Count	Probability
0	0
1	0
2	0.349112426035502
3	0.439235320892124
4	0.172788067644697
5	0.0344121758287876
6	0.00411659131322629
7	0.000318525456147501
8	1.63350474083679E-05
9	5.46466155182669E-07
10	1.11926365891922E-08
11	1.22756869715455E-10
12	5.5798577143389E-13
Total	1

From here, you can calculate the probability of getting a specific number of red cards. For instance, in the case of 4 cards, the probabilities are as follows:

Red Card Count	Formula	Probability
0		0
1		0
2	(.5)^2*.5	0.125
3	(.5)^3*.5	0.0625
4	(.5)^4	0.0625

When I do all the calculations, I get a return of 0.702 if the pays are "for 1" and 0.952 if the pays are "to 1" (that is, you have your original bet returned).

Dude the program is freaking awesome if the percentages are correct! I had no idea how to even approach that side of the problem. I tried to write out all possible Hit/Stands for two cards then move to three cards but the chart gets exponentially big on you after you have to consider basic strategy.

Right so my plan was to take
Take the chance of getting "X" red cards in a row * chance that dealer hits to "X" amount of cards

The only part i figured out was the red cards in a row part

chance of getting "X" red cards in a row
2 Card Red 156/312 155/311 = 0.2491961414790997
3 Card: = 0.1237942122186495
4 Card: =.06129616333156433
5 Card: = 0.03025005463116162
6 Card: = 0.01487869136581565
7 Card: = 0.007293476159713552
8 Card: 0.00356304245179449
9 Card: 0.001734639088373633
10 Card: 8.415575775278022e-4
11 Card: 4.068457162882752e-4
12 Card: 1.959888002053153e-4

When I ran your Card Count Probability Times my Red Count Probability

I got 36.57% House Edge any conformations?

Using my dealer hand distribution (which is based on infinite decks) and your red count probability (which is based on 6 decks), I get a house edge of 30.3% if you do not get your original bet returned. It has a house edge of 5.4% if you do get your original bet returned.

When you calculate your red count probabilities, you need to take into account that the following card needs to be black in order to end the streak. Maybe this is where we are different.

If I get a chance (probably tomorrow), I can do this with 6 decks.

Do you know if you get your original bet returned or not?

You do get your bet back so if its 2x1 you have a total of three units in front of you