1. ## Bobby: Bonus BJ

Hello to all Gurus: Don, Parker and Viktor,

Have a question, posted on "International Scene", but no real response.

I saw a very interesting Bonus BJ in one of underground casinos in Istanbul. You can bet on bonus box from table min to your box bet.

1) You get paid 2:1 if one of your cards has the same suit like dealer and bigger. F.e. dealer Th, you Jh8s. Jack is bigger then ten. Ase is the biggest card.

2) You get paid 5:1 if your both cards has the same suit as dealer has and at least one of your cards is bigger. F.e. dealer - 7d, you Ad2d

3) You get 10:1 paid if you have a pair and one of your cards has the same suit like dealer and bigger. F.e. dealer Jc, you AcAs

4) You get paid 25:1 if you have a pair and your both cards have the same suit as dealer and bigger. F.e. dealer 6s, you - TsTs

What kind of count you suggest? Is just red/black good enough? Should it be connected with normal count?

Thanks

Bobby 2. ## Dog Hand: I show a player edge...

Bobby,

I spent some time analyzing this sidebet today and, assuming it's a 6 deck shoe, I found the following results showing a player's edge of over 1.58%:

```
Player			Dealer			Payout	Probability	Return
Suited Pair		Suited Lower		25	0.00186703	0.04667566
Unsuited Pair		Suited Lower		10	0.01400270	0.14002697
Suited non-Pair		Suited Lower than High	5	0.05489555	0.27447775
Unsuited non-Pair	Suited Lower		2	0.16131107	0.32262213
Lose						-1	0.76792366	-0.76792366
Total							1.00000000	0.01587885
```
.

As an example of my calculations, the probability that the player will receive a suited pair is 5/311 = 0.016077..., because once the player has his first card, say 9C, then the shoe has 5 more 9C's out of 311 cards.

The probability that the dealer's card will be the same suit but a different rank is then (5/311)*(12*6)/310 = 0.003734..., because the club suit has 12 other ranks besides the 9C times 6 decks, divided by the remaining 310 cards.

Finally, overall the player will have the higher rank half the time (the other half of the time, of course, the dealer will have the higher rank), so we multiply by one-half to get the win probability: (5/311)*((12*6)/310)*(1/2) = 0.001867...

For an 8-deck game, the player's edge is even higher: just over 1.75%.

Perhaps I'm misunderstanding the rules. When you say "25:1", do you mean that, for a \$1 bet, you get \$25 PLUS you get back your original \$1? That's the usual meaning of the phrase "25 to 1", and so that's what I assumed in my calculations. If instead you get \$25 but the dealer keeps your \$1, then we say that is "25 for 1". If the odds you quoted are actually "for" odds, then the house has a gigantic edge of over 21%.

Can you show how you calculated a house edge of 4%? Also, can you confirm that the odds are "to" odds?

Dog Hand 3. ## Bobby: Re: I show a player edge...

Hey Dog,

I Have a problem only with the last part of your calculation.
"Finally, overall the player will have the higher rank half the time". It cannt be true cause the successfull cards by player are (12+11+10+9+8+7+6+5+4+3+2+1)*6. Total combinations are 13*77 (i calculate 77=13*6 - 1). So I will get 46,7%.

What do you think about red/black count?

Cheers

Bobby 4. ## Dog Hand: I think you missed a subtle point...

Bobby,

You said:
I Have a problem only with the last part of your calculation.
"Finally, overall the player will have the higher rank half the time". It cannt be true cause the successfull cards by player are (12+11+10+9+8+7+6+5+4+3+2+1)*6. Total combinations are 13*77 (i calculate 77=13*6 - 1). So I will get 46,7%.

First of all, you erred slightly here. If the player is dealt a suited pair, that leaves 76 more suited cards, not 77. Thus, your prob of the dealer's card being lower should be (12+11+10+9+8+7+6+5+4+3+2+1)*6/(13*76) = 0.47368...

What you missed is that when I calculated the probability that the player will be dealt a suited pair and the dealer a suited lower card, I showed the result as follows:

(5/311)*((12*6)/310)*(1/2) = 0.001867...

Here, as I explained in my previous post, the middle term: ((12*6)/310), is the probabilty that the dealer's card will be suited but NOT of the same rank: that's why the numerator is 6*12 (six decks times 12 non-matching ranks).

Instead we could calculate the overall probability as follows:

prob = (prob of suited pair for player)*(prob of suited card for dealer)*(prob that dealer's card is smaller)

Here, we'd get this:

prob of suited pair for player = 5/311 (as before)

prob of suited card for dealer = (13*6 - 2)/310, because now we have 13*6-2 - 76 suited cards left in the 310-card shoe

prob that dealer's card is smaller = ((76 - 4)/2)/76 = 0.47368..., because of the 76 remaining suited cards, 4 are the same rank as the player's pair, and of the other 72 suited cards, on average half are smaller and half are larger.

This gives the overall prob as

prob = (5/311)*(13*6 - 2)/310)*(((76 - 4)/2)/76) = 0.001867...

which is the same value I reported in my original post.

Have you confirmed that the odds are in fact "TO" rather than "FOR" odds?

As for the red/black count, I'll try to work on that to see how valuable it would be.

Dog Hand 5. ## Dog Hand: CORRECTION! I erred...

Bobby,

I found a similar sidebet analyzed by the Wizard of Odds called "Block": you can find the webpage at this URL:

wizardofodds.com/games/blackjack/appendix/8/#block

His game has a better 6-deck paytable than yours, and even then he shows a house edge of about 5%.

Somehow, I erred in my calculation of the probability of the 5:1 payout.

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