Dragon 8's side bet

I'm not soo sure how beatable it is.. The pay tables and limits don't look like they are "stout" enough at first glance. It'll be fun to work through though.

Just looking at the pair of 8s pay at 7:1 -- 0.004136*7 = 0.02895 return.(approx) (assuming your first 2 cards aahve to be 8's)

Triple unsuited 8s pay 50:1 -- 0.000369*50 = 0.01845 return. (approx)

Hmmm... check my math.

I think that a pair would include if you had a card and the dealer had a matching card, because with quick calculation, there seems to be an absurd HE if that was not the case.

If such is the case, then with the 8s, I get a (6*(12/13)*(1/13)^2)*7 ~= 0.2294, which seems reasonable.

PS: These are my results for an infinite deck for the returns for the specific hands (in percents). Fewer decks would mean a higher HE due to high dependence on pairs.
-888s 0.569
-888 1.707
-567 6.827
-triple 8.193
-8s 22.940
8, 18, 28
-pair 39.326
------------------------------------
-Total 79.562 + Baccarat Return

Couldn't figure out the baccarat 8 returns... Probably would need a sim. Anyone?

Pairs isn't right. In an infinite deck model you get a pair 1 out of 13 two-card sets. With 3 2-card combinations (player1/player2, player1/dealer, player2/dealer) you should see a pair about 23% of the time, no?

If your first two cards and the Dealer's top card contain any of the following combinations, you win!

"Baccarat" 8(Totals 8,18 or 28) 2-1

Here's a new side bet I came across. Did not see any info about it on the WOO. Beating this one would require somekind of specialized count. Here's the rules:

Must make regular blackjack bet, then you can make additional Dragon 8's bet.
Minimum bet $5
Maximum bet $50

If your first two cards and the Dealer's top card contain any of the following combinations, you win!

Three Suited 8's 200-1
Three 8's 50-1
Straight 18 (5,6,& 7) 25-1
Any Triple 15-1
Pair of 8's 7-1
"Baccarat" 8(Totals 8,18 or 28) 2-1
Any Pair 1-1

Ok, so working through them all using info from WOO:

Three suited 8s:
p=0.000016, return=200*0.000016=0.003191

Three 8s:
p=0.000388, return=50*0.000388=0.019388

567:
p=0.002757, return=25*0.002757=0.068936

Trips (not 8s):
p=0.004845, return=15*0.004845=0.072670

Pair of 8s:
p=0.015855, return=7*0.015855=0.110988

Baccarat:
p=217/2197=0.098771, return=2*0.098771=0.197542

Any Pair (not 8s):
p=0.190264, return=0.190264

So I get returns of:
0.003191
0.019388
0.068936
0.072670
0.110988
0.197542
0.190264

for a total return of 0.662979 or a HE of 33.7021%.

Clearly there is a big difference between my numbers and SOH's numbers. The first couple of differences are explained by my assumption of a 6 deck shoe. In fact, I pulled a couple of my numbers directly from the WOO chart for the Lucky Lucky side bet, since trip 7s is the same odds as trip 8s, etc.

Our biggest difference is obviously the return on pairs, both a pair of 8s and any pair. I figured these wrong myself the first time through, but we discussed our methods in the chat. Sonny helped us figure out our problem and we arrived at the numbers I've listed above.

At this point, I'm pretty confident in all of the numbers in this post, but as always, I'd love for someone else to look them over and see if they concur.

Everything seems okay with your analysis except the return for a pair you forgot to multiply by 2, and the house edge was not calculated correctly. House edge= Percentage loss - Percentage win, so you need to calculate of the probability of a no-winner hand, which is 1 -(sum of probabilities of winner hands) 1-(sum of p's you have listed)= 1- 0.31289= 0.6871

the return on that is -1, so HE= 0.6871-(total return from winning hands) = 0.6871-0.6630= 0.0241= 2.41%

Can you show your math for the pairs? Here is mine (based heavily on Sonny's single deck math from the chat last night). This is for a pair of 8s:

24*23*288+24*288*23+288*24*23=476928 total ways to get a pair of 8s.
312*311*310=30079920 total ways to choose 3 cards
476928/30079920=0.015855 probability of getting a pair of 8s.

For any other pair, simply multiply by 12.



For the house edge, HE is of course simply the house's EV, expressed as a percentage of the bet. The house EV in this side bet (or any, I suppose) depends on whether or not the initial bet is retained on a win. I assumed it was not, since that's how the side bets near me work. In other words, around here, 200-1 would imply 200 for 1, not 200 to 1, if I've got my terminology right. If that's the case, the EV for the house is simply 1-0.662979=0.337021, or 33.70%.

If the initial amount of the side bet is retained on a win, the math for the house EV is 1*(1-0.000016-0.000388-0.002757-0.004845-0.015855-0.098771-0.190264)-0.662979=0.024125, or 2.4125%, as you stated.

Clearly this is a huge difference. Looking again at the payout table, you are probably correct that the initial bet is retained. It is unlikely that a pair would be a push, for example. SystemsTrader can probably confirm one way or the other for us. Thanks for catching that.

Using Excel and assuming infinite decks, I made a quick attempt at simply listing all of the possible combinations of 3 cards and checking the totals. I then checked to see if the total added up to 8, 18, or 28. I came up with 217/2197 chance of hitting this payout, so it returns an average of 217 / 2197 * 2 = 0.1975 bets. If SleightOfHand's numbers are correct (didn't look through them myself) and if my numbers are correct, that puts the total return for this side bet at 99.316% with no side count. If that is the case, then it looks easily beatable to me.

I get 207. You are counting 666(a triple), 567 (straight), and 883( a pair of 8's), all of which have a higher payout.

Everyone seems to be getting too many for any pair for the same reason ie 224 would be a Baccarat.

Edit to add:
Ok I assumed (and you know what that means), that someone would have looked at wizofodds. straight 8's Just change some payouts. I'll still have fun with excel so I can get the eor's.

Well i didnt actually look at your probability calculation, but i just wanted to mention that you forgot to multiply by the payout for a pair which is 2 to get the return for getting a pair .

You can calculate the probability using combinatorials (nCr formula), just like you do it when calculating frequencies of poker hands

Frequency of getting a pair of eights with 3 cards in a six deck shoe = nCr(24,2)*nCr(288,1)=79488

Total number of possible combinations for three cards with a six deck shoe = nCr(312,3)=5013320

p of getting a pair of 8's = 79488/5013320= 0.015855,

your frequency and number of combinations are not correct but the error cancels out when you divide the numbers.

The house edge for 6 decks is 8.29% and the pay freq. is 28.58%.

I get the same. I get the following eor's a-10

+0.001205441132009
+0.001497881488375
+0.001205441132009
+0.002152851212138
-0.001129247994023
-0.001740714193699
-0.001462775012028
-0.006815319716431
-0.000682732656514
+0.001442293652041

Edit to clarify that these are the changes in the player's expected value . For example: A new 6deck shoe. Dealer burns an 8. The house edge is now: 8.97%