Hypothetical Side Bet (Is my math right?)

So here's a side bet I came up with, mostly for fun, so if anyone wants to steal it... go ahead. (It's easily countable, so I doubt it'd do well except on a CSM anyway) It was more of a math exercise for me and I want to see if I did the math correctly. I'm confident it's approximately correct, but I also think it may be slightly off, as I don't know if I have to compensate for something or if it would simply average out.

So it's a simple side bet. Basically if the Dealer or Player gets a Blackjack, it pays 8.5 to 1, and if both do, it pays 21 to 1.




I calculated it like this (in excel)

A3 is 312, for the number of cards in deck. A5 is the odds of both Player and Dealer Blackjack (the second formula)

8.5 to 1 was =((((24/A3)*(96/(A3-1)))*2)*2)-A5

21 to 1 was =(((24/A3)*(96/(A3-1)))*2)*(((23/(A3-2))*(95/(A3-3)))*2)



Since that's pretty cluttered and messy I'll just explain it. The first one formula is the simple "odds of a blackjack" formula, but multiplied by 2 because either the dealer of player can receive it, and minus the chance of both receiving it. The second is the odds of both receiving a BJ. (I had no trouble with the second formula.)

Here's where I'm second guessing my math. Are the odds of either player or dealer getting a blackjack slightly higher based on the fact that the other didn't get it? If so, how do I calculate that effect? Or is it as simple as "odds of blackjack multiplied by 2"?

Blackjack hands are anticorrelated, since given one hand is a blackjack, the odds of the other hand's blackjack is reduced. How much depends on the deck size. If there are 416 cards, a single hands blackjack probability is p=2*(16/416)*(64/415).

The probability of a hand's blackjack, given that another hand has also a blackjack is q=2*(15/414)*(63/413).

So the probabilities you are looking for are: 2*p*(1-q) for your 8.5 to 1 payout, and p*q for your 21 to 1 payout. All numbers for 8 decks.

I understand that, but that's not the issue. I don't think I was clear enough about the rules.

The bet is that the Player or Dealer gets a blackjack. It's a single bet that will win if either happens, and if both happens. It loses if neither gets a blackjack. So I was thinking there would be a slight positive correlation because one of the hands is not a blackjack, which would raise the odds of the other one being a blackjack. I also ran into the problem that there are many combinations of different hands which have different effects on those odds, so I couldn't easily think of a formula.

So given that the bet is for either (Dealer OR Player BJ) and a bonus (21 to 1) for both, how could I calculate the odds for the 8.5 to 1 payout? I understand how to calculate both and that you'd have to subtract the odds of that happening from the 8.5 to 1. What I can't quite wrap my head around is how to calculate the slight positive correlation I'm predicting.

Hi Boney526

Read Double Shot Blackjack

Quick Rules
The player can wager on the outcome of their 2-card
hand, the dealers 2-card hand or both hands at the
same time. By aiming for a ‘SingleShot‘ or ‘DoubleShot’
Blackjack the paytable changes accordingly.

Of course, the probability is increased for a blackjack, if the other hand os a non-blackjack hand.
It is simpler to calculate the inverse (probability of blackjack if the other hand is a blackjack), because a non-blackjack hand may still contain a ten or ace. You can however calculate it using Bayes formula.

Quote: MangoJ

Of course, the probability is increased for a blackjack, if the other hand os a non-blackjack hand.
It is simpler to calculate the inverse (probability of blackjack if the other hand is a blackjack), because a non-blackjack hand may still contain a ten or ace. You can however calculate it using Bayes formula.

I looked up the formula. It seemed to confuse the tar out of me, but I know it can't be that hard. It seems logical enough but I just can't wrap my head around the math right now. Thanks for pointing me in the right direction, though.

I pretty much understand what it's (Bayes Theorem) trying to do, I think, but ATM it's just too much for me to process.

Player Blackjack: 416C2 / (32*128) = .0474513
Dealer Blackjack if the Player has blackjack 414C2 / (31*127) = .0460516
Dealer Blackjack if the Player does not have a Blackjack AND does not have an ACE or 10-Value: 414C2 / (32*128) = .0479115
Dealer Blackjack if the Player has ONE ACE in the hand: .0464142
Dealer Blackjack if the Player has TWO ACES in the hand: .04491701
Dealer Blackjack if the Player has ONE 10-VALUE in the hamd: .0475372
Dealer Blackjack if the Player has TWO 10-VALUES in the hand: .0471628

Chance that the Player has TWO 10-VALUES: 416C2 / 128C2 = .0941613
Chance that the Player has ONE 10-VALUE: 416C2 / (128*256) .3796107
Chance that the Player has TWO ACES: 416C2 / (32C2) = .00574606
Chance that the Player has ONE ACE: 416C2 / (32*256) = .0949027
Chance that the Player has NO ACE AND NO 10-VALUE: 416C2 / 256C2 = .3781279

From here you can determine the Dealer Chance of Blackjack under all conditions dealt to the Player.

I am not sure about the formulas but the probabilities for 6 decks are:
PROBABILITIES
Player BJ 4,74895%
Dealer BJ 4,74895%
Player AND Dealer BJ -0,21665% Deduct Player+Dealer BJ Prob because is double counted.

Player OR Dealer BJ 9,28125% Includes Player+Dealer BJ Prob

Player AND Dealer BJ -0,21665%

Player OR Dealer BJ 9,06460%
BUT not both

EV
Win 21 0,21665%
Win 8.5 9,06460%
Lose -1 90,71875%

EV = -9,12002%

If you put then normal win at +9 the Ev becomes -4,58772%

I think I get it. Hell, I'm pretty sure I could now do the math properly, it would just take a little while to work though. Anyway, my estimate is at very least, very, very close. If I didn't have a seperate payout for both winning on BJ, common sense tells me that the conditional probabilities would cancel eachother out in a way that the formula could be simplified to "2*(24/312)*(96/311)"

The only reason that formula isn't accurate is because the Payout for both being BJ is seperate, so you have to consider the odds of one blackjack under every other possible arrangement of cards.

AceTwo, I'm pretty sure your math is wrong. Mine is very close, although I don't know how you calculated -9.12%, I got -7.06%. It's likely a little higher, due to conditional probabilities (that of a BJ between 2 hands should be slightly more than twice that of single hand getting BJ, right?)

IDK maybe I'm just really confused, but I'm 99% sure that the odds of any blackjack out of two hands (player or dealer) has to be more than twice the odds of a single hand getting a BJ, b/c of the effect of removal of cards that don't make up a BJ. A win of 8.5 assumes that only one hand received a BJ.... the other did not.

The probability of Player BJ in 6 decks is Combin(96,1)*combin(24,1) / Combin(312,2). That is 2304/48516 = 4,74895%
The Dealer BJ Probability is exactly the same = 4,74895%
Upto here the calculations are pretty straightforward and cannot be disputed.

The probability of Both Player and Dealer BJ are = { (Prob of Player BJ) * combin(95,1)*combin(23,1) / Combin(310,1)*combin(309,1) } * 2 OR
PROB = COMBIN(96,1)*COMBIN(24,1)*COMBIN(95,1)*COMBIN(23,1)/(COMBIN(312,2)*COMBIN(310,1)*COMBIN(309,1))*2
This Prob = 0,21665%

This 0.21665% is already included in the above 4,748945% for Player BJ and 4,748945% Dealer BJ, so it must be deducted because of Double counting.

So the Prob of Player BJ or Dealer BJ (meaning either or both) = 4,748945% + 4,748945% - 0.21665% = 9,28125%

But we need to Calculate he Prob of Player BJ or Dealer BJ (But not both), so we need to deduct this again to get:

Player and Dealer BJ: 0.21665% with Payout +21
Player or Dealer BJ (but not both)=9,28125% - 0.21665% = 9,06460% with Payout +8.5
No BJ = 1 - the above = 90,71875% with Payout -1

The sumproduct of the above is 9,12002%

So where is my mistake?

Quote: Boney526

IDK maybe I'm just really confused, but I'm 99% sure that the odds of any blackjack out of two hands (player or dealer) has to be more than twice the odds of a single hand getting a BJ, b/c of the effect of removal of cards that don't make up a BJ. A win of 8.5 assumes that only one hand received a BJ.... the other did not.

I think you are confused.
The odds of Player getting a BJ is 4,74895%. (or if you diagree with thi snumber call it X%). The dealer probability is exactly the same.
Some times when the Player gets BJ, The Dealer will also get a BJ.
So the Probability of Player or Dealer BJ (either or Both) = Player BJ + (Dealer BJ excluding those cases where Player has already BJ)
This (Dealer BJ excluding those cases where Player has already BJ) will always be a smaller number.
So the Probability of Player or Dealer BJ (either or Both) will always be smaller than Twice the Prob of Player BJ.

Think Venn Diagram with the overlapping region as Both Dealer and Player having BJs.

Quote: AceTwo

The probability of Player BJ in 6 decks is Combin(96,1)*combin(24,1) / Combin(312,2). That is 2304/48516 = 4,74895%
The Dealer BJ Probability is exactly the same = 4,74895%
Upto here the calculations are pretty straightforward and cannot be disputed.

The probability of Both Player and Dealer BJ are = { (Prob of Player BJ) * combin(95,1)*combin(23,1) / Combin(310,1)*combin(309,1) } * 2 OR
PROB = COMBIN(96,1)*COMBIN(24,1)*COMBIN(95,1)*COMBIN(23,1)/(COMBIN(312,2)*COMBIN(310,1)*COMBIN(309,1))*2
This Prob = 0,21665%

This 0.21665% is already included in the above 4,748945% for Player BJ and 4,748945% Dealer BJ, so it must be deducted because of Double counting.

So the Prob of Player BJ or Dealer BJ (meaning either or both) = 4,748945% + 4,748945% - 0.21665% = 9,28125%

But we need to Calculate he Prob of Player BJ or Dealer BJ (But not both), so we need to deduct this again to get:

Player and Dealer BJ: 0.21665% with Payout +21
Player or Dealer BJ (but not both)=9,28125% - 0.21665% = 9,06460% with Payout +8.5
No BJ = 1 - the above = 90,71875% with Payout -1

The sumproduct of the above is 9,12002%

So where is my mistake?

For 6 Decks the answer can be found using 312 instead of 416 cards... THus,

Player BJ = 312C2 / (24*96) = .0474895
Dealer BJ = 310C2 / (23*95) = .0456206 (This presumes the Player has a BJ... as in my previous post the possibilities of Aces and 10-values in the Player hand has to be accounted, and how often they occur to realize the OVERALL chance for a Dealer BJ, just convert from 8-decks to 6 when doing out the math)
When the Player DOES NOT have an Ace or 10-Value in the hand the chance of a Dealer BJ is 310C2 / (24*96) = .048105

By the above two figures for 6 Decks, the Chance of a BJ-Tie are .0474895*.0456206 = .0021665 or about 1 in 461.6

Quote: AceTwo

The probability of Player BJ in 6 decks is Combin(96,1)*combin(24,1) / Combin(312,2). That is 2304/48516 = 4,74895%
The Dealer BJ Probability is exactly the same = 4,74895%
Upto here the calculations are pretty straightforward and cannot be disputed.

The probability of Both Player and Dealer BJ are = { (Prob of Player BJ) * combin(95,1)*combin(23,1) / Combin(310,1)*combin(309,1) } * 2 OR
PROB = COMBIN(96,1)*COMBIN(24,1)*COMBIN(95,1)*COMBIN(23,1)/(COMBIN(312,2)*COMBIN(310,1)*COMBIN(309,1))*2
This Prob = 0,21665%

This 0.21665% is already included in the above 4,748945% for Player BJ and 4,748945% Dealer BJ, so it must be deducted because of Double counting.

So the Prob of Player BJ or Dealer BJ (meaning either or both) = 4,748945% + 4,748945% - 0.21665% = 9,28125%

But we need to Calculate he Prob of Player BJ or Dealer BJ (But not both), so we need to deduct this again to get:

Player and Dealer BJ: 0.21665% with Payout +21
Player or Dealer BJ (but not both)=9,28125% - 0.21665% = 9,06460% with Payout +8.5
No BJ = 1 - the above = 90,71875% with Payout -1

The sumproduct of the above is 9,12002%

So where is my mistake?

Why did you subtract .21665% twice?

Visualise a Venn Diagram with 2 Overlapping circles.
Circle A has are of 4,74895.
Circle B is identical and has area of 4,74895.
The overlapping area is 0.21665.

To find the are of Circle A + Circle B (excluding the overlapping area) = Area Circle A + Area Circle b - 2 x Overlapping area.
Becuase the ovrlapping area is included in both circle A and circle B area.

It is exactly the same as the above analsyis. BJ on both is included in Player BJ and in Dealer BJ so it must be deducted twice.

Quote: AceTwo

Visualise a Venn Diagram with 2 Overlapping circles.
Circle A has are of 4,74895.
Circle B is identical and has area of 4,74895.
The overlapping area is 0.21665.

To find the are of Circle A + Circle B (excluding the overlapping area) = Area Circle A + Area Circle b - 2 x Overlapping area.
Becuase the ovrlapping area is included in both circle A and circle B area.

It is exactly the same as the above analsyis. BJ on both is included in Player BJ and in Dealer BJ so it must be deducted twice.

I gotcha, thanks.