What is the general consensus on Side Bets (ex. perfect pair)?
Whether you won or lost the last hand has no bearing on how the next one will play out. If you are referring to a streak side bet then it would depend on the exact payouts and win requirements but it is unlikely the modest increase in % chance of winning the next hand from a higher count would make the bet +EV.
Last edited by Meistro123; 10-14-2017 at 07:17 PM.
When the count is high that doesn't change your Win Rate very much. So when the count is super high, you're not a big favorite to win the hand (i.e. to go on a winning streak). The money comes from doubles, splits, dealer busting slightly more often, index plays, etc, etc, etc.
When referring to side bets, in my opinion, it matters what level of a player you are. If you're a low roller, side bets are almost a must as they can easily double your hourly EV, with a safe spread, at a lower level table. However, since the mass majority of them have low betting limits (such as a $25 max) then they're not worth as much % in EV wise to someone playing at a higher level (amount). For example, if a side bet can make you $15/hour, to a low roller making $15/hour that's double the EV... however to someone making $100/hour, that's only a 15% boost (not saying it's not worth it, etc, just showing an example of importance). Most side bets are worthless, but some can be counted using your regular blackjack count. Tons of info out there on these...
Don't think you have a winning game; know you have a winning game.
Progressions are poor performers in rare event payouts. Too long between wins. Plus adding a luck factor into bet size when you hit a big payout just increases variance. Just say no to side bets until it is and advantage bet. Watch out, some side bets are used to identify advantage play.
Perfect Pairs was invented by an Australian couple who made money licencing it to local casinos. One of its features is that the House Edge decreases as the number of decks increases so if you theoretically played with an infinite deck then the player would have the advantage because there would be no card removal effect. In practice it cannot be beaten with any known count. The house edge for this side bet with a 6 deck Blackjack game paying 5:1 on Mixed Pairs, 10:1 on Coloured Pairs and 30:1 on Suited Pairs is as follows.
AnyPair: (4*6-1)/(52*6-1) = 23/311
Coloured Pair: (2*6-1)/(52*6-1) = 11/311
Suited Pair: (1*6-1)/(52*6-1) = 5/311, (5/311)*30 = 0.482
Mixed Pair: (23/311) - (11/311) = 12/311, (12/311)*5 = 0.193
Coloured, Unsuited Pair: (11/311) - (5/311) = 6/311, (6/311)*10 = 0.193
No Pair: 1- (23/311) = 288/311, (288/311)*-1 = -0.926
Totalling: 0.482 + 0.193 + 0.193 + (-0.926) = -0.058
Therefore, the House Edge on this side bet is 5.8%
Last edited by davethebuilder; 10-18-2017 at 06:44 AM.
Casino Enemy No.1
davethebuilder,
If you found this SB on an 8D game, and then noticed that all the Kings were missing, you'd have an edge of 18/383, or nearly 4.7%, on this SB.
By the way, even on an infinite-deck game, the house still has an edge of 2/52.
Hope this helps!
Dog Hand
Dog Hand,
I always learn something when I talk to you but I am confused about the house/player edge for the infinite deck model that you have quoted. Perhaps I should have said the player has an edge if they play with an infinite number of decks. The key lies in the Suited Pair calculation. At some point the calculation crosses over to a player advantage. So to calculate the number of decks that would be needed to create the player edge:
(x-1)/(52x-1)*30 = -0.193 - 0.193 + 0.926
x =15, so if a player was theoretically playing with 15 decks or more they would have the advantage?
Also, Kings have no more value than any other pair with this side bet and to remove a whole rank from play would change the initial calculations to being based on a 48 card deck(8*48 - 1 = 383) which still maintains a house edge based on adjustments to the above model which does not allow for suit and rank depletion, just totals? If the Kings were removed during play of a standard BJ game then the same applies. Still, I have noticed that this side bet predominates on CSM tables and I guess there is a reason for that.
Last edited by davethebuilder; 10-18-2017 at 05:08 PM.
Casino Enemy No.1
davethebuilder,
If we had an 8D game, but all 32 Kings were gone, then we'd have 384 cards remaining.
Say for the sake of argument that our first card is a Spade 8. The deck has 7 more Spade 8's, 8 Club 8's, and a total of 16 Heart & Diamond 8's.
Thus, our Suited Pair probability is 7/383, our Colored Pair probability is 8/383, and our Mixed Pair probability is 16/383: if you sum these probabilities, you get a total of 31/384 for our overall win probability, so our loss probability is 1 - 31/383, or 352/383.
The SB return is then:
(30*7 + 10*8 + 5*16 - 1*352)/383 =
(210 + 80 + 80 - 352)/383 =
(370 - 352)/383 =
18/383 = 0.047 = 4.7% (approximately).
The infinite deck is exactly equivalent to drawing a card from a single deck, then returning that card to the deck, shuffling, and drawing again: this is because, in an infinite deck, we always have a probability of exactly 1/52 of drawing a given card.
If again we assume our first card is a Spade 8, then our Suited Pair probability is 1/52, our Colored Pair probability is also 1/52, and our Mixed Pair probability is 2/52: if you sum these probabilities, you get a total of 4/52 (which equals 1/13) for our overall win probability, so our loss probability is 1 - 4/52, or 48/52.
The infinite deck SB return is then:
(30*1 + 10*1 + 5*2 - 1*48)/52 = +2/52 = +3.85%
(approximately)... oops, I guess I subtracted 52 instead of 48 when I did the math: that's what I get for answering before coffee this morning!
Hope this helps!
Dog Hand
Thanks, the calculations make sense now.
Just to make sure I'm understanding the Probability calcs in your example correctly:
Probability of Coloured Pair: (2*8 - 1)/(48*8 - 1) = 15/383
Probability of Coloured(Unsuited) Pair = Probability of a Coloured Pair - Probability of a Suited Pair = 8/383 which is then multiplied by the Payout ratio, in this case 10.
I guess a casino that paid 25:1 instead of 30:1 on the Suited Pair would have an edge of around 4.4%
Casino Enemy No.1
davethebuilder,
If we have this SB on a game with "n" regular decks, then the house edge can be calculated as follows.
Suited Pair prob= (n-1)/(52*n-1)
Colored Pair prob= n/(52*n-1)
Mixed Pair prob= 2*n/(52*n-1)
Summing gives the Win prob= (4*n-1)/(52*n-1)
Thus, Lose prob= 48*n/(52*n-1)
Therefore,
EV= (30*(n-1)+20*n+20*n-48*n)/(52*n-1)
Simplifying gives
EV= (2*n-30)/(52*n-1)
This confirms your earlier assertion that a 15-deck game would give a zero house edge for this side bet.
Hope this helps!
Dog Hand
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