"Optimal EV/Minimal Risk" - ALL you need to know about spreading to 2 or 3 hands

J

jee_pack

Well-Known Member
Hello everyone! Just spent 3 hours finishing this new chart, so I'm going to make a brief intro, then I'll just post the formula, and in the first reply after, I'll post all my math deductions for those who want to debate this or verify if my calculations are true or not. Although please don't waste your time reading the first reply (the math deductions) if you are not interrested in my formulas.

OKAY! So here we go, ever heard you could maximize your EV by spreading to more than 1 hand? Or you might have heard that you should not spread to more than 1 hand because it will cut down on the amount of rounds you'll play in positive TC.

Well... both of these statements are true, Optimal Betting (using the kelly forumla to find the Optimal bet) with 2 or 3 hands does increase the EV and does NOT increase Risk. Spreading to 2 or 3 hands could also REDUCE the risk while keeping the SAME EV. And yes! Spreading to 2 or 3 hands WILL cut down on the amount of rounds you'll play in good counts because you are eating up more cards. But neither of these 2 statements are exact unless they are both taken into account in 1 formula.

What I mean by that is that there is a threshold level at which the deck's faster depletion will outweight the EV increase, just like there's a threshold level at which the increased EV will outweight the deck's faster depletion.


So here is the chart:

HEADs-UP WITH THE DEALER: PLAY 1 HAND
1 OTHER PLAYER: PLAY 2 HANDS - MinBet: 66.67% - MaxBet: 72.68% (9% EV increase)
2 OTHER PLAYERS: PLAY 2 HANDS - MinBet: 65.50% - MaxBet: 72.68% (16% EV increase)
3 OTHER PLAYERS: PLAY 3 HANDS - MinBet: 46.67% - MaxBet: 57.08% (22% EV increase)
4 OTHER PLAYERS: PLAY 3 HANDS - MinBet: 44.44% - MaxBet: 57.08% (28% EV increase)
5 OTHER PLAYERS: PLAY 2 HANDS - MinBet: 57.15% - MaxBet: 72.68% (27% EV increase)

Only exceptions (if you want to minimize the Risk AS MUCH AS POSSIBLE):
1 OTHER PLAYER: PLAY 3 HANDS - MinBet: 55.56%
2 OTHER PLAYERS: PLAY 3 HANDS - MinBet: 50.00%

You can always bet ANYWHERE between the MinBet and the MaxBet. The MaxBet will maximize the EV and the MinBet will minimize the risk. If you bet the MAXBET, your risk will not be lowered, it will remain the same as if you were playing only 1 hand. If you bet the MINBET, the EV will not be changed, but as soon as you are betting over the MINBET, you start improving the EV and unless you reach the MAXBET, you will still be lowering the risk to a certain extent.

PS: the calculated EV increase is only accurate if you bet the MaxBet. Also, it's important to understand that the 27% EV increase you get from spreading to 2 hands when playing with 5 other players still gives you a much smaller EV than if you were heads-up against the dealer. In fact the less players at the table, the higher your EV, no matter how much hands you spread to.


Any questions or comments?
~Jee~
 
Last edited by a moderator:
J

jee_pack

Well-Known Member
Math deductions and draft for all the calculations

Kelly formula for the optimal bet when in an advantage situation:

bet = bankroll x edge / variance

With this formula, we can find the optimal bet for 1, 2, or 3 hands.

Spreading to more than 1 hand will not increase the Risk as long as you use the kelly formula to

find the correct bet amount for each hand.

The variance in BJ is (1.33 + (0.5 x number of extra hands you play))

So lets say we use 100$ for the 1 hand bet.

1 hand: bankroll x edge / 1.33 = 100$
2 hands: 100 x 1.33 / 1.83 = 72.68$ per hand (total per round = 2 x 72.68$ = 145.36$)
3 hands: 100 x 1.33 / 2.33 = 57.08$ per hand (total per round = 3 x 57.08$ = 171.24$)


TBP = TRP x TBPR
_TBP: Total bets placed:
_TRP: Total rounds played
_TBPR: Total bet per round

TRP = THD / THPR
_TRP: Total rounds played
_THD: Total hands dealt
_THPR: Total hands per round

TBP = (THD / THPR) x TBPR

1 hand_
_0 other player hands: TBP = (THD/2) x 100 = 50 x THD
_1 other player hand: TBP = (THD/3) x 100 = 33.33 x THD
_2 other player hands: TBP = (THD/4) x 100 = 25 x THD
_3 other player hands: TBP = (THD/5) x 100 = 20 x THD
_4 other player hand: TBP = (THD/6) x 100 = 16.67 x THD
_5 other player hands: TBP = (THD/7) x 100 = 14.29 x THD

2 hands_
_0 other player hands: TBP = (THD/3) x 145.36 = 48.45 x THD
_1 other player hand: TBP = (THD/4) x 145.36 = 36.34 x THD
_2 other player hands: TBP = (THD/5) x 145.36 = 29.07 x THD
_3 other player hands: TBP = (THD/6) x 145.36 = 24.23 x THD
_4 other player hand: TBP = (THD/7) x 145.36 = 20.77 x THD
_5 other player hands: TBP = (THD/8) x 145.36 = 18.17 x THD

3 hands_
_0 other player hands: TBP = (THD/4) x 171.24 = 42.81 x THD
_1 other player hand: TBP = (THD/5) x 171.24 = 34.25 x THD
_2 other player hands: TBP = (THD/6) x 171.24 = 28.54 x THD
_3 other player hands: TBP = (THD/7) x 171.24 = 24.46 x THD
_4 other player hand: TBP = (THD/8) x 171.24 = 21.41 x THD


Now if we compare these results.

_0 other player hands:
1 hand: 100 % EV
2 hands: 97 % EV
3 hands: 86 % EV

_1 other player hands:
1 hand: 100 % EV
2 hands: 109 % EV
3 hands: 102 % EV

_2 other player hands:
1 hand: 100 % EV
2 hands: 116 % EV
3 hands: 114 % EV

_3 other player hands:
1 hand: 100 % EV
2 hands: 121 % EV
3 hands: 122 % EV

_4 other player hands:
1 hand: 100 % EV
2 hands: 125 % EV
3 hands: 128 % EV

_5 other player hands:
1 hand: 100 % EV
2 hands: 127 % EV


Minimal Risk Bet:
1 hand: 100$
2 hand: 50$
3 hands: 33.33$

The risk is minimized because your are placing less of your bankroll on each hand and EV is kept

the same because the total bets per round is still 100$.

Playing the 50$ bet times 2 or 33$ bet times 3 will decrease the EV in all situations because

while betting the same amount per round, you are playing less rounds since you are playing more

hands. So to find the real Minimal Risk Bet, we need to find when a 2 hand's bet gives the same

EV as a 1 hand bet, and we need to find when a 3 hand's bet's EV equals the EV of a 1 hand bet.

And this btw has to be done only in the situations where playing more than 1 hand has the

potential of increasing the EV.

If the EV is equal, the TBP (total bets played) are the same.

TBP playing 1 hand = TBP playing 2 hands (and the same for playing 3 hands)

So lets look at this again:

1 hand_
_0 other player hands: TBP = (THD/2) x 100 = 50 x THD
_1 other player hand: TBP = (THD/3) x 100 = 33.33 x THD
_2 other player hands: TBP = (THD/4) x 100 = 25 x THD
_3 other player hands: TBP = (THD/5) x 100 = 20 x THD
_4 other player hand: TBP = (THD/6) x 100 = 16.67 x THD
_5 other player hands: TBP = (THD/7) x 100 = 14.29 x THD

2 hands_
_0 other player hands: TBP = (THD/3) x 145.36 = 48.45 x THD
_1 other player hand: TBP = (THD/4) x 145.36 = 36.34 x THD
_2 other player hands: TBP = (THD/5) x 145.36 = 29.07 x THD
_3 other player hands: TBP = (THD/6) x 145.36 = 24.23 x THD
_4 other player hand: TBP = (THD/7) x 145.36 = 20.77 x THD
_5 other player hands: TBP = (THD/8) x 145.36 = 18.17 x THD

3 hands_
_0 other player hands: TBP = (THD/4) x 171.24 = 42.81 x THD
_1 other player hand: TBP = (THD/5) x 171.24 = 34.25 x THD
_2 other player hands: TBP = (THD/6) x 171.24 = 28.54 x THD
_3 other player hands: TBP = (THD/7) x 171.24 = 24.46 x THD
_4 other player hand: TBP = (THD/8) x 171.24 = 21.41 x THD


So if we can isolate the TOTAL BET PER ROUND on the right side, we found our Minimal Risk Bet.

(THD / THPR) x TBPR ---> for 1 hand = (THD / THPR) x TBPR ---> for more than one hand

_1 other player hand:
2 hands: (THD / 3) x 100 = (THD / 4) x TBPR-2
3 hands: (THD / 3) x 100 = (THD / 5) x TBPR-2

_2 other player hands:
2 hands: (THD / 4) x 100 = (THD / 5) x TBPR-2
3 hands: (THD / 4) x 100 = (THD / 6) x TBPR-2

_3 other player hands:
2 hands: (THD / 5) x 100 = (THD / 6) x TBPR-2
3 hands: (THD / 5) x 100 = (THD / 7) x TBPR-2

_4 other player hands:
2 hands: (THD / 6) x 100 = (THD / 7) x TBPR-2
3 hands: (THD / 6) x 100 = (THD / 8) x TBPR-2

_5 other player hands:
2 hands: (THD / 7) x 100 = (THD / 8) x TBPR-2
1 other player hand at the table:

Now isolating TBPR:

_1 other player hand:
2 hands: TBPR = 133.33$
3 hands: TBPR = 166.67$

_2 other player hands:
2 hands: TBPR = 125$
3 hands: TBPR = 150$

_3 other player hands:
2 hands: TBPR = 120$
3 hands: TBPR = 140$

_4 other player hands:
2 hands: TBPR = 116.67$
3 hands: TBPR = 133.33$

_5 other player hands:
2 hands: TBPR = 114.29$


Now calculating the bet for each hand:

_1 other player hand:
2 hands: TBPR = 133.33$ / 2 = 66.67$ or 66.67% of the 1 hand bet
3 hands: TBPR = 166.67$ / 3 = 55.56$ or 55.56% of the 1 hand bet

_2 other player hands:
2 hands: TBPR = 125$ / 2 = 65.50$ or 65.50% of the 1 hand bet
3 hands: TBPR = 150$ / 3 = 50.00$ or 50.00% of the 1 hand bet

_3 other player hands:
2 hands: TBPR = 120$ / 2 = 60.00$ or 60.00% of the 1 hand bet
3 hands: TBPR = 140$ / 3 = 46.67$ or 46.67% of the 1 hand bet

_4 other player hands:
2 hands: TBPR = 116.67$ / 2 = 58.34$ or 58.34% of the 1 hand bet
3 hands: TBPR = 133.33$ / 3 = 44.44$ or 44.44% of the 1 hand bet

_5 other player hands:
2 hands: TBPR = 114.29$ / 2 = 57.15$ or 57.15% of the 1 hand bet


And now you have the 2 thresholds percentage you needed for each possible situation:

MIN BET (FOR MINIMAL RISK):

_1 other player hand:
2 hands: 66.67%
3 hands: 55.56%

_2 other player hands:
2 hands: 65.50%
3 hands: 50.00%

_3 other player hands:
2 hands: 60.00%
3 hands: 46.67%

_4 other player hands:
2 hands: 58.34%
3 hands: 44.44%

_5 other player hands:
2 hands: 57.15%

And Max Bet for Optimal EV:
72.68% x 2 hands
57.08% x 3 hands


"Optimal EV / Minimal Risk" hand spread beting formula (spreading from 1 to 3 hands):

Heads up: Play only 1 hand

1 Other Player:
_MinBet
__(2 hands) = 66.67%
__(3 hands) = 55.56%
_MaxBet
__(2 hands) = 72.68%
__(3 hands) = 57.08%

2 Other Players:
_MinBet
__(2 hands) = 65.50%
__(3 hands) = 50.00%
_MaxBet
__(2 hands) = 72.68%
__(3 hands) = 57.08%

3 Other Players:
_MinBet
__(2 hands) = 60.00%
__(3 hands) = 46.67%
_MaxBet
__(2 hands) = 72.68%
__(3 hands) = 57.08%

4 Other Players:
_MinBet
__(2 hands) = 58.34%
__(3 hands) = 44.44%
_MaxBet
__(2 hands) = 72.68%
__(3 hands) = 57.08%

5 Other Players:
_MinBet
__(2 hands) = 57.15%
_MaxBet
__(2 hands) = 72.68%


("x%" means "bet x% of the normal 1 hand bet on each hand")
(assuming other players play 1 hand each)



The following #s are the percentages of the "playing 1 hand EV" that the "playing with more than

1 hand EV" represents. (this is of course, taking into account the depletion of the cards). The

OPTIMAL CHOICE is the one with "######" next to it.

_0 other player hands:
1 hand: 100 % EV ######
2 hands: 97 % EV
3 hands: 86 % EV

_1 other player hands:
1 hand: 100 % EV
2 hands: 109 % EV ######
3 hands: 102 % EV

_2 other player hands:
1 hand: 100 % EV
2 hands: 116 % EV ######
3 hands: 114 % EV

_3 other player hands:
1 hand: 100 % EV
2 hands: 121 % EV
3 hands: 122 % EV ######

_4 other player hands:
1 hand: 100 % EV
2 hands: 125 % EV
3 hands: 128 % EV ######

_5 other player hands:
1 hand: 100 % EV
2 hands: 127 % EV ######
 
J

jee_pack

Well-Known Member
good morning everyone! So I finished the calculations last night, but I didn't take time to explain the theory behind what I did for those of you who want to know without reading all the equations...

So it was fairly simple. First thing I assumed is that there was a number of hands "x" left to be played before THE TC CHANGES or before the SHOE ends. So assuming you are playing with an edge, lets say you are at TC+4, the calculations are made assuming that the rest of the game will be played at +4. This in my head can give us the optimal move in the present time where there is no way of knowing if the TC will go up, down, or stay the same after the next round. But the TC does not always remain the same! Right. But if you look at the maths of it, you could play 3 hands at +4 to then find the count going to +3. Or you could find that the TC was going to jump to +5. Either way, playing 3 hands at +4 has the same chance of being a bad move as a good move assuming the TC will only remain +4 for the next round. But, if the TC remains at +4, then your move was optimal. So this means, you will get times where the 3 hand spread was not optimal which will even out with lucky times where it was (in the case where the TC drops right after you played your 3 hands). And after those 2 situations cancelling eachother out, you are left to add to your EV all the optimal moves you played when the TC remained the same.

So with these assumptions made, you can use algebra to calculate how much money you will be betting in total BEFORE THE TC CHANGEs or THE SHOE END. Lets say you are at the table with 2 other players. You say Hands per round = (1 for the dealer, lets say 2 for me, and 2 for the others) = 5.
So you do (x = hands left in Shoe or before tc changes) x/5 = rounds left to be played. Multiple this by the amount of $ you bet per round. (with 2 hands, the optimal bet is about 73% of your unit bet times 2 hands (using the kelly formula), so if the one hand bet is "y". Then you total bets for playing 2 hands with 2 other players will be (x/5) * 2(0.73y) = 0.292xy (x = # of rounds left and y = normal 1 hand bet). Now you have to calculate what would be your total bets places by playing only 1 hand. So total hands per round = (1 for the dealer, 1 for me and 2 for the others) = 4. The x/4 = rounds left. And you are betting the 1 hand bet so bet per round = y. Total bets playing 1 hand will equal x/4 * y = xy/4. or if you convert 1/4 to decimal, y = 0.25xy

Now you are left with comparing 0.292xy with 0.25xy, you see your total amount of bets placed while at a certain postitive TC will be exactly 17% higher if you spread to 2 hands. If we didn't take into account the fact that we are eating up more cards, we would have said the total bets are increased by 45%, but after using algebra, we see that the fact that more cards will be eaten up reduces this EV increase to 17%. With this done you see that the EV increase outweighs the hand depletion rate.

Next: we now know that in this situation, you max bet will be 117% of the 1 hand bet divided in 2 hands evenly, so how do we find out the minimum be that will give the the same EV as playing 1 hand while minimizing risk? I'm not going to show all the maths because this part was the longest, but after all the algebra has been simplified (starting with an equation that assumes the Outcome result in total bets with 2 hands will be the exact output result as playing with 1 hand), then we just a simple division and multiplication with the # of hands per round, and I isolate the Bet Per Round (of playing 2 hands) and it gives us what to bet on each of the 2 hands in order that if you were to play both 1 hand and 2 hands the entire shoe, in both choices, you will still have bet the same amount of cash on a certain TC....

That's pretty much it....
 
Last edited:
Sonny

Sonny

Well-Known Member
Nice work. And your conclusion seems to agree with Don Schlesinger's analysis, which is always a good sign. :) He disagrees with you when there are 5 other players (he says to play 3 hands) but that is probably a minor point.

-Sonny-
 
J

jee_pack

Well-Known Member
Sonny said:
Nice work. And your conclusion seems to agree with Don Schlesinger's analysis, which is always a good sign. :) He disagrees with you when there are 5 other players (he says to play 3 hands) but that is probably a minor point.

-Sonny-
Click to expand...
thanks a lot for validating these results for me

Well I said 2 hands with 5 other people cuse I thought there was only 7 available spots to play from at the table (2+5). Does that mean you could actually fit 8 hands somewhere on a standard blackjack table?
 
Last edited:
You must log in or register to reply here.
Top