is this math correct

[Meistro123]

it is 1 and 1/4 decks into a six deck shoe. you have seen no aces.

the ratio of aces to non aces is 24:223 or 1:9.29.

you have a 9.71% chance of getting an ace.

YOu are playing at a casino that offers 10:1 insurance on a ten.

You have a 6.81% edge on this insurance bet.

[21forme]

YOu are playing at a casino that offers 10:1 insurance on a ten.

Sorry to sidetrack right off the top, but is there a name for this sidebet?

[Meistro123]

Yes, they call it insurance.

[stopgambling]

but if you only have 9.71% chance of catching an ace why would you have an edge? don't you need over 10% chance of getting an ace to be positive Ev?

[Freightman]

I looked at it a different way. Your percentages came out the same as mine.

You repeat the identical process 1000 times with a $10 sidebet. Your total wager is $10000.
You lose 902.9 times losing $9029.00. You win 97.1 times retaining your $10 sidebet plus win $9710.00.

Your gross is $681 on $10000 wagered, I.e you have $10681 versus original $10000.

[Freightman]

but if you only have 9.71% chance of catching an ace why would you have an edge? don't you need over 10% chance of getting an ace to be positive Ev?

Because you have to calculate 10-1 payoff versus 3-2

[Meistro123]

but if you only have 9.71% chance of catching an ace why would you have an edge?

The break-even point is 9.09%. If the odds were 9:1 you would need to win 1 time in 10 (10%) in order to break even.

For example, you could lose 9 times of 1 bet (-9) and win 1 time for a 9 payout (+9), and you would end up breaking even.

[DSchles]

it is 1 and 1/4 decks into a six deck shoe. you have seen no aces.

the ratio of aces to non aces is 24:223 or 1:9.29.

you have a 9.71% chance of getting an ace.

YOu are playing at a casino that offers 10:1 insurance on a ten.

You have a 6.81% edge on this insurance bet.

You have a 6.88% edge -- 17/247.

Don

[lij45o6]

Let:

A = # of Aces left
C = # of non-Aces left
D = SUM(A, C)
E_A = Expectation of Ace underneath

A = 24
C = 288
D = 312
E_A = 10

EV = E_A(A/D)-(C/D)

= 10(24/(312 - 1.25*52)) - ((312 - 24 - 1.25*52)/(312 - 1.25*52))

= (240-223)/247

= 17/247

=0.068825911

[James989]

let:

A = # of aces left
c = # of non-aces left
d = sum(a, c)
e_a = expectation of ace underneath

a = 24
c = 288
d = 312
e_a = 10

ev = e_a(a/d)-(c/d)

= 10(24/(312 - 1.25*52)) - ((312 - 24 - 1.25*52)/(312 - 1.25*52))

= (240-223)/247

= 17/247

=0.068825911

ev = (24(10)-223 )/ 247 = 0.06882591

[Freightman]

Let:

A = # of Aces left
C = # of non-Aces left
D = SUM(A, C)
E_A = Expectation of Ace underneath

A = 24
C = 288
D = 312
E_A = 10

EV = E_A(A/D)-(C/D)

= 10(24/(312 - 1.25*52)) - ((312 - 24 - 1.25*52)/(312 - 1.25*52))

= (240-223)/247


= 17/247

=0.068825911

Your numbers are wrong.
A=24
B=223
D=247
E/A=10

Remember, 1.25 decks have been played. What am I missing. I still come to 6.81%.

[stopgambling]

If the odds were 9:1 you would need to win 1 time in 10 (10%) in order to break even.

so 10 for 1 is what i was thinking ,right ?

[lij45o6]

Your numbers are wrong.
A=24
B=223
D=247
E/A=10

Remember, 1.25 decks have been played. What am I missing. I still come to 6.81%.

I did, in fact, account for the missing cards from the shoe.

A standard 6D shoe has 312 point-value cards. 24 of them are Aces. That leaves us with 288 non-Ace cards. The expectation for each can be compute as the following:

A = 24 * 10
N = 288 * -1

The sum of the magnitudes of A and N are SUM(|A|, |N|) = 312. The 312 standard point-value cards represented as 6 unique French decks.

Now, OP specified that 1.25 decks have been depleted with no Aces removed. That means we have a D value of 312 - 1.25 * 52 = 312 - 65 = 247

The EV(X) can now be computed as the ratio of the total Expectation of X divided by the remainder of the total non-Ace cards (240 - (247 - 24)) / (247) = 17/247 ~ 0.0688

I still get 6.88%

One can reconstruct the new Insurance payout by following the old Insurance function:

2* (T_n/(D_n)) - ((D_n - T_n)/(D_n))

Where T_n is the number of Tens left in a pack/shoe, D_n is the number of cards left in a pack/shoe.

For single deck:

2 * (16/52) - (36/52) = (32 - 36)/52 = -4/52 = -1/13 = -0.076923 or -7.69%

Just replace T_n with A_n, the 2 with 10, and recompute D_n to reflect the missing 1.25 decks from the 6D shoe.