Insurance Calculation

Hello,

I am reading K-O Blackjack, and I have a question as regards the calculation of the disadvantage of taking insurance assuming that one only follows basic strategy. The authors provide the following expected outcome: (4/13)(2) + (9/13)(-1) = -(1/13). The authors explained that 4/13 is the probability of the hole card being a 10-value card and 9/13 is the probability of the hole card not being a 10-value card. My question is what do the numbers 2 and -1 represent?

Furthermore, is "expected outcome" another way of saying "expected value"?

Thank you.

Quote: Plato

Hello,

I am reading K-O Blackjack, and I have a question as regards the calculation of the disadvantage of taking insurance assuming that one only follows basic strategy. The authors provide the following expected outcome: (4/13)(2) + (9/13)(-1) = -(1/13). The authors explained that 4/13 is the probability of the hole card being a 10-value card and 9/13 is the probability of the hole card not being a 10-value card. My question is what do the numbers 2 and -1 represent?

Furthermore, is "expected outcome" another way of saying "expected value"?

Thank you.

Expected outcome probably means expected value....although I've never heard of it before.

The 2 is the payoff and the -1 is how much you lose.

4/13'ths of the time you win 2 units and 9/13'ths of the time you lose 1 unit.

Quote: RS

The 2 is the payoff and the -1 is how much you lose.

Do these numbers derive from the fact that insurance pays 2 to 1?

Quote: Plato

Do these numbers derive from the fact that insurance pays 2 to 1?

Yes, the first number does, i.e., the 2 represents the payoff of 2 times the insurance bet. The -1 represents the loss of your insurance bet.

Thank you for your explanations.

I have another question as regards the mathematics of "even money". The authors of K-O Blackjack write the following:

"Let's say you've made a $100 wager, are dealt a natural, and the dealer has an Ace showing. If you take even money, you'll wind up $100 richer. On the other hand, if you insure the natural for $50, one of two things can happen: 1) the dealer has a natural, in which case the naturals push, but you win $100 on the insurance bet; 2) the dealer doesn't have a natural, in which case your natural wins $150, but you lose $50 on the insurance bet. In either case, you wind up $100 ahead.

My confusion arises from the fact that they write "on the other hand". Did they merely intend to elaborate on their first sentence, or are there really 2 opposed situations between the first sentence and the two examples that they give?

Furthermore, they continue by explaining why it is not profitable, on average, to take even money:

"The expected outcome for insuring a natural is to profit $1 for every $1 wagered on the main bet [what do they mean by this?]. But when we don't insure our natural, our expected outcome is ... (4/13)(0) + (9/13)(1.5) = (27/26)."

From what I understand from the equation, the 0 represents the situation in which the dealer does not have a 10-value hole card and the 1.5 is the payoff that the player receives where the dealer has any other value card, thus not a natural. Am I right? Am I missing something?

1) When they say "on the other hand" they're simply pointing out the other situation that could happen. There's 2 situations: one where you don't insure your blackjack, and "on the other hand" where you do. That's all it means, you're overthinking it.

2) If you take even money, it's the same as insuring... so if you always insure your blackjack you're always going to get paid "even money" which is 1 to 1 (or $1 to $1). If you always insure, you're never going to win more than $1 to every $1 you bet. I think you're over thinking it again.

3) The 0 represents your payout. In the equation above the (4/13)(0) is the EV of the situation the dealer DOES have blackjack and what you're going to get as a result. 4 out of 13 times (King, Queen, Jack, and Ten) the dealer will have blackjack which will result in you losing your bet, i.e. getting a 0 payout. 9/13 times (all the other cards) the dealer will not have blackjack, resulting in you getting a 3:2 payout, or 1.5 units for every 1 unit bet.

You're quite overthinking insurance. It's a sucker bet unless you're counting cards. So NEVER take insurance unless you're counting cards, and then take it if the TC is +3 or greater. That's that =).

I always take even money on BJ vs A, regardless of the count. The difference for that hand is pretty small, I think, as the variance overcomes the EV.

Value is $150 * 9/13 = $103.84 for not insuring, vs guaranteed $100. The rarity that is this hand (BJ is every 21 hands times a 1/13 for a dealer's ace...every 275 hands you get this match up?).


Technically, you should never insure for a BS player or never insure below a TC +3 (or your counts equivalent) to maximize EV.

So nearly $4 per blackjack where the dealer also shows an ace. A few times per session perhaps? That could get a bit costly.

I'd rather NEVER insure a blackjack if I were going to take one of the two extremes. Taking insurance at the right time is more of a red flag than not taking it.