Funny Story - Insurance Vs. "Even Money"

[MJGolf]

Thank you. I've read explanations before and it slipped by me. I just accepted it as true and moved on. Your explanation really made sense and now I understand it(although I'm a bit ashamed that I didn't work it out on my own).

Although the math should be basic to us, don't feel bad in the least. As this thread first started, you have DEALERS who think that "even money" on a BJ is DIFFERENT than insurance and will TELL you that. It's because for some reason, and this is just opinion or "limited" personal experience, that they think it's taking an "automatic win" v. taking a chance on the dealer's BJ or some other convoluted reasoning. It's amazing. Somewhere along their lines of ALWAYS claiming to know basic strategy...............

[Renzey]

Although the math should be basic to us, don't feel bad in the least. You have DEALERS who think that "even money" on a BJ is DIFFERENT than insurance and will TELL you so.

That belief is pervasive among blackjack dealers. In a pitch game, having a face-down blackjack against a dealer's Ace, throw out a very small token Insurance bet and say, "For less." Then watch what happens after the Insurance bet has been settled when you flip up your natural. Often, mass confusion -- particularly if the dealer has blackjack.

Incidentally, for those who can't bear to run the risk of not being paid on their blackjacks, insuring for far less is a superior play to taking even money.
I.E., You take $5 worth of Insurance on a $30 blackjack. The worst thing that can happen is you win ten bucks. Yet, your overall average payback is $30.78 rather than $30 for even money. Declining Insurance altogether yields $31.16.

[TeamMoney]

You may get push back from the pit. By procedure, you must show your blackjack prior to to the completion of the insurance bet offering in order to get paid even money. Additionally, having an insurance bet out there is still a bet against the dealer hand and it is at risk.

So the outcomes could be:

1. you get even money on your blackjack and also paid on your insurance bet.
2. You get even money on your blackjack and lose your insurance bet.

Now youre not understanding what im saying as well as bigdaddy. Im talking about two separate hands. Getting paid even money on my blackjack and then insuring my OTHER hand. Im aware you cant insure the same hand you get even money on. And my point is by taking even money on the blackjack hand and winning the insurance bet on the non-insurance hand, it beats out the insuring of both hands, would it not?

[Renzey]

Im talking about two separate hands. Getting paid even money on my blackjack and then insuring my OTHER hand. My point is by taking even money on the blackjack hand and winning the insurance bet on the non-insurance hand, it beats out the insuring of both hands, would it not?

Yes, taking even money on B/J and NOT insuring hand number two beats out insuring BOTH hands. But DECLINING even money and NOT insuring hand number two beats them both out. That's because Insurance on ANY hand pays 2-to-1 on a bet that is basically 2.25-to-1 against winning. ALL hands do better without it -- except in high-ten-density shoes.

[Bigdaddy]

Now youre not understanding what im saying as well as bigdaddy. Im talking about two separate hands. Getting paid even money on my blackjack and then insuring my OTHER hand. Im aware you cant insure the same hand you get even money on. And my point is by taking even money on the blackjack hand and winning the insurance bet on the non-insurance hand, it beats out the insuring of both hands, would it not?

Sorry - still not sure that I fully understand. Are you saying that you would take even money on the blackjack and not insure the other hand? If so, then I agree with Renzey's response in post #30. If that is not what you are referring to, then the only way you're going to get through to me on your point is to provide an example with numbers.

[TeamMoney]

Sorry - still not sure that I fully understand. Are you saying that you would take even money on the blackjack and not insure the other hand? If so, then I agree with Renzey's response in post #30. If that is not what you are referring to, then the only way you're going to get through to me on your point is to provide an example with numbers.

NO. Im saying take even money on the blackjack hand, and insure the other non-blackjack hand. If you win the insurance on the non blackjack hand and take even money on your blackjack hand, you actually MAKE money. If you were to just insure both hands from the start you would just break even.

Only way I can see im wrong, is how the payout works when you insure both hands and have a blackjack there, do you win insurance on both and get paid 3:2 on the blackjack?

[TeamMoney]

Yes, taking even money on B/J and NOT insuring hand number two beats out insuring BOTH hands. But DECLINING even money and NOT insuring hand number two beats them both out. That's because Insurance on ANY hand pays 2-to-1 on a bet that is basically 2.25-to-1 against winning. ALL hands do better without it -- except in high-ten-density shoes.

You're not understanding what im saying. Im saying insure the non blackjack hand and take even money on the other hand. That would be the most optimal scenario in all the scenarios would it not? Most people say to just insure both of them? How do the payouts work if you insure both of them and win? Do you win both insurance bets and get paid 3:2 on the natural?

[Gronbog]

Only way I can see im wrong, is how the payout works when you insure both hands and have a blackjack there, do you win insurance on both and get paid 3:2 on the blackjack?

WInning insurance means the dealer had blackjack, so your blackjack pushes.

You keep saying that you would come out ahead by taking even money on the blackjack instead of insuring it, but you haven't backed that up. Please show us your numbers.

[Stealth]

Only way I can see im wrong, is how the payout works when you insure both hands and have a blackjack there, do you win insurance on both and get paid 3:2 on the blackjack?

No, You will show the dealer your hand with the blackjack prior to the insurance bet and announce you want even money on that hand. While I am 99% sure that the taking of even money removes the ability to take insurance on that hand.

You can then decide if you want to insure the other hand (that you know is not blackjack). If the dealer has blackjack you win the insurance bet and lose the hand bet.

[Renzey]

You're not understanding what im saying. Im saying insure the non blackjack hand and take even money on the other hand. That would be the most optimal scenario in all the scenarios would it not?

OK, now I understand what you want to do. The answer is, YOU ARE COMPLETELY WRONG! Now follow along to see this. Three basic scenarios are possible:
1) The dealer has blackjack (31 times out of 100).
2) The dealer does not have blackjack, but beats random hand number two (*44 times approx.).
3) The dealer does not have blackjack and is beaten by random hand number two (*25 times approx.).

If you take even money on a $10 blackjack and insure $10 random hand number two for $5;
Scenario #1) You win ten dollars 31 times.
Scenario #2) You lose five dollars 44 times.
Scenario #3) You win fifteen dollars 25 times.
In total, you win $465 -- or $4.65 per time.

If you decline both even money and Insurance;
Scenario #1) You lose ten dollars 31 times.
Scenario #2) You win five dollars 44 times.
Scenario #3) You win twenty-five dollars 25 times.
In total, you win $535 -- or $5.35 per time.

And the difference is all because you chose to make $10 worth of Insurance bets 100 times at a negative expectation of about 7% per time -- depending upon whether random hand number two contained any Tens.

*taken from table of expected values for all hands against an Ace for illustration purposes only. Actually, the difference between the two strategies will always be the same since the math simply nets out the difference for making two Insurance bets vs. not making them.

[TeamMoney]

OK, now I understand what you want to do. The answer is, YOU ARE COMPLETELY WRONG! Now follow along to see this. Three basic scenarios are possible:
1) The dealer has blackjack (31 times out of 100).
2) The dealer does not have blackjack, but beats random hand number two (*44 times approx.).
3) The dealer does not have blackjack and is beaten by random hand number two (*25 times approx.).

If you take even money on a $10 blackjack and insure $10 random hand number two for $5;
Scenario #1) You win ten dollars 31 times.
Scenario #2) You lose five dollars 44 times.
Scenario #3) You win fifteen dollars 25 times.
In total, you win $465 -- or $4.65 per time.

If you decline both even money and Insurance;
Scenario #1) You lose ten dollars 31 times.
Scenario #2) You win five dollars 44 times.
Scenario #3) You win twenty-five dollars 25 times.
In total, you win $535 -- or $5.35 per time.

And the difference is all because you chose to make $10 worth of Insurance bets 100 times at a negative expectation of about 7% per time -- depending upon whether random hand number two contained any Tens.

*taken from table of expected values for all hands against an Ace for illustration purposes only. Actually, the difference between the two strategies will always be the same since the math simply nets out the difference for making two Insurance bets vs. not making them.

You and others are still not understanding what I'm saying. I never mentioned or asked about if my hands would win outright after dealer not having insurance or whether or not I can insure a blackjack after taking even money. What I'm simply asking about is this:

I'm playing an 8 deck shoe at 2x100. I get a blackjack on one hand and a 15(for example) on the second hand. I take even money on the blackjack hand and then place a $50 insurance bet on the second hand. If I win insurance I break even on the second hand, but by taking even money on the blackjack, I now come out ahead on the overall round, whereas if I had just originally insures both hands at the onset and not taken even money, and proceeded to win the insurance bet, I would ONLY break even and not have a net profit for the round.

So my point is, by taking even money on the blackjack hand and insuring the other non-blackjack hand, in this case a 15, I would come out ahead.

Only way I can see that I'm wrong is if somehow after insuring both hands at the onset and not taking even money and winning the insurance, you get paid on insurance and on top of it get paid 3:2 on the blackjack, which I doubt is possible.

[Banjoclan]

Why do you keep insisting taking even money and insuring your BJ are two different things?


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[Gronbog]

whereas if I had just originally insures both hands at the onset and not taken even money, and proceeded to win the insurance bet, I would ONLY break even and not have a net profit for the round.

This is where your logic is flawed.

If you take insurance on both hands and the dealer has blackjack, you are correct that the overall result is a push on the non-blackjack hand. However for the blackjack hand, you get paid 2 x half your bet (i.e. even money) and your blackjack hand pushes leaving you with an even money profit on that hand. Exactly the same as if you had just taken even money in the first place.

Now if the dealer does not have blackjack, on your blackjack hand you lose your insurance bet, which is 1/2 of your original bet, but you immediately win 1.5 x your original bet for a profit of (1.5 - .5) = 1 x your original bet. Just as if you had just taken even money.

So, for the blackjack hand, taking even money is exactly the same as taking full insurance.