Insurance Math

I know insurance is extremly bad bet, and it should never be taken, unless we count cards!

This question is pure math. It seems to me that house advantage is more than 7 percent when players take insurance.

To help make my point here is the quote from "Play BlackJack like Pros" by Kevin Blackwood.

"If you took insurance thrteen times at fifty bucks a pop, on average you'd win the bet four times (since there are four face cards for every thrteen cards) and lose the other nine times ( all the none face cards). Your wins would net you an extra $400, but your losses would total $450."

We all know that insurance pays 2:1 so if we win 4 out of thrteen times that means if we bet 50 bucks 4x50=200 and since insurance pays 2:1 that would be 200x2=400.

We also lose nine times out of 13 therefore 50x9=450

Well if we bet 100 buck that means that our insurance is $50. Yes we get 2:1 payoff when we win for insurance bet but also we lose our original bet so it is overall push. Those 7 percent would only work if there is no original bet on table than that is 7 percent , but when we have bet on the table that is push so that is lot more than 7 percent because we dont make anything with payoff of 2:1 because we lose our bet!

Well this is how I see it, and I am probably wrong so I would really appreciate your opinion!

my 2 cents

bj,

I won't go into any long explanations about how much of a percentage improvement insurance adds or subtracts, depending on how you use it because frankly I don't know. what I do know is that like you say, unless you count, it is a sucker bet for the basic player. also it is an individual bet and should not be looked at in any other way. whatever happens with your insurance wager does not affect your original bet so to call it insurance is really misleading. thats why it's more important in my opinion to understand why it's a sucker bet and then to understand what makes it a good bet when your count calls for you to take insurance. that can easily be explained with math. if thats your question let me know and I will be happy to share.

The insurance bet must be considered completely separately from the original bet, because the decision to take insurance is separate from the decision to make the original bet.

Your original bet is lost already if the dealer has the blackjack, so don't let it interfere with your analysis of the math of insurance.

Hopefully this makes sense.

Math

Yes I know that house has 7% advantage on insurance bet,{but that is only considering insurance bet and insurance pay off and posibility of dealer getting ten (4 in 13)}. I know the math of insurance bets but that is not what I am interested in.

That is only advantage on separate bet, but I am interested what is advantage that house gets from insurance bets overall on the game not just the separate bet.

I am just trying to solve that little thing that buggs me all day long, I gues I have a lots of time to waste since we all know that we shouldn't take insurance, well if anybody know theory and math (graphs, formulas, %, explanations, examples.....) of insurance bet I would really like to know!

Thanks!!!!

yeah....

.....you are right in books they always say that house advanatge on that particular bet is 7%. Nothing is mentioned what is advantage on game overall, because that is just separate bet.

Well thanks guys for answers anyway, now I can go to sleep!

If you want to know how much taking insurance every time would cost you, you can figure it roughly like this:

The dealer will have an Ace up 1/13th of the time. When he does, you'll bet half your original bet on insurance at -7.69%.

Half that (because you're betting half as much on insurance) is 1/13 * 1/2 * -7.69% = -0.30%

So, take the normal basic strategy house edge, and add another 0.30% for the effect of insuring every time.
For a typical six deck S17 game, the house edge would increase from 0.44% to 0.74%.

Thanks Ken

That's What I wanted to know!

Great explanation!!