xxrenegadexx
Well-Known Member
Does anyone know at what point(true count) or if at all it is the correct play to take the even money option for a blackjack?
even money option for blackjack
- Thread starter xxrenegadexx
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Even money is the same as insurance, so take it at the same TC you'd take insurance if you didn't have a blackjack.
Automatic Monkey
Banned
To nitpick, taking even money at a slightly lower count decreases variance, which makes it a more profitable play. I'll usually take even money if the insurance count is close, or if I have a big bet out and the count has dropped during the hand.
No, even money IS insurance.
Automatic Monkey
Banned
Yes, see above. Any decrease in variance is the same as being able to bet a higher unit with the same RoR, which means a higher win rate.
ihate17
Well-Known Member
Costs you near nothing
At a borderline count, the cost to you is insignificant in my opinion. I weight this situation with another factor, what has happened in that pit in the past. If I had recently taken even money in a higher count, now have a borderline count, I will always take it again. If the opposite is true, then I would decline.
ihate17
At a borderline count, the cost to you is insignificant in my opinion. I weight this situation with another factor, what has happened in that pit in the past. If I had recently taken even money in a higher count, now have a borderline count, I will always take it again. If the opposite is true, then I would decline.
ihate17
It doesn't matter if the variance decreases or not, that doesn't make the actual bet of even-money more valuable. I'm not talking about lifetime EV here, I'm talking about the insurance/even money bet EV.
Let's say me and you are sitting at the same table and we both bet 100 dollars. You get a blackjack and I get a 20. You take even money, and I insure my 20. That means we both have bet 50 dollars the dealer will have a BJ. Well, how does your 50 earn more money than my 50? It doesn't, and never will earn more than mine, because it's the same bet.
So, the EV of the ACTUAL BET of insurance or even money is always the same. One will not have a slightly higher EV in the longrun.
Automatic Monkey
Banned
All right, that's true, the individual bet will not. But in terms of a strategy decision for a card counter (who always has a finite bankroll) they are not equal.
Related to this is the fact that there are a lot of double and split index plays that do have a positive EV, but when actually applied to a counting strategy they have a negative value, because they increase your risk more than they increase your EV.
Yeah, I understand that one might be slightly better in the long run due to reasons relating to variance. I just think that insurance and even money are both the same thing. They are the same thing (insurance) just in different situations. So yeah, maybe even money is worth slighly more in the long run, but that doesn't mean it's different from insurance.
So my conclusion seems to be that even money is the same as insurance, but even-money is slightly more valuable because it decreases variance more than insurance because your hand is a BJ.
So, the only point to knowing this is that it might be better to take insurance at a lower count when you have a blackjack. Is it enough difference to matter? I don't think so.
It could be something like this. Take insurance at +3, but take even money at +2.95. Maybe taking even money at +2 is taking the idea too far. Someone would have to sim it to find out the true number. It would also depend on your bet size at the time too. Also, the increase in EV is probably miniscule. So... is it even worth it to try take this idea into effect?
We both have a $100 wagered.I have BJ and take even money. I have wagered $100 and won $100.You have a 20 and take insurance for an additional $50.Dealer has a BJ and you collect $100. But you ended up putting $150 at risk.The end payouts are the same,but the overall amount wagered is different.
Could this matter in the long run? I don't see how it could,but can it?
Automatic Monkey
Banned
It does make a difference, a small one, but a difference nonetheless. We are usually talking about a swing of 10-20 units here, and that represents the EV of a full evening of play. Here's a rule of thumb I'll pull out of my arse: take even money at one true count less than you would take insurance.
Not only that, most ploppies take even money on naturals so it gives us cover worth more than the variance reduction.
Snyder did a nice insurance cover article that demonstrates taking insurance at anything over your minimum bet costs you jack-diddly-squat compared to taking insurance at the right place and is excellent cover.
The overall amount wagered is still the same, because you bet 100 dollars to play the hand, and bet 50 on insurance. It's the same with the person who had the 20. Both players will have wagered 150 dollars.
So, no, the overall amount wagered is not the same. Maybe you meant the overall risked money is different, and that is where the difference lies. Since it's assumed you have a blackjack you can't lose any money on the hand, but I have a 20 so I can. That means the variance is not equal in both situations, and that is why there is a difference in value (not actual bet EV) in even money and insurance.
So the difference between even money and insurance is variance, and as AutoMonkey said, variance affects your winrate, but it doesn't affect the EV of the bet.
I'm not sure I understand what everyone is saying here but, without counting, it'll cost you a heck of a lot more to take insurance on all hands other than BJ compared to only taking even money on BJ.
Like always taking even money with a BJ will cost u less than 5% of what always taking even money or insurance will cost u every time the dealer has an Ace.
Like always taking even money with a BJ might be an OK cover play but always taking insurance on all other plays would be very costly as a cover play.
Like always taking even money with a BJ will cost u less than 5% of what always taking even money or insurance will cost u every time the dealer has an Ace.
Like always taking even money with a BJ might be an OK cover play but always taking insurance on all other plays would be very costly as a cover play.
While the bets do have the same EV, the EV varies with the count in a way that is not even, this is because the NEUTRAL cards are cards that do NOT form a blackjack, and neutral cards are not counted in a hi-lo count. The expected value of even money is dependant on the ratio of Neutral+Low vs High, the EV of insurance is dependant on two ratio's: Neutral+high vs Low and Neutral+Low vs High. Thus the EV's are dependant on the count in different ways!
So even though insurance and even money appear to have same EV, at different counts the EV changes!
This is getting pretty advanced but here is the explanation:
The reason is not variance. Variance has nothing to do with expected value (every roulette table bet has the save EV but different variance).
The EV's in general appear to be the same, because the even money bet is tied into our original bet, the insurance bet is completely independent.
Assume chance of blackjack is 4/13 (10 j q k), and non blackjack is 9/13 (23456789A), the EV for not taking even money (9/13*1.5 + 4/13*0)= 103.8%, for even money it is obviously 100%.
As for taking normal insurance, Well it is an extra bet, on top of our original bet. The EV is thus (4/13*2 + 9/13*-1) = -7.7%. Compared to no insurance. The EV is 0% since we neither win or loose, since we are not betting.
Now if 4/13 increases, and 9/13 decreases (as the count goes high), and you will find that -7.7% increases to 0% at the exact same time that 103.8% decreases to 100% (at 4.333/13 and 8.666/13), as expected. Thus it at first appears the count should effect the decisions for insurance/even money bets at the same time!
However this is not so, lets see why:
As the count changes 4/13 varies by an unknown amount (lets call it x), and 9/13 changes by an unknown amount (lets call it y), such at ((4+x)/(13))+((9-y)/(13)=13. With the relations ship between x and y being x+y=0.
But look at the EV calculations again. The EV for even money does not depend on x!
The change over point (when you take insurance) is at EV = (4+x)/(13)*2 + (9-y)/(13)*-1
while the change over for break even ends up being at EV = (9-y)/(13)*1.5.
But both of these EV's occur at the same count!
These change over points are different, thus the choice to take even money happens at a different count compared to taking insurance.
So even though insurance and even money appear to have same EV, at different counts the EV changes!
This is getting pretty advanced but here is the explanation:
The reason is not variance. Variance has nothing to do with expected value (every roulette table bet has the save EV but different variance).
The EV's in general appear to be the same, because the even money bet is tied into our original bet, the insurance bet is completely independent.
Assume chance of blackjack is 4/13 (10 j q k), and non blackjack is 9/13 (23456789A), the EV for not taking even money (9/13*1.5 + 4/13*0)= 103.8%, for even money it is obviously 100%.
As for taking normal insurance, Well it is an extra bet, on top of our original bet. The EV is thus (4/13*2 + 9/13*-1) = -7.7%. Compared to no insurance. The EV is 0% since we neither win or loose, since we are not betting.
Now if 4/13 increases, and 9/13 decreases (as the count goes high), and you will find that -7.7% increases to 0% at the exact same time that 103.8% decreases to 100% (at 4.333/13 and 8.666/13), as expected. Thus it at first appears the count should effect the decisions for insurance/even money bets at the same time!
However this is not so, lets see why:
As the count changes 4/13 varies by an unknown amount (lets call it x), and 9/13 changes by an unknown amount (lets call it y), such at ((4+x)/(13))+((9-y)/(13)=13. With the relations ship between x and y being x+y=0.
But look at the EV calculations again. The EV for even money does not depend on x!
The change over point (when you take insurance) is at EV = (4+x)/(13)*2 + (9-y)/(13)*-1
while the change over for break even ends up being at EV = (9-y)/(13)*1.5.
But both of these EV's occur at the same count!
These change over points are different, thus the choice to take even money happens at a different count compared to taking insurance.
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Actually, I would think in the long run it would be exactly the same thing, since as time goes on and the number of hands played approaches infinity, the actual EV (does that make sense? actual expected value?) should converge to the theoretical EV. In other words, variance plays less and less of a role in your winnings the more hands you play.
However, in the short term, you might be a little bit farther ahead due to the decreased variance of taking even money if the count is borderline. It's my opinion that taking insurance and taking even money is the exact same bet and over the long haul have the same EV.
QFIT
Well-Known Member
I think terms may be confusing the issue a bit. There is no difference between taking even money or Insurance in the situation that even money is possible. That is, a natural against an Ace. Thus most people say there is no difference between Insurance and even money. This is not the same as saying even money and taking Insurance against any upcard have the same variance. The difference is not between Insurance and even money; but insuring a BJ against an Ace (where the outcome is unknown) vs. a non-Ace (where the outcome is known.) So, over all cases including where even money is not possible there is a difference.
Hope I haven't confused the issue further
Hope I haven't confused the issue further
Mimosine
Well-Known Member
if ever you're playing a game where the BJ payout is greater than 3:2, then a definite deviation arrises, where you should not take even money, but should insure or insure for less (i forget which and i think it depends on if it is 2:1 vs. 3:1).
but i'm not sure that is what "this guy" was talking about.