theoriemeister
theoriemeister
Joined: Jul 4, 2015
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June 30th, 2018 at 9:50:44 PM permalink
Tonight for the first time I played Free Bet BJ, as the $5 table of regular BJ was full. Since I have never played this variant of BJ, I watched it for 15 minutes or so and asked a couple of questions about how the "free bet" worked. (This is the version wherein a dealer's 22 is a push and not a win for the table.) I quickly figured out that I should double (for free) on every 9-11 and split every pair (for free) except 10s (not allowed) and 5s (free double instead).

I didn't fare well (I busted on 12s almost every time. . . grrr), but it was different (to me) and kind of fun. I wasn't entirely sure of all the BS for this variant, so I played the BS for the 6D, H17, DAS game, which I know by heart. I also practiced my counting and that's when I began to wonder how counting indices such as the I18 would effect the BS of this variant. (I also didn't win even once when I had more than my minimum bet at stake. Curse you, variance!) I couldn't find anything of the WoO site about it.

So my questions are: first, should I follow the I18 indices regarding deviations from BS, and second, since the HE in this variation is 1.04%, should I wait until TC = +3 before raising my bet?
ars longa vita brevis
Romes
Romes
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July 2nd, 2018 at 10:02:28 AM permalink
Quote: theoriemeister

...So my questions are: first, should I follow the I18 indices regarding deviations from BS, and second, since the HE in this variation is 1.04%, should I wait until TC = +3 before raising my bet?

The Wizards page on Free Bet Blackjack has the basic strategy: https://wizardofodds.com/games/free-bet-blackjack/

Yes, similar to regular blackjack you need to figure the House Edge "off the top". The counting will work exactly the same (same values, etc) but with a higher house edge. Insurance would remain the same, and I'd assume you'd follow most of the other deviations EXCEPT you would always continue to double more when it's free. There may or may not be slight difference when the dealer has a 2 up (perhaps more likely to hit 22 or something - just thinking off the top of my head) but I'd think for the most part the deviations would play out the same, again except for the doubling ones (you would more than likely always want to double, even if the deviation says stand).

I do enjoy encouraging "new" counters to make it easier on them to think of each TC as .5% for the player, but the nitty gritty is TC +1 is actually worth over .7%, so after TC +2 in a 1% game you would have a small slight edge. So if you want to get a solid edge, TC +2.5 or TC +3 would pretty good to start betting a bit more.
Playing it correctly means you've already won.
theoriemeister
theoriemeister
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July 13th, 2018 at 1:03:34 PM permalink
Romes,

Sorry for the delay. I've been traveling.

Thanks for the reply! Good to know to wait until +3 to increase the bet. From an optimal play standpoint, isn't it still better to follow BS with respect to doubling and splitting? For instance, I know to double 9 v. 2 when the TC = +1, but since it doesn't cost me any more to double against a dealer 2, why not do it when TC =< 0? And what about splitting, say, 4s v. dealer 2 - 4 (or other splits not normally included in BS)?

My limited experience says BS is still the optimal way to play the hand and to resist unnecessary splits or double simply because they are free.

Theorie
ars longa vita brevis
charliepatrick
charliepatrick
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July 13th, 2018 at 3:32:39 PM permalink
Doubling costs you the inability to take a 4th card (or more) but doesn't cost you any more money (as it's free you still only stand to lose one unit) while you stand to win twice as much. Thus you actually do it even if you're about a 2 to 1 underdog. If I recall the worst split is 4-4 vs A (no peek) yet it's still correct to take the free split. I can't imagine taking free money changes according to the count.
theoriemeister
theoriemeister
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July 14th, 2018 at 11:15:52 AM permalink
But I thought the whole point of BS is to have the best chance of winning the hand, right? So if you're dealt 4-4 v. dealer, say, 7, isn't the chance of winning the hand by hitting the 8 better than by splitting the 4s? Or, does the possibility of a bigger payout at no cost to the player outweigh BS?
ars longa vita brevis
charliepatrick
charliepatrick
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July 14th, 2018 at 10:40:50 PM permalink
Say (I don't have the exact numbers to hand) you had a 60% chance of winning with an 8 but only 40% with a 4, then you'd be better off having two chances at 40%. Similar logic applies to the doubles.

FreeBet has a different strategy than BS and it also depends on whether real money or free money. For instance you hit Hard 17 vs 7 with free money as a standoff doesn't pay you anything.
theoriemeister
theoriemeister
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July 15th, 2018 at 3:25:17 PM permalink
Quote: charliepatrick

Say (I don't have the exact numbers to hand) you had a 60% chance of winning with an 8 but only 40% with a 4, then you'd be better off having two chances at 40%. Similar logic applies to the doubles.



Are you saying that by splitting the 4s, you have a better than 60% chance at winning one of the hands? That'd result in a push, no?

Quote: charliepatrick

FreeBet has a different strategy than BS and it also depends on whether real money or free money. For instance you hit Hard 17 vs 7 with free money as a standoff doesn't pay you anything.



Can you explain in a bit more detail? The only way to get a hard 17 with 'free' money would be after a split. So let's say I split 7s v. dealer 7. At this point on the table I have my original wager and a 'free bet' to cover the split. Then I'm dealt a 10 for my first 7. I should hit that?? And then dealt a 10 for my second 7. Hit that as well? This doesn't make sense to me.
ars longa vita brevis
gordonm888
gordonm888
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July 15th, 2018 at 4:25:07 PM permalink
Quote: theoriemeister

Are you saying that by splitting the 4s, you have a better than 60% chance at winning one of the hands? That'd result in a push, no?

In charlie's example, if you split the pair of 4s, you are playing two hands each with a 40% chance of winning, so the total expectation of winning a payout is 0.4+0.4=0.8 or 80% (the math only works that way if the second bet is a free bet).

Quote: theoriemeister

Can you explain in a bit more detail? The only way to get a hard 17 with 'free' money would be after a split. So let's say I split 7s v. dealer 7. At this point on the table I have my original wager and a 'free bet' to cover the split. Then I'm dealt a 10 for my first 7. I should hit that?? And then dealt a 10 for my second 7. Hit that as well? This doesn't make sense to me.



No, you can get a hard 17 when your first two cards are 10-7 or 9-8, or your first 3 cards are 10-4-3, etc.

The reason for hitting a hard 17 is not related to having a free bet, it is due to the fact that dealer pushes on 22.

Normally, Standing on a hard 17 is a -EV option because it only pushes against a dealer 17 or wins when dealer makes a 22, 23,24 25 or 26. By converting the instances when dealer makes a 22 from a Win (in standard BJ) to a push (In Free Bet), it reduces the EV of Standing on a hard 17 so much further that Hitting a hard 17 vs 7 becomes a better option.

Again, you need to look at the Basic Strategy for FreeBet Blackjack on the WOO site -you will find it on the url link that was posted a little earlier in this thread. You are giving away large amounts of money if you are not using this basic strategy. And again, some of the changes in Basic Strategy are due to the push 22 rule, which makes standing on 12-17 less attractive.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
SM777
SM777
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July 15th, 2018 at 4:30:05 PM permalink
Simple strategy: Take any and every free bet allowed.
BlackjackGuy123
BlackjackGuy123
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July 18th, 2018 at 2:59:50 PM permalink
The reason you hit hard 17 v 7 on your free bet hand is because a push is essentially a loss.

No, you cannot use blackjack indexes on freebet. You have to simulate various true counts and test whether hitting / standing etc has the highest expectation. Failing that, just use basic strategy.

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