Dealer shows next card

[Gamblor]

Originally Posted by moo321

Your logic on staying is correct, and this was the right play.

Your logic about an "index" for surrendering given that a ten is coming, is a bit off.

You have, using Hi-Lo, a ratio of aces+tens to lows. The problem is, you know that the dealer does not have an ace in the hole, because he has checked for blackjack. So, 1/4 of your count is out the window. Also, the uncounted 7-8-9 are against you, so hi-lo will tend to give a poor index.

I'd guess the "index" would be closer to +10 to 12.

Thanks moo, yeah I didn't think through all the nuances of the exact index number. But its pretty high, and situation is rare enough, where its not even worth memorizing it, even for 16 v 10.

[tthree]

Originally Posted by Gamblor

Thanks moo, yeah I didn't think through all the nuances of the exact index number. But its pretty high, and situation is rare enough, where its not even worth memorizing it, even for 16 v 10.

Hence the need for an ace neutral count with a side count of aces.

[21gunsalute]

Originally Posted by moo321

No, you and 21 gun are both missing the fact that you know a ten is coming.

Normally, you surrender because your EV of hitting is lower than 25%, and the dealers bust chance is under 25%.

Now, the dealer's bust chance is greater than 25% (dealer automatically busts with any low card in the hole due to the ten coming).

Again I disagree. If you know you have 66% or greater chance of losing the hand, surrender, especially since the option of hitting to improve your hand doesn't apply here.

[moo321]

Originally Posted by 21gunsalute

Again I disagree. If you know you have 66% or greater chance of losing the hand, surrender, especially since the option of hitting to improve your hand doesn't apply here.

It's not 66%. It's 75%.

If you have a $100 out, with a 33% chance to win, your EV equals .33*200= $66 (your bet + the win).

Your EV for surrender is $50, because you keep half your bet.

66>50

[21gunsalute]

Originally Posted by moo321

It's not 66%. It's 75%.

If you have a $100 out, with a 33% chance to win, your EV equals .33*200= $66 (your bet + the win).

Your EV for surrender is $50, because you keep half your bet.

66>50

Let's try this again. Not sure what I was thinking, but according to Wizard of Odds the dealer busts about 23% of the time with a 10 showing, meaning he will make hand 77% of the time (at an average count of zero I assume). I believe then that he has a greater than 77% chance of making a hand at a TC of 3. So my figures were off but surrender is still the best option.

[newbctr]

21gunsalute - you don't know what you are talking about and are confused. What you read on Wizard of Odds is irrelavent because in this case the next card (10) is known. Surely, the dealer will make a hand a lot less now. The 77% is so high because when the dealer has a 2-6 in the hole, he/she will still make a hand a lot of the time (unless we know with 100% certainty that a 10 is coming)

Get some rest

[Sucker]

Originally Posted by 21gunsalute

Let's try this again. Not sure what I was thinking, but according to Wizard of Odds the dealer busts about 23% of the time with a 10 showing, meaning he will make hand 77% of the time (at an average count of zero I assume). I believe then that he has a greater than 77% chance of making a hand at a TC of 3. So my figures were off but surrender is still the best option.

Yes, NORMALLY the dealer WILL bust 23% of the time when he has a ten up and has already checked for BJ - BUT; that's because of the fact that when he IS stiff, he'll STILL often be able to hit with a small card and make a hand anyway.
In the example under discussion, that CAN'T happen - the dealer will bust every SINGLE time that he has a stiff (1/3 of the time); because we know that the hit card's a TEN.

Not a surrender.

[21gunsalute]

Okay, then what % of the time will the dealer bust at a TC of 3 in this scenario?

I still think surrender is the best option because this is a special situation not likely to repeat itself and the player is likely to lose this hand if he doesn't surrender.

[tthree]

Originally Posted by 21gunsalute

Okay, then what % of the time will the dealer bust at a TC of 3 in this scenario?

I still think surrender is the best option because this is a special situation not likely to repeat itself and the player is likely to lose this hand if he doesn't surrender.

I dont have the ability to run sims but the deck composition that would produce a +3 TC have so many possibilities Im a little curious how a sim guy would attack this problem. I guess knowing a 10 is coming simplify it greatly but you get the point. This problem is simple enough I think my hand calculation that I very quickly produced would be pretty accurate. I may have botched terminology but I dont think the speed of production caused a math error. It showed a 34.5% dealer bust rate.

[Sucker]

Originally Posted by 21gunsalute

Okay, then what % of the time will the dealer bust at a TC of 3 in this scenario?

I still think surrender is the best option because this is a special situation not likely to repeat itself and the player is likely to lose this hand if he doesn't surrender.

Simply put, for THIS particular scenario; what you need in order to surrender is; in the unseen cards, the ratio of 2-6 to non-aces has to be less than 1:3 (one in four).

In order to ACCURATELY determine the surrender point by the count, you would actually have to use a special count in which 2-6 counts as +1, 7-K counts as -1; and you would have to keep a side count of aces, so that you could subtract them from the total number of unseen cards.

Obviously, none of us are going to be keeping such a bastard count, so we would have to use the count at hand in order to estimate. On MY simulator, which uses the RAPC; that index number is +28. Converting that to Hi-Lo, the TC would have to be something like +10 to +12, which just happens to coincide exactly with the OP's original estimate. Don't forget to subtract that exposed 10 from your count!