What Shall You Do

Here is the sitution

Last hand only 1 advances there are 2 players left

You are BR 1 with 500 bet and 1400 held back

BR 2 has 500 bet and 1000 held back


BR 2 acts first and stands on 18

You have A-A

Dealers up card is 8


What do you do with your A-A ?


This tournament offers surrender

 

Clarification on Ace-Splitting Rules

Just so we cover everything, what are the rules on Split-Aces? Also, can we surrender any of the hands from the split-aces? Depending on the rules, it could be like starting the hand with an Ace and holding back 1150.

 

One Card

You can only draw 1 card on each Ace .

Surrender is available on your first 2 cards only


Joep2

 

Analysis I can't do while actually playing!!!

What do you do with your A-A ? SURRENDER

Without the rules for splitting and doubling, and therefore assuming typical rules and late surrender allowed, I would opt to set myself up to lose less rather than to win too much, given the other up cards and the bets. Splitting is not necessary in this case, so it's a risk you should not take. "Always split aces and eights... but not if you don't have to". If BR1 splits, he only gets one card for each hand and has to win one of them just to break even. This isn't good enough if BR2 pushes or wins the hand. Pushing would be more likely. Losing both of the split hands would be bad because BR1 would lose the round even if BR2 loses the hand. Given the bets placed and the amounts held back, BR2 cannot win the round unless BR1 loses or pushes and BR2 pushes or wins the hand respectively, because he locked himself into that position by not doubling or doubling for less. With a dealer 8 and BR2 having 18, the likely outcome is that BR2 loses or pushes. Neither of these will make him a winner if BR1 surrenders. BR1 should minimize his risk rather than trying to maximize his gain, by surrendering. BR2 should have doubled and made BR1 really work for the win.

 

Correction - "This isn't good enough if BR2 pushes or wins the hand".

Should read - "This isn't good enough if BR2 wins the hand". If BR2 pushes and BR1 only wins one hand or pushes both, BR1 still wins. This doesn't change my opinion that surrender is the best move. The key is that BR2 loses if he loses the hand or pushes, if BR1 surrenders. I would still surrender.

 

Rules Are Posted

Mr Powerman the rules are there .You are allowed one card on each ace if you split and if you chose to Double Down one card on that also .Now Doubling Down on A-A should only be done by you if you are on the show " King Of Las Vegas " then its encouraged

 

Hit

Best Option - Hit ('til soft 18 or hard 16)
Over 76.4%

 

One other piece of information.

I like it. Opps, I need to write 10 characters.

 

solution

Monkeysystem, here's the chart of probabilities for playing as a dealer (H17 - infinitive deck):

...............17.......18.......19.......20.......21.......Bust

A, A........8.29....16.25...16.25...16.25...16.25...26.71
8 up.......12.86...35.93...12.86....6.94.....6.94...24.47

As you can see, here we Hit on hard 16 instead of stand as offered, so the optimum play is even slightly better. (details later)
Now we have the following logical chain:

* Dealer busts - we need to make the hand......... 24.47 * 73.29 = 17.93%
* Dealer makes 17 - we need to win....................12.86 * 65.00 = 8.36%
* Dealer makes 18 - we need to win or push.........35.93 * 65.00 = 23.35%
* Dealer makes 19, 20, 21 - we are OK................26.74 *100.00 = 26.74%
Finally Total is........................................................................76.38%

Details:
Here is the example which helps to understand that Hit is better than Stand when we have hard 12 thrugh 15.
We have hard 12, opponent has 18, Dealer is 8 up.
Lets consider one-card Hit and compare with Stand.

-We Stand - to advance we need the dealer bust or make 19,20,21.
P= 51.21%

-We Hit only once (for simplicity) - we advance when:
* Dealer busts - we need not to bust................24.47 * 9/13 = 16.94%
* Dealer makes 17 - we need 18,19,20,21..........12.86 * 4/13 = 3.12%
* Dealer makes 18 - we need 18,19,20,21..........35.93 * 4/13 = 11.06%
* Dealer makes 19,20,21 - we always advance....26.74 * 100 = 26.74%
Finally total is.....................................................................57.86%
Same technic you can apply to hard 13,14,15.
Actually you have to start from hard 16 to see where the turning point is, so once you make sure that Stand is better, you go to hard 15 considering only one hit etc. Now if it gives you Hit that means that hard 14,13,12 are also Hit and may be even more than once 'til the turning point.

 

A, A questions

tirle_bj,

Nice job of analysis. There are 2 questions I'd like to ask:

1) Above, I reproduced the chart you used for "A, A". Where did you get that chart? I can't find it in any of my references.

2) Again referring to the "A, A" chart, can you explain why the probability is the same (16.25%) for having a final total of 18, 19, 20, or 21? This seams unusual to me.

 

explanation

Toolman1, thank you for interest in this topic.
I've never seen these numbers at any sources. All the charts are developed by me using combinatorial method (not simulation) for infinitive decks. Without going through the details, let me show you just why 18,19,20,21 are the same and 17 is different.
The difference comes from the rule of H17. As you can see there is no way that we can get to 17 by one step. All others are possible to obtain by one step only. Now if we'll take any chain of cards, for given point (say 18), we can create identical chain of cards for other points (19,20,21) by changing only the last card, which proofs that the number of possible ways for 18,19,20,21 is same. Example:
Say, we have the following chain: (A, 2, T, 3) for total of 18.
To get to 19,20,21 we simply need to change 3 (last link in chain) to 4,5 or 6. Same logic works for all (18,19,20,21). As we can see now the number of combinations for those are same.
Again this logic does not work for soft 17, because we hit it and stay on soft 18, 19, 20 and 21.

 

Thanks tirle_bj

tirle_bj:

Thanks for the response.

I kind of thought that the reason for the same probabilities percentages was what you said but couldn't be sure until you confirmed it.

Since you calculated the "A,A matrix" yourself, you must have a whole bag of tricks that are not published anywhere. Good for you!

Looking forward to chatting with you again when the appropriate subject comes up.

 

Wow!

Tirle - BJ:

You've made me feel like the rank amatuer I am Can I ask whether in the heat of a tourney that would be the decision you would have made or did you analyse the probelm for an hour or 2 before responding to the post? Just interested really.

Joep:

Is this based on actual events? If so what happened?

Ps. I recognise this link from Ken Eineggers forum. Funnily enough I came on today to ask that you put it here as well as this site seems to get alot more traffic.

 

BIG Decision

I asked Tirle_BJ the same question and his honest answer was that he would SPLIT the Aces.
And he explaination was - Because dealer's 18 is the most probable outcome in that cituation, which means BR1 need to at least push, and SPLITTING Aces is a very good way to PUSH. But if you know exactly that BR2 is going to push his/her hand, than the best thing is to SURRENDER.
Here help comes from second most probable outcome for dealer - BUST. In that case you want to stay in the game.
He did the calculations in very short time and came to surprising outcome - HIT.
And he admitted that he could not calculate that kind of close decision during the game.

 

Interested to hear what some of the other regular contributors to the forum think. Come on ZG, come on Ken, what would you do?

Cheers

Reachy

 

Think Quick

I will tell you this when I witnessed this hand at the Million Dollar Tournament my first reaction was not to split .Did I know at that time that is was the mathematical correct play NO.

But I will tell you this that in the short period of time that players are given in their hand decision time you must try to see as much of the situation and its probable outcome .

When I saw the hand live I believed that the 18 was at best a push and maybe a loser vs. the dealers up card of 8 .So if the BR 1 was to split the aces he would give up the low to BR 2 never a good thing to do on the last hand.

If he hit the hand and made any hand from 18 on up he locked out BR2 if he didn't make a hand or busted he could still win if the dealer made 19-20-21 .

So as you see he had more ways to win by treating it like a 12 but a 12 he could hit and not break on his first hit .

6-7-8-9 on his first hit locks it up

2-3-4-5 and he can hit again without breaking

Just the 10-J-Q-K puts him in a breaking situation .But a situation he can still win with .Remember at of all the stiffs that you have to hit a 12 is the best one, as 9 cards out of 13 do not break you



Joep2

 

tirle_bj, where were you when I needed you?

About 2 years ago, I was BR1 and last to bet at the Final table.

With a maximum bet in the box, I got a pair of aces, which I split. I was dealt two stiffs. The dealer made a hand. I think 17 or 18. I plummeted from BR1 to BR7. Ironically,the sum of the four cards added up to 21.

I think I qualified for the Terry Malloy Award:"I coulda been a contenda":laugh: