Three choices

I was recently faced with the following -

Code:

6D  S17  Min:100  Max:1000  Surr.

BR1 (Me) : $1951.50    Bet : $900     Cards : AA
BR2      : $1425       Bet : $1000    Cards : TT

Dealer   : T

In the twenty seconds I had to act, I contemplated splitting, hitting and surrendering, and was fairly clueless about which might be best!

I surrendered, which I've now worked out was quite a lot worse than hitting would have been -

Code:

Surrender :     40.61%  [dealer 20 or 21]
Hit to h17s19 : 49.39%  [If I've got my calculations right]

Can anyone work out how splitting would compare? I'm guessing it would be worse than hitting, but the calculation keeps defeating me.

I think there were a couple of psychological factors that led me to surrender -

1. Having sized my bet so that a surrender would beat a push, I had the chance to show how clever I had been to show such foresight.
2. I was gripped by a fear of the dealer's 'ten in the hole', giving that possibility more weight than it deserves.

A happy footnote to the story is that the dealer did indeed have a ten in the hole.

 

choices

LONDON, I seldom play in tournaments that offer the surrender option but I do realize that surrender is a potent weapon to have in your arsenal of tournament play. With that being said, I often think some players commit themselves to play for a possible surrender, when other choices of bets may be better than a surrender bet.


The numbers may show that 900 is the better bet here but my old style of play suggest that 525 would be a good bet and also cover a single bet and double down by Br 2.

Again, I don't know the numbers, but I think I would have split the aces.
With 3 tens already on the table, there is less chance of the dealer having 20 and a higher probability that the dealer would be required to hit and possibility of a bust.
I guess I must leave the definitive answer to the number crunchers.
Good luck

 

Thanks for responding, Mike.

It has to be a good idea to give yourself as many options as possible. The difficulty then comes in working out which is the best one to take, once you've seen the cards. That's where I tend to struggle sometimes.

Actually, I described this rather badly. Even a max bet of 1000 would have offered the possibility of surrender, and the choice of 900 rather than 525 is to cover an all-in DD from BR2. Even without the surrender option, my understanding is that this is the better bet. 1000 might actually have been better still, to also cover a BR2 natural.

In a 6-deck game, I doubt it's worth worrying about the number of tens on the table. At the time, I actually ruled out splitting first, and then had a choice between hit and surrender. (I picked the wrong one out of those two, and would not be surprised to learn that I picked the worst of the three! )

 

Well I've had another stab at working out the numbers for splitting, and arrived at the answer 54.75%. So, if that's right, I did indeed pick the worst option, by some margin.

 

It's meant to be the chance of winning the table. For fear of embarrassing myself, I'd like to check my working for mistakes in logic and/or arithmetic, before I risk posting it here.

I was surprised to get an answer that beats hitting. In some ways, I hope I do find a mistake. As things stand, my at-the-table evaluation seems to have been as wrong as it was possible to be, as I've now arrived at probabilities to win of -

Code:

Surrender:     40.61%
Hit to h17s19: 49.39%
Split:         54.75%

 

That was in my mind, but I figured that a net push would lose out to a BR2 win more than it would gain from matching a BR2 push [i.e. when dealer makes 20 and one of the split hands is 21.]

Anyway, for what it's worth, here's how I arrived at the 54.75% ...

Probability BR1 loses table after splitting :-

Code:

Dealer   BR1 loses table if                   Prob
------   -----------------                    ----
17       Either hand < 17 [B]or[/B] both = 17        1 - (8/13 * 8/13) + (1/13 * 1/13)
18       Either hand < 18 [B]or[/B] both = 18        1 - (7/13 * 7/13) + (1/13 * 1/13)
19       Either hand < 19 [B]or[/B] both = 19        1 - (6/13 * 6/13) + (1/13 * 1/13)
20       Neither hand = 21 [B]and [/B]NOT both = 20  (9/13 * 9/13) * (1 - (1/13 * 1/13)
21       Neither hand = 21                    (9/13 * 9/13)

Multiplying by the dealer outcome probabilities (and writing the above probs as decimals) gives -

Code:

17       0.1213 * 0.62722 = 0.07608
18       0.1210 * 0.71598 = 0.08663
19       0.1213 * 0.79290 = 0.09618
20       0.3684 * 0.47645 = 0.17552
21       0.0377 * 0.47929 = 0.01807
                            ---------
                            0.45249
                            ---------

Therefore prob BR1 [B]wins [/B] =  1 - 0.45249  =  54.75%

 

Colin, this was a very elegant way of figuring out chances for the split playing option.
Many a player would get lost trying to come up with all different configurations of two split hands vs. various dealer’s outcomes.

For absolute precision it would be interesting to know if your hit cards to the split Aces were shown face down or up.
BR2 would have to surrender his 20 if you had gotten 20 and 21, or 19 and 21– his only hope would be dealer making 21. If both hit cards to your Aces were Nines or Tens, BR2 would have to hit (or double for less, keeping back at least $152) wishing for the dealer’s 21, or hitting an Ace and the dealer ending up with either 20 (for BR1 twenties) or 21 (in both cases).

This calculation shows that splitting is the best option, perhaps it may be helpful for developing a “better feeling for the game”, to accent that two singular situations that benefit BR1 the most are when the dealer breaks or makes a total of 20 and BR1 receives at least one Ten on the split Aces. Just those two outcomes (advancing BR1 with split Aces) occur 44.4% of the times

S. Yama

 

Thanks. I have a waste-paper basket full of failed attempts!

The cards were face up, so there would be surrender and hitting possibilities for BR2. No double for less, however (other than all-in).

Yes, that's helpful. While I enjoy working through these puzzles at my leisure, I need to develop some techniques to help me arrive at better decisions when the clock is ticking.

 

Small Mistake

I doesn't make any real difference, but I think I made an error -

Code:

Dealer   BR1 loses table if                   Prob
------   -----------------                    ----
20       Neither hand = 21 and NOT both = 20  (9/13 * 9/13) * (1 - (1/13 * 1/13)

I think it should be -

Code:

Dealer   BR1 loses table if                   Prob
------   -----------------                    ----
20       Neither hand = 21 and NOT both = 20  (9/13 * 9/13) - (1/13 * 1/13)

That makes 0.47337, instead of 0.47645, and therefore -

Code:

17       0.1213 * 0.62722 = 0.07608
18       0.1210 * 0.71598 = 0.08663
19       0.1213 * 0.79290 = 0.09618
20       0.3684 * [B]0.47337[/B] = [B]0.17439[/B]
21       0.0377 * 0.47929 = 0.01807
                            ---------
                            [B]0.45135[/B]
                            ---------

Therefore prob BR1 wins  =  1 - 0.45135=  54.87%