A close decision

London Colin

London Colin

Top Member
I was faced with the following on the penultimate hand of a heads-up game at GPC -

BR2 572.50
BR1 612.50 (me)

BR2 acts first (so it will be me to act first on the final round). Bets 500.
I match the 500 bet.

Cards -
Dealer: 10
BR2: 16
Me: 9

BR2 stands on 16.

So I was faced with the choice of surrendering, with a very good chance of winning the game outright on this hand - (If the dealer makes a hand [77%?], I win; If he busts I lose.), or playing out the hand using a don't bust strategy, which offers -
  • a small chance that I win on this hand by getting a swing. (push or win)
  • a small chance we both win and go into the last hand with me as BR1, acting first.
  • the most likely outcome that we both lose and I go into the last hand as a big favourite (able to take the high and low against an all-in bet).
I chose the latter, and made 16. We lost to the dealer's 20 and I did indeed win the last hand. I was not at all sure I was doing the right thing by not surrendering, and of course was peeved when the dealer showed 20.

I've tried some back-of-an-envelope calculations and I'm arriving at the conclusion that I did stumble upon the right action, but it looks very marginal and I'm as unsure about my calculations as I was about my at-the-table decision making.

Would anyone care to comment? ;)
 
You Made The Right Decision

The playing decision you made guaranteed you would have the lead on the last hand. This was critical to your success because you had the button on the last hand. If you had surrendered and the dealer busted you were dead meat. Same thing if you had busted your hand and the dealer double busted you.

Having any lead is much more important than having a bigger lead, especially if you have the button on the last hand.
 
Yes, but ...

A surrender would have been for the whole thing. If the dealer made a hand I would have an unassailable lead of 362.50 vs. 72.50 and could min bet the last hand.

So my chances of winning the table would be the same as the dealer's chances of not busting with a 10, which I believe is 77%.

I always make mistakes when trying to figure out conditional probabilities, but I make my chances of winning the table the way I played it 83%. (Not that I knew that at the time! :) )
 
Very interesting situation. A surrender by BR1, in effect, guarantees that this will be the last meaningful hand of the game since one player will have an insurmountable lead over the other going into the last hand. Since BR1 has a 77% probability of having that insurmountable lead over BR2 with a surrender by BR1 then this is the way to go. It's unlikely that BR1 will get a better win probability in the last hand if he doesn't surrender this hand.

Would I have made this decision in the heat of battle? I doubt it. Ain't hindsight great. :)
 
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Surrender

Hmm. If you surrender your chance of winning is 77%.

That's hard to beat unless you're certain that your opponent will bet the max after you lead off with the max on the last hand.

If you had to make a minimum bet on the last hand your chance to win was only 56%.

You could almost call this a conditional lock. It's conditional on the dealer making a hand.

London's Lock?
 
Hit that 9!

I'd think that hitting your 9 was easily the correct choice. Since BR2 stopped at 16, he cannot swing you, but you've got a chance to swing him and lock it up. If you both lose (as you did), you're sitting at 112.50 vs. 72.50. This allows you to bet 35 and cover both high and low to his all-in.
 
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London,
I know you're looking to see if you made the proper play, but perhaps you are looking at the wrong bet....

BR2 572.50
BR1 612.50 (me)

He bet 500
If you bet 465.....

Both win....he has 1072.50
............you have 1077.50
Both lose...he has....72.50
............you have..147.50 which will cover his all-in win on the final hand..
just a thought
 
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Swog,
Good point. Don't know how I missed that!

LeftNut,
I don't know about 'easily', but you've accurately described the thought processes that led me to play as I did. (and accurately predicted my bet on the last hand :) )

toolman1,
I was afraid you would ask that! I'd be surprised if there aren't some mistakes in the logic and/or the numbers, but for what it's worth my reasoning was -

P(Win Table) =
P(Swing BR2 on 1st Hand)
+
P(Don't Swing) x
(
P(Both Win) x P(Win On Next Hand by Taking Low {BR2:1072.50, BR1:1112.50})
+
P(Both Lose) x P(Win On Next Hand by Taking High and Low {BR2:72.50,BR1:112.50})
)

I used Ken's 6-deck dealer probability tables and the probabilities of me reaching totals of 17 to 21, to arrive at a figure of 0.21 for P(Swing BR2 on 1st Hand).

The whole thing is
0.21 + (0.79 x (0.23 x 0.56 + 0.77 x 0.86)) =
0.21 + 0.62 =
0.83

One thing I've realised since is that BR2 could have won the last hand with a BJ, which I haven't accounted for in my calculation. (I also wonder if it would be better or worse to bet 70 instead of 35, to cover the BJ, with the option to surrender back to a BR of 77.50 if the cards dictate it.)
 
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fried brain

Off the top of my head I would say you did the right thing. If you correlate you guarantee the lead in the final hand, and there is the small possibility of a swing. If you both lose you have >81% of success on the final hand. It could be a bit more difficult for you if the dealer busts because then BR2 may be able to take either the high or low from you on the final hand depending on how you play it, to shorten your odds.

I would have played it the way you played it.

Cheers

Reachy
 
London Colin said:
One thing I've realised since is that BR2 could have won the last hand with a BJ, which I haven't accounted for in my calculation.
Click to expand...

Yes, with a bet as I suggested (and apparently that's you did! :p ) then a BJ for your opponent would put you in a bad spot. But a split or DD of your own could overcome it.
 
In order to avoid long phrases, this post contains the following abbreviations:
LH = Last Hand
LH-1 = Second from last hand

What we have here is a situation where BR1 or BR2 will completely lock up the win on LH-1 if BR1 surrenders on LH-1. The contest is virtually over if BR1 surrenders on LH-1 because the chip leader then need only make a minimum bet on LH to lock up a win. On the other hand, if BR1 hits the 9 on LH-1 then the probabilities for BR1 winning wining the table on the LH depend on whether the dealer busts or makes a hand on LH-1.

The important thing to keep in mind is:
The objective is to win the table and that means that sometime concentrating on the best probability to win the last hand will not necessarily give one the best over all probability to win the table. On the question raised at the start of this thread, assuming BR1 hits the LH-1 then BR1 has that 83% probability of winning the table on LH only if the dealer makes a hand on LH-1. If the dealer busts on LH-1 then BR1’s probability of winning the table on LH plummets.

Seems to me, after a few days of thought, that this is one of those situations that require some analysis seldom applied in this forum because we are looking at LH-1 not LH. How BR1 handles LH-1 is critical to the outcome of the game. LH-1 is much more important than LH in this situation. Concentrating on the fact that BR1 can end up with an 83% probability of winning the final hand does not tell the whole story. How he gets there is the important question.

So what does this mean? It means that both hands must be considered in the calculation of the correct play for LH-1. I have made some calculations below that I think are right but not 100% sure.

The assumptions I made for LH-1 are:
1) Dealer has a 77% probability of making a hand on LH-1.
2) Dealer has a 23% probability of busting on LH-1.
3) BR1 has a 83% probability of winning LH if the dealer makes a hand and BR1 hits his 9 to a stiff on LH-1.
4) BR1 has a ??% probability of winning LH if the dealer busts on LH-1.


Now my calculations for BR1 winning the table:
BR1 hits to a stiff on LH-1 and dealer makes a hand on LH-1:............. .77 x .83 = .64
BR1 hits to 17 or higher on LH-1 and dealer makes a hand on LH-1:..... .?? x .83 = something less than .64
BR1 hits on LH-1 and dealer busts on LH-1: ………….......................…. .23 x .?? = something much less than .64
BR1 surrenders on LH-1 and dealer makes a hand on LH-1……........…… .77 x 1.00 = .77
BR1 surrenders on LH-1 and dealer busts on LH-1 …………………......……. .23 x 0.00 = .00
NOTE: I have not factored in BJs because it is too damn complicated. After taking into account that a BJ by BR1, BR2, or the dealer will have a cancelling effect on the over all percentages, BJs should not affect the overall “best choice”.

As you can see, how LH-1 plays out is critical to the final outcome. Clearly BR1 surrendering on LH-1 is the best choice with a 77% probability of winning the table.

Oh if we could only have 3 or 4 days to make a decision at a table.
 
toolman1 said:
On the question raised at the start of this thread, assuming BR1 hits the LH-1 then BR1 has that 83% probability of winning the table on LH only if the dealer makes a hand on LH-1. If the dealer busts on LH-1 then BR1’s probability of winning the table on LH plummets.
[...]
Concentrating on the fact that BR1 can end up with an 83% probability of winning the final hand does not tell the whole story. How he gets there is the important question.
[...]
So what does this mean? It means that both hands must be considered in the calculation of the correct play for LH-1. I have made some calculations below that I think are right but not 100% sure.
Click to expand...
In the formula I presented, 83% is my bottom-line answer after considering both hands.


toolman1 said:
3) BR1 has a 83% probability of winning LH if the dealer makes a hand and BR1 hits his 9 to a stiff on LH-1.
Click to expand...
I actually had 86% as my equivalent to this (arrived at by adding Wong's 'A wins and B Pushes' and 'A Wins and B Loses' , and subtracting from 100.)

toolman1 said:
Now my calculations for BR1 winning the table:
BR1 hits to a stiff on LH-1 and dealer makes a hand on LH-1:............. .77 x .83 = .64
BR1 hits to 17 or higher on LH-1 and dealer makes a hand on LH-1:..... .?? x .83 = something less than .64
BR1 hits on LH-1 and dealer busts on LH-1: ………….......................…. .23 x .?? = something much less than .64
BR1 surrenders on LH-1 and dealer makes a hand on LH-1……........…… .77 x 1.00 = .77
BR1 surrenders on LH-1 and dealer busts on LH-1 …………………......……. .23 x 0.00 = .00
Click to expand...
Have you checked the logic in my previous post? Substituting your 83% prob of winning on LH1 after both players lose on LH-1, for my 86% would give -
0.21 + (0.79 x (0.23 x 0.56 + 0.77 x 0.83)) = 0.82

I can't see how you are intending to combine the elements you've listed. I may have misunderstood, but it looks like you've overlooked the fact that you need to add the separate probabilities of winning the table on LH-1 and LH. (0.21 + ....)
 
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London Colin said:
I can't see how you are intending to combine the elements you've listed. I may have misunderstood, but it looks like you've overlooked the fact that you need to add the separate probabilities of winning the table on LH-1 and LH. (0.21 + ....)
Click to expand...
Adding the separate possibilities? Don't understand. Example: If the probabilities of winning a hand are 44% then the probability of winning 3 in a row is 9% (.44 x .44 x .44) not 132% (.44 + .44 + .44).

As to the 83%, I thought you were talking about the last hand only in a previous post. I'll review when time permits but off the top of my head that seems awfully high as a probability for winning the table. I'll take a look later. Like I said, I wasn't sure of my numbers and relied on other postings for some of them.

As to the formula you wrote: "0.21 + (0.79 x (0.23 x 0.56 + 0.77 x 0.83)) = 0.82"
Lets recalculate this:
0.21 + (0.79 x (0.23 x 0.56 + 0.77 x 0.83)) = 0.21 + (0.79 x (0.08)) = 0.21 + (.06) = 0.27
How did you arrive at 0.82??? I'm not sure where the numbers came from in the formula, I'm talking about the arithmetic.
 
toolman1 said:
Adding the separate possibilities? Don't understand. Example: If the probabilities of winning a hand are 44% then the probability of winning 3 in a row is 9% (.44 x .44 x .44) not 132% (.44 + .44 + .44).
Click to expand...
A better analogy would be to ask what is the probability of winning one of the three hands. As you say, you don't add .44 + .44 + .44; but you do you have to add separate probabilities - conditional probabilities:
0.44 +
( 1- 0.44) x 0.44 +
(1- 0.44) x (1- 0.44) x 0.44
= 0.82
i.e.
P(Win 1st) +
P(Don't win 1st) x P(win 2nd) +
P(Don't win 1st or 2nd) x P(Win 3rd)

(Wong's figure of .86 for the 3-step progression is slightly larger because it allows for pushing hands and then trying again, thus playing more than three hands.)

toolman1 said:
As to the 83%, I thought you were talking about the last hand only in a previous post. I'll review when time permits but off the top of my head that seems awfully high as a probability for winning the table. I'll take a look later. Like I said, I wasn't sure of my numbers and relied on other postings for some of them.

As to the formula you wrote: "0.21 + (0.79 x (0.23 x 0.56 + 0.77 x 0.83)) = 0.82"
Lets recalculate this:
0.21 + (0.79 x (0.23 x 0.56 + 0.77 x 0.83)) = 0.21 + (0.79 x (0.08)) = 0.21 + (.06) = 0.27
How did you arrive at 0.82??? I'm not sure where the numbers came from in the formula, I'm talking about the arithmetic.
Click to expand...
It's 0.21 + 0.61
 
Probabilities

In this problem we're operating under some assumptions.

The probability of the dealer busting LH-1 is 23%. Therefore the probability of the dealer not busting is 1 - 23% = 77%.

There are two possibilities if you surrender. The probability of winning if you surrender and the dealer busts is zero. The probability of winning if you surrender and the dealer makes a hand is 100%. Therefore the probability of winning if you surrender is 100% * 77% + 0% * 23% = 77%.

If you stand on a stiff there are two possibilities for LH-1; the dealer busts and pays you both or the dealer makes a hand and beats you both.

If the dealer busts, your lead is such that you have to take the low. Your chance of winning in that case is 56%. Your chance of winning that way is 23% * 56% = 13%.

If the dealer makes a hand your lead is such that you can bet 35 on LH and have the high and low, as well as winning the lose-push. Your chance of winning is 86%. Your chance of winning that way is 77% * 86% = 66%.

Your chance of winning if you stand stiff is the chance of winning if the dealer busts LH-1 plus your chance of winning if the dealer makes a hand on LH-1. It's 13% + 66% = 79%.

This 79% figure represents the worst case scenario if you hit instead of surrendering. If you had hit your 9 and made a pat hand that pushed or beat a dealer pat hand, your chance of winning was 100%.

79% > 77% so you made the right decision.
 
Thanks Monkeysystem

That's an excellent point about the worst case scenario.

All the expressions you quoted are contained within my convoluted formula. In trying to work out the overall probability, and being uncertain about some of the reasoning involved in that, I didn't spot that the worst case could be more easily and clearly calculated. And if that's > 77% then the rest is academic.

(Though there is still the issue of how much BR2's chance of a BJ on the LH depresses the 86% chance of winning with a 35$ bet.)
 
BJ chances

London Colin said:
(Though there is still the issue of how much BR2's chance of a BJ on the LH depresses the 86% chance of winning with a 35$ bet.)
Click to expand...

Actually, both BR1 and BR2 have the same chance of getting a bj. If BR1 gets a bj it's game over. If BR2 gets a bj BR1 can still recover with a forced double down.

There's a 4% * 70% = 3% chance that BR2 will win with a bj that BR1 can't cover with a forced double. There's a 1% chance that BR1 will cover a bj with a forced double. There's a 4% chance br1 will win with a bj of his own. There's a negligible chance both players will get bj's. This represents a 5% probability of BR1 winning if either player gets a bj.

There's a 92% chance neither player will get a bj. To figure in this bj possibility we use these equations:

BJ's don't count if the dealer busts LH-1 because you take the low on LH. So the 13% figure still applies.

If the dealer makes a hand on LH-1 it's 77% * ((92% * 86%) + 5%) = 65%. 65% + 13% = 78%. Factoring in the possibility of bj's on LH reduces your worst case scenario by 1% but still is higher than if you surrender LH-1.

The 78% figure is still the worst case scenario.
 
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I think London Colin and I are having a hard time on this because we are having a hard time communicating our calculating processes to each other. Now with Monkeysystem's comments we have the answer we were trying to get.

The bottom line here is that I think Monkeysystem came up with the right answer although I had a hard time understanding his 1:48pm post because he used to term "winning" without saying if he meant winning the "hand" or "table". To restate Monkeysystem's answer:

Standing on a stiff on LH-1 = 79% probability of winning the table
Surrendering on LH-1 ........= 77% probability of winning the table

The numbers are close enough to show me that either choice would be OK in the heat of battle but of course standing on a stiff is the preferred choice.

Now I can sleep tonight knowing math still works. Speaking of math, does anyone know fgk42's post count for today? Is he up to par for a Friday or is he slacking off?
 
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2 cents more

By surrendering Br1 also gives up a possible push with the dealer.

Great posts guys.

tgun
 
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