Aleph_O play

S

S. Yama

Active Member
Aleph_O' play

Here's the situation. Two players advance, $25 to $500 limits, four players remain on the last hand with the following bankrolls:
Spot 1: 2,525
Spot 2: 1.650 (me)
Spot 3: 1,000
Spot 4: 1,100
Unfortunately I have to bet first and here are the bets and each player's first two cards versus the dealer Ace
Spot 1: 25, Cards: I forget
Spot 2: 500, Cards: T,7
Spot 3: 500, Cards: 3,3
Spot 4: 500, Cards: 9,2

Here's my analysis.
I like to quantify tournament situations by using numbers (expressed in percentages). In the heat of playing a round it is usually just an approximation I can muster at the time. It comes down to comparing and choosing better chances signified by numbers. It may seem like hard work at first but with some practice (okay, a lot of practice) it is the most gratifying part, other than collecting winnings:D, of course.
Having a piece of paper, pen and couple of hours to kill I worked out the following numbers.

Deciding the bet:
My gap to BR1 (player with the most money) is 875, conclusion: I can’t get him by winning a single bet but I can if I win doubled bet (or swing him if he bets big) - -bet at least 450.
The lead over players that have less money is 550 and 650 - -it is safe to bet max bet of 500.
I know that it is easier to protect my lead over BR3 (spot 4) and BR4 (spot 3), because of relatively big correlation between hands, than to try to catch BR1 – however, bet of 500 covers it all.
My bet, as was yours, would be 500.

Acting:
The way of making bet would be to discourage other players from making good bets. Not much can be done here. Nevertheless, I would make my bets in a way saying to BR3 and BR4 (without actually saying it) you have no chance, give up, your bet is not important, bet minimum 25. At the same time I would look confidently at BR1, as to saying: you made it man, you have lock (not true), bet minimum 25.

Playing decision:
Surrender doesn’t do me any good, I need to choose between standing on seventeen, hitting it once or doubling.
I don’t need to double because winning single bet increases my bankroll to 2150- which is less than BR1 will have but it is beyond the reach of either BR3 or BR4 even if they win their doubled bets. BR4 doesn’t have money for the third bet and BR3 can’t make the third bet.
It looks that both players BR3 and BR4 will split and double their hands regardless whether I stand or hit (and bust).

If I hit and win my hand (hit hard seventeen) which has about 20% chance of success (same as if I doubled) -I am guaranteed to advance. Learning percentages of winning for doubling (same as hitting once) on all different totals is one of the most important skills.
If I push my hand against dealer’s Ace (no bj) BR3 or BR4 can’t win their doubled bets –this adds a few more points to my chances (4%).
If I lose- both BR3 and BR4 have to lose or push their bets. Depending on style of play of BR4 (he may choose to stand on a high stiff hand after getting a good first hand to his split treys) and BR3 (he may choose to still double instead of just hitting his 11 after my busting and BR4 busting on at least one hand) one of them wins 40% to 50%.
If I hit seventeen my total chances of advancing are about sixty percent.

Now, if I stand on my seventeen and win (17%) –they simply can’t catch me.
If I push -neither BR3 nor BR4 can win their doubled bets in order for me to advance. I push my seventeen against dealer’s Ace almost 19% percent and one of my opponents would win almost half the time (slightly more if BR4 plays to swing me).
My seventeen loses to the dealer almost two third of times. In this instance BR3 and BR4 play influences the outcomes significantly. BR4 needs to hit both of split threes to at least eighteen and BR3 should hit his eleven to at least nineteen instead doubling. One of them wins his hand 20% to 40% of time.
If I stand on seventeen my total chances of advancing are about seventy percent.

Soooo...:confused:
don’t hit the seventeen, just STAND:cool:.

Regards,
S. Yama
 
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Excellent analysis, as always, S.
I appreciate the time and effort that went into your post!

Further supporting your conclusion, as you point out, BR3 and BR4 can make your chances better yet if they fail to realize they should play their hands to 18 or 19.
 
another opinion

I like S. Yama's approach because this is really all about the numbers. Here's my analysis:

For simplicity let's consider just three players when only one advancing (BR1 has been ignored).

Suppose all of them are PERFECT players and all bet max $500 bet.

Let's consider the case when BR2 ($1,650) STANDS on hard 17:

Now the turn is for BR4 (3,3) with $1,000. It is clear that if BR2 will win the hand there is no chance for BR4 to advance. So now there are two ways:

- SPLIT and HIT each hand 'til 18
- HIT 'til 19 (I'm surprised that this case was never been considered).

DD doesn't work because the best he can get is 17 which wins only together with BR2.

The Probabilities to overcome BR2 are: SPlT - 11.75%, HIT - 15.76% a huge difference (4% over 11.75%). Notice that SPLIT doesn't make any better to overcome also BR3, since BR3 can protect himself by DD.

Now the turn is for BR3 (9,2) with $1,100.

The Probabilities to overcome BR2 are: DD - 28.4%, HIT 'til 19 - 30.4%.

So finally we have the PERFECT play for BR4 (HIT 'til 19) and BR3 (HIT 'til 19). Taking this into consideration we can find the probability for BR2 to advance by STAND - 62.3%.

By hitting once I estimated the probability for BR2 to advance - less than 50.8%.

As you can see, STAND is more than 12% better than HIT, considering PERFECT opponents.

All the calculations are available upon request.

My regards, Tirle_BJ
 
Why bet $500 in the first place?

With a $550 lead for second place and two advancing, why bet more then $25, with a $500 max. betting limit? I would have made the other two "Have" to double or split to beat me on the last hand.
 
Dear TXtourplayer:
By betting $25 you're forcing the runner(s)-up to double regardless what. And if he(they) will win you're out. The probability of that is around 30% for one particular player. So your probability to advance against one runner-up is around 70%, against two - less since none of 'em should win the double.
Now when you bet $500.00, your opponent(s) should try to swing you (win-lose situation), because if you win, then their double is useless. That means in our case, for example, that the dealer should make the hand and they have to try to beat that hand, which means hit until better point. The probability for one runner-up when the lead is more than max bet is about 12%, so your probability to advance is about 88% (compare with 70%). And with the number of chasers increased, this difference will be increased dramaticly.
Questions are welcome, thanks for reply.
 
Tirle_BJ, as usual, excellent analysis. My hat’s off to you.

But… for the entertainment of other readers, let’s find some friendly buts (a conjunction, not a noun).

Analyzing perfect plays is very important but analyzing multiple possibilities of most likely plays (each having specifically assigned probability of happening) is even more crucial.
It is a must for a serious player to understand and recognize a perfect play, but bj tournaments, at least in my mind, are about “playing my hand well” and are not about playing it computer perfect. Ruthless utilization of not so perfect plays by other players is the top priority for me to succeed, other than not making a butt of myself first.

Let’s get specific.
I believe that most (two thirds?) players as BR4 (a guy with pair of treys and $1000 bankroll) would split them and at least two thirds of them would just hit to seventeen. Most of the players (two thirds again?) that didn’t split would just hit to seventeen. That leaves us with about ten percent of players who would do something else. And that something else may include double down, staying on less than seventeen, as well as hitting to more than seventeen.

You are absolutely right that hitting the pair of threes to nineteen is the best play for BR4.
You wrote: Probabilities to overcome BR2 are: SPLIT - 11.75%, HIT - 15.76% a huge difference (4% over 11.75%)
But probabilities of overcoming BR2 are insignificant -- it is both BR2 and BR3 that need to be overcome.

BR4 hitting to 19 overcomes both BR2 and BR3 7.4% of times if BR3 hits to nineteen. The same play gets 8.7% success if BR3 doubles down. It is exactly 6% if BR4 splits and hits to eighteen and BR3 doubles.
You can look at this difference (7.4% over 6%) in threee different ways.
1.) A huge one, as it is a 23% improvement.
2.) Simply 1.4 percent difference,
3.) Since it was a small tournament, a measly, literally, penny worth gain.

Happy number crunching to all,
S. Yama
 
No arguement from me.

Dear tirle_bj

Don't get me wrong I am not disagreeing with you, that might be the higher percintage play. But for me I'm more comfortable sitting back with the lead making the other players have to catch me. What I was telling S. Yama is what I would have done.

On a "No Cash Value Chip" tournament I could see going for it, on an live money game no, but reguardless I would have made them have to double or split to beat me.

I look at it like this if you beat me and had to catch a double or split to do so, well then you beat me. I have been in that situation more times then I care to think about (having to double or split on the last hand) for me I would guess I hit only 10% of the time to advance, (this is for me personally).

I know hitting a twelve against the dealers two is the better play also, but for me it is the kiss of death. I also take insurance everytime (in tournament play) when I have a Blackjack. That is just for me personally not saying it is the highest perintage play by any means.
 
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Dear S.Yama:

Thanks for reply, it was a pleasure to check the numbers and make sure that 7.4% is matching exactly.
However instead of 6% I got 6.8% which is even closer to 7.4%.
Actually:
1) 9% better
2) just .6% better

I'll have my calculations with me in Frontier. See you Soon.
Best regards, tirle_bj
 
Yeah, will do that if I get to go to Vegas at that time.

Talking about, as you said:
Suppose all of them are PERFECT players…
A perfect player should know the actual composition of cards to be dealt. In situations where effectiveness of playing decisions seems to be similar – the “c” word is really what determines the best play, often significantly.

For BR2 (spot 2, hard 17) extra small cards (and Aces) increase chance of dealer making a pat hand (about 2.7% more at TC +5) but at the same time having one extra Ace through four (per deck) increases the chance of winning his dd from 19.7% to 25.1%, while at the same time decrease good dd for BR3 (spot 4, total eleven)- important in case of ties.

For BR4 (spot 3, pair of threes) the key cards are 6, 7, and 8. These cards not only help dealer to beat (or push) BR2 standing on seventeen, but also make a nice two-card totals to split threes.

S. Yama
 
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