Last hand at Biloxi

Monkeysystem beat me to the punch

I was going to post something similar. Not sure it's how I would actually have bet when put on the spot; I suspect I might have gone with the sort of bet I implied in my previous post ( > 85 , < half BR).

However, the first half of this discussion has shown that a big bet from BR2 not only makes it painful for BR1 to match it, but also makes matching it potentially the wrong choice for BR1, if BR3 is considered a good player.

Bizarrely, it seems that to guesstimate the likelihood that BR1 will match you, you need to assess not only BR1, but also BR1's assessment of BR3!

As a minor refinement, I think the minimum bet should be 290, rather than 235. This would mean that, if the bet is matched, a BJ for us would lock out BR1.

 

I see the reasoning in the Monkey and Colin posts above. However, I think my little bitty $15 bet would also cause some pain for BR1, as he would have to choose between matching that bet and pushing out the larger stack, thereby giving up the low to me and opening the door to a possible full swing to BR3. Obviously I'd be giving up a lot of options with such a small bet but I like the idea of virtually locking out BR3 - making it a two-man battle.

 

Between $85 to $210

Any bet between $85 to $210 will work here as BR2's bet (having to bet first). As far as assuming Ken would be to your left as BR1, there is NO assuming here it was posted that is the situation.

The $85 to $105 bet does give you the benefit of DD if needed against BR1 and still cover's BR3's should you lose the DD and they push.

I understand leftnuts $15 play, but I'm not crazy about this play because it allows BR3 another chance to advance with BJ/win and win/loss. By simply betting $85 to $210 bet you remove the BJ/win option's.

Addressing Monkeysystems $285 rainbow bet, again I understand, but I don't want to give away another chance for BR3 to beat me with a win/push or win/loss. I prefer to try and lock out BR3 if possible. Plus by making this bet I am giving up my option of spliting if needed.

Bottom line is you need to get lucky on the last hand to win. We've all seen how strange things happen in tournament play and depending on who get's the luckest is who ends up winning. Of course increasing your odds helps with getting lucky.

 

My logic was that the $15 bet removes the roughly 12% chance of BR3 getting a swing, even though it exposes me to the roughly 4% chance of his getting a BJ and sending me to the rail. Since the choice is one or the other, I like 4% better than 12%.

 

Br3

Actually I hope I'm BR3 in the first four rounds at Winstar next Sunday, make it to the finals and then have this problem to deal with...lol.

 

Its interesting that no one thought about the table minimum 5$ bet.
I'm not saying that I would bet 5 but think about this scenario where you bet 5.
Lets compare with both the players separatly
5$ bet:
BR1 Wins - BR2 lost eitherway
BR1 Lost - BR2 wins eitherway
BR1 Push - BR2 is still in

I dont run any simulation softwares, but looking at this the probabilities are 4/6 favorable to BR2
BR3 is completly insignificant when you make that bet
6/6 favorable to BR3

so the total would be 10/12

210$ does the same thing but does not cover the BR1's Push. The reason I thought about this is because there are so many instances where I had lost the final hand because the other player pushed and I lost.(on 17's and 20's mostly)

Also considering the fact that in any blackjack game the probability of the dealer's wins is more when compared to the player

 

My bad, I didnt read the minimum bet (10) correctly. But I still would like to hear any opinions from the experts if the table minimum is 5 instead of 10

 

I think in the spur of the money I would quickly narrow down to two choices: $10 and $210. I like to think I'd pick the $210, but I've made bigger mistakes than that one! Obviously if BR3 has just a few more dollars then $210 is the easy choice. But as it is, $10 is pretty appealing bet when you are thinking quickly. It certainly gives you >50% chance of winning, which is pretty attractive on a final hand with 3 people left.

I wouldn't put BR3's odds of going all in at 20%. I've played with a lot of terrible players -- usually high rolling Asian baccarat players that have no idea what they are doing in terms of blackjack, let alone tournament strategy. But I think 98% of people would be smart enough to min bet there, or at least keep more chips than BR1 and BR2 if they mostly go all in. It would be pretty obvious to anyone except someone extremely dense. They'd be thrilled, actually -- since they would feel "out of it" up until that point being dwarfed by the larger bankrolls.

 

Just saw this thread. So post my opinion here.
-----------------------------
BR2 has $420, bets $420
BR1 has $425 (Ken)
BR3 has $200
-----------------------------
I think the best bet in this situation is $20. Sounds weird? Let me explain.

The optimal bet here should be that you can make sure you lock out one opponent, while taking either high or low against the other.

1) If you match BR2's bet here by betting $420, you will have both the high and the low against BR2. But what about BR3? It is 100% sure he will secure a low by holding one more unbet chip than both BR2 and you.

Obviously betting $420 can not accomplish the aforementioned optimal bet.

2) If you bet $210, you are going to have two problems: A. If you can not double or split, BR2 will secure a high against you. B. Meanwhile, BR3 can play for a swing against you.

Obviously betting $210 can not accomplish the aforementioned optimal bet either.

Now get back to the point: a bet of $20.

With this bet, you are almost certain to lock out BR3, no matter what he bets behind you and what the outcome is. His best possible hand is a Blackjack if he goes all-in, which we simply ignore.

Meanwhile, with this bet you have effectively secured a low against BR2.

Therefore, a bet of $20 accomplishes our optimal bet: locking out one opponent while securing a low against the other. This is the best possible choice you can make under this condition.

* On a sidenote: we always consider the most likely situation, that is, we put a single win probability ahead of double, split or blackjack.

 

In a perfect world maybe

Garygo I understand your thinking, but you leave yourself wide open for so many possible up sets and NO WAY to recover if you need to.

You need to factor in all the possibilities and make your highest percentage play based on your situation. Your $20 bet paints you in a corner and unless things play out perfectly for BR1.

Lets face it any one of the three players can still win the table and advance if they get lucky, which is what we need on the last hand 96% of the time.

Even a bad bet could win for you if the situation plays out in your favor, but what we are looking for is the highest percentage play to help you advance in the situation described.

All of us that have played tournaments long enough understand it isn't over till it's over. Going into the last hand you can have first high and low and the last bet and still lose to the lowest chip count on the table if you get swung. This is the beauty of the tournaments blackjack, all we can do is play the highest percentage with our bets and actions, but it is left up to the fate of the cards we are dealt to how we finish up.

 

Thanks for your thoughts, TXtourplayer.

A bet of $20, as explained above in detail, is the highest percentage, or optimal, play in my view.

In this situation, no matter how you bet it is certain one of your opponents can have either the low or the high against you, then you simply accept it while making sure at least you are a lock to one of them. It is a better choice.

 

Not a lock

Garygo a $20 bet doesn't lock out BR3, should they hit a blackjack all in you're screwed! This is my point, you cannot lock out BR3 with the $20 bet. However with a bet of a minimum of $80 to could cover BR3's possible BJ.
Now knowing you need to bet at least $80 to lock out BR3. Why not bet $210, that will cover wins by BR1 & BR3, losses by BR1, BR2, & BR3, push/win by BR1 & BR3, and allows BR1 the chance to DD or split if need to against BR2.
Against one other player taking the low isn't a bad play, but when two players can beat you (or more) your best play is to take the high, or in this case give yourself the low and a chance at the high as well.

 

TXtourplayer - I am afraid this is not true. I think I have explained clearly in my preceeding comment.

1) The probability of a Blackjack is low so when you place the $20 bet, you are ALMOST a lock. If you bet $80 instead, you will risk a swing by BR3 (when he wins his SINGLE all-in bet and you lose). A SINGLE swing win by your opponent is MORE likely than a BJ win. That's why I said we simply ignore the BJ win possibility and treat it as a LOCK (unless BR3 gets VERY lucky).

2) Betting $210 only ENSURES that BR2 has the high against you, that is, if he wins you are done (Of course you can double/split to take the high back, but like I said, we always put the SINGLE win probability in the first priority). And guess what? Even if you take your high back, you are STILL exposed to BR3's low.

Also, betting $210 exposes you to a possible SINGLE bet swing by BR3 (any bet above $20, more accurately, $25, will do the same to your fate).

So a bet of $210 offers you TWO worries. You suffer not only a BR2's high, but also a BR3's swing.

3) Now we can easily see the advantage of betting $20: in this case, you can simply IGNORE BR3 as if he was dead. The only concern left is BR2. Once he loses his bet, you advance. You have only ONE worry. Simple as that.

When two players can beat you and you can not lock out any of them nor have both high and low over all of them, your best play is to take the high. This is right. But when you can lock out one of them while STILL having the low against the other, this is your best play.

I think this is the CORRECT and LOGICAL thinking process in this case.

Thank you for your thoughts.

 

Ok, now for this situation, if we use the same logic and thinking process, it is not hard to work out the best bet for BR2: $15.

1) This bet locks out BR3 (unless he gets VERY luck to receive a Blackjack for his all-in, which we ignore). If we place a higher bet, we will risk a SINGLE bet swing by BR3.

2) This bet opens an opportunity for BR2 to double/split to secure a high when BR1 matches his bet (Betting behind, BR1 will be able to match BR2's bet anyway).

You are BR2 and must bet first, and even worse, only 1 player advances. This is the best choice you can make as an underdog - at least you take the full advantage of another underdog: BR3.

 

Regarding the first part, can I refer you back to an earlier post of mine ... https://www.blackjacktournaments.com/posts/40858

I calculated that 20 is better than 420, if you assume that BR3 is 100% certain to take the low if offered. But nothing is 100% certain, and Ken estimated that this particular player was only 80% likely to take the low ...

As for the choice between 210 and 20 (if you assume that BR3 is a strong player and so the 420 bet is excluded from consideration), I think the factor you are not considering, Gary, is the benefit you gain from seeing how BR3 bets after you, BR2's final hand total, and the cards BR3 receives, before deciding whether or not to double down.

It's true that there may be times when you curse the fact that you are compelled to double down on a hard 19, but the overall picture, the sum of all the ways the round could play out, probably gives you more ways to advance than you get by just taking the low against BR2 (and BR3's possible BJ).

 

welcome to the board

Garygo, welcome to the board.
Your enthusiasm and convictions are appreciated.
In this thread some experienced and very sophisticated players expressed their opinions. It is a privilege and great help to be able to get their estimations. Though, we all make mistakes, when dealing with specific situations, real and concrete numbers can be applied.
I would be careful stating that a play is “optimal” based on one aspect of the game, though it may be the most common, or the most obvious. Optimal play is one that is the best when you add up all possibilities all together.
Your thinking is very logical, but not necessary correct. Many tournament plays have very close outcomes and only exact analysis, that includes all elements, can determine “the optimal play”.
In the heat of the game, when we grasping for time and can’t add up all elements finding the most common occurrence, or basing play on one “strongest” result is often the best we can do. But here, on this board, we can be a bit more precise if one has the interest and time to do so.

In the specific case there were three quite popular bets. Betting 20, 210, and 420. We may estimate pretty close the end results for all three players (db win, bj, win, push, loss, and db loss), looking for all combinations that would advance BR1.

One advances, last hand, no surrender
BR2 has $420, bets $420
BR1 has $425 (Ken)
BR3 has $200

(your preferred) bet 20
BR1 advances:
BR2 loss, BR1 any, BR3 no bj,
also, BR2 push, BR1 push or win, BR3 no bj
within less than half percent, depending on players expertise, but “an okay” play = .51*.955= 48.5%

Here are all combinations that advance BR1 for two other bets. Try to add up all the chances and come up with an approximate number expressed in %. Others and I will be glad to give you some specific percentage numbers if you want them.

bet 420
BR1 advances:
BR1 win, BR2 no bj (except BR1 also bj), BR3 any,
also BR1 push, BR2 push/loss, BR3 no bj
also BR1 loss, BR2 loss, BR3’s bet is 200 and loss

bet 210 (this one would be most fun to play)
BR1 advances:
BR2 win, BR1 win double bet, BR3 any,
also BR2 push, BR1 win or push (if pushed then BR3 no bj), BR3 any,
also BR2 loss, BR1 win or push (if pushed then BR3 no bj), BR3 any
also BR2 loss, BR1 single loss, BR3 push or loss

And one more addition, if you bet small, bet 10 instead of 20.
Betting 10 allows you to “offensively” split, to end up with a total money bet pushed, which advances you if BR2 pushes and BR3 wins. Losing 10 bucks bet split when BR2 loses but BR3 wins still advances BR1, but losing 20 chips bet split when BR3 wins does not.

S. Yama

 

Thanks for your comments London Colin.

Only 80% likely? It is a very high percentage. We ONLY consider high percentage probability when placing our bet and ignore that 20% chance. Yes most reasonable players in BR3's situation will definitely take the low here, when seeing BR1 and BR2's big bets. I don't think this is a hard call.

Since BR3 will secure an absolute low if you bet big (i.e. matching BR2's all-in bet), you should bet small to lock him out first, while securing an absolute low against BR2.

I am afraid you've complicated situations here. Let's think about it in a simple way: we are trying to make an optimal bet based on the infomation available and the most probable outcomes.

Now get back to my proposed bet of $20: 1) with this bet we don't need to worry about BR3 - he is ALMOST done (most probable). 2) with this bet we don't need to worry about BR2's final hand total - he is done once he loses. In other words, we only need BR2's loss to advance. Is there anything simpler than that?

If BR1 bets $210, he will be out if BR2 wins, or if he loses and BR3 wins. Both opponents pose a very probable threat to him, no matter he wins or loses. No lock.

I think even a bet of $420 is superior to $210 in any situation, as BR1 only needs to win his hand to advance. If he wins, he is ALMOST a lock. However, the weak points of $420 bet are: 1) it is not a pure high as BR2 can get a blackjack to win; 2) it is slightly more likely you will lose your hand due to the imbeded House Edge in every hand; 3) BR3 is sure to secure a pure low - no lock.

A bet of $20 offers both a pure low against BR2 and a lock to BR3. That's the beauty of simplicity.

 

garygo,
it is not about "almost", threats, ignoring things, or being exposed to something, etc..

I gave you all things to consider. Look at it and have fun dealing with the specifics...

Actually, there is one more aspect benefiting bets 210 or smaller, who will be first to find it?

S. Yama