What would you do?

TXtourplayer

TXtourplayer

Executive Member
This is the situation:

You are in 3rd place at the MD III tournament, last hand and you have the puck. Here are the players totals and bets for hand 30, how would you play out the hand in seat number one (BR3). The question is do you take the lead and stay on the Blackjack or DD?

Seat...........Totals............Bets.............Cards

1 (BR3)........8,900...........2,500.............10/A

2 (BR5)........7,800...........2,500..............6/2

3 (BR1).......10,300.............500.............10/4

4 (BR2).......10,250...........2,500.............Q/3

5 (BR4).........7,400...........2,500.............3/K

Dealers up card is a 3.
 
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Close call

I'm not sure what I would have done at the table, with limited time.

Using the Dbl Outcomes chart and the Dealer Outcomes chart, this looks like a close call. Please double-check my process here, and see if I'm off track anywhere.

If you stand, seat 2 should double and seat 4 should stand. You advance unless the dealer busts. (62.6%) There's also a chance that seat 4 might double even though he shouldn't. Since he's already going to have a higher total than you, anything he does to misplay the hand adds to your percentage.

If you double, you're sure to advance if you win (59%), but you can also win with a push of 19,20, or 21. They would be sure wins except for the fact that when you get a good card BR2 is likely to double instead of stand. Even if the pushes with 19-21 were sure wins, that would add only another 6/13 X 7.5% = 3.2%, making doubling a 62.2% proposition for you.
It appears standing with blackjack was slightly better than doubling.

Any errors or oversights in my thinking?
 
Limited time is right!

My thoughts were the same as yours.

I was counting BR1's chips as the cards were being dealt out as I looked down I saw I had hit a BJ. Excited I quickly looked around the table to find BR1 had made a HORRIBLE bet. Know by staying on my BJ I over took BR1 for 1st place and everyone esle had to beat me from that point.

If I DD I could take myself out of the top top, so I found it to be a NO BRAINER to just keep my BJ.

The dealers Busted and I ended up in 3rd ($300) behind.

Woulda, Coulda, Shoulda.
 
How to play hand!

As a non-pro, my thought's based on all players being experts and the count being neutral:

Seat #2 has to use Curt's revenge and therefore double down.
With dealer showing a 3, all or most of the other seats will also
double.

Therefore, I would double the BJ vs 3.

How bad is my logic?
















:
 
Totals means the bankrolls including the bets, right?


Scenario 1:
You stand on your BJ and have sure 8900 + 2500 * 1.5 = 12650 chips.

Player 2 has to double his 8 and win in order to end up with 12800 chips. The probability for this is about 43.2%.

Player 3 is locked out even with a won double (11300).

Player 4 can stand on his 13 and has 12750 if the dealer busts (probability about 37.2%)

Player 5 is locked out even with a won double (12400).

You only need to be among the first 2, so your chance of advancing ist 62.8%.


Scenario 2 - you double

If you win your double, you're through. This will happen if the dealer busts (37.20%) or the dealer has 20 and you 21 (0.1222*0.3056=3.73%) or the dealer has 19 and you 20 or 21 (0.1248*0.3854=4.81%) or the dealer has 18 and you 19, 20, or 21 (0.1313*0.4651=6.11%) or the dealer has 17 and you 18,19,20, or 21 (0.1330*0.5449=7.15%).
That's 37.20+3.73+4.81+6.11+7.15=59%.

If you have a standoff with the dealer, it'e getting real complex. This will happen with 21-standoff (0.3056*0.1167=3.57%), a 20-tie (0.0797*0.1222=0.97%), a 19-tie (0.0797*0.1248=0.99%), a 18-tie (0.0797*0.1313=1.05%), and a 17-tie (0.0764*0.1330=1.02%). 3.57+0.97+0.99+1.05+1.02=7.60%.

Player 2 will probably double and - if he wins - is locked (43.2%).

Player 3 has you locked out even if he loses. And he won't double. :-)

Player 4 will stand on his 13 and has you when dealer busts (37.2%).

Player 5 will double and beats you when he wins (about 30% - too lazy to calculate exactly right now).

So you advance with a tie only in 56.8*62.8*70=25%.


That makes your overall advance chance with a double about 59% + 7.6*25% = 60.9%.


Your first thought not to double was right. The only problem is, how long do you have time to make your decision and how good are you when it comes to calculating probabilities without computer support?
 
Selecticom assumptions

Not a bad process selecticom, but I have two issues with your post...

"Player 4 can stand on his 13 and has 12750 if the dealer busts (probability about 37.2%)"

Player 4 won't, or at least shouldn't, stand if both players 1 & 2 double.

The other problem is that you unfortunately can't multiply out the probabilities like this: "So you advance with a tie only in 56.8*62.8*70=25%."

The 56.8, 62.8 and 70 numbers are not independent events. They're all correlated because of the common dealer that all these players are competing against.

For example, your percentage of success for player 2's double (43.2%) and
player 5's double (30%) are mostly overlapping when the dealer busts. You cannot determine the probability of neither one winning their double like this:(1.0 - .432) * (1.0 - .30) = 39.8%
Nor could you determine how likely they are to both win the double as .423*.3. Neither is accurate.

Instead, you must look at the probability of each dealer outcome and determine the effect on both players simultaneously. Tournament probabilities get unwieldy in a hurry.
 
Quick fix

I first added my total and knew that BR1 couldn't catch me with DD, next I tried to fiqure what BR2 had to do to beat me. That is about all the time you have in which to make you play.

My main thought was I have the lead, BR1 is out of the running so I had to ride it out (the BJ). Not that I want to lose, but I'd rather get beat this way then have the lead and give it away on a DD when I didn't need to.

It was a bad beat, it happens to all of us. It just wasn't meant to be this time. Just goes to show how much luck plays in tournament play.
 
math excess

OK, Ken, you got me!

Player 4 won't, or at least shouldn't, stand if both players 1 & 2 double.
Click to expand...
Of course you are right - no matter what outcome the doubles have.

The other problem is that you unfortunately can't multiply out the probabilities like this: "So you advance with a tie only in 56.8*62.8*70=25%."

The 56.8, 62.8 and 70 numbers are not independent events. They're all correlated because of the common dealer that all these players are competing against.
Click to expand...
True again, I admit.

For example, your percentage of success for player 2's double (43.2%) and player 5's double (30%) are mostly overlapping when the dealer busts.
Click to expand...
Even better - when the dealer busts, the whole calculation is obsolete. We are considering the case that Player 1 has a standoff!

So is goes this way:

Player 3 has me inevitably. I need to be ahead of the all doubling Players 2, 4, and 5.

Dealer 21 - Player 1 21:
Only Player 4 can beat me and only with doubling to 21 (92.0% advance chance).

Dealer 20 - Player 1 20:
If at least 1 out of the 3 players 2, 4, and 5 double to 21, I do not advance. Also if player 4 doubles to 20.
Player 2 has 8, needn't be considered.
Player 4 20 or 21: 15.9% -> player 4 below 20 or busted: 84.1%
Player 5: 21: 8.0% -> player 5 below 21 or busted: 92%
-> Advance chance ((player 4 below 20 or busted) and (player 5 below 21 or busted)): 77.4% (These events /are/ independent, aren't they?)

Dealer 19 - Player 1 19:
Almost same as above.
Player 4 19 to 21: 23.6% -> player 4 below 19 or busted: 76.4%
Player 5 20 to 21: 15.9% -> player 5 below 20 or busted: 84.1%
-> Advance chance ((player 4 below 19 or busted) and (player 5 below 20 or busted)): 64.3%

Dealer 18 - Player 1 18:
Almost same figure, but player 2 can also beat me with an ace...
Player 2 19: 7.6% -> player 2 below 19: 92.4%
Player 4 18 to 21: 31.6% -> player 4 below 18 or busted: 68.4%
Player 5 19 to 21: 23.6% -> player 5 below 19 or busted: 76.4%
-> Advance chance ((player 2 below 19) and (player 4 below 18 or busted) and (player 5 below 19 or busted)): 48.3%

Dealer 17 - Player 1 17:
Player 2 18 or 19: 38.2% -> player 2 below 18: 61.8%
Player 4 17 to 21: 39.2% -> player 4 below 17 or busted: 60.8%
Player 5 18 to 21: 31.6% -> player 5 below 18 or busted: 68.4%
-> Advance chance ((player 2 below 18) and (player 4 stiff or busted) and (player 5 below 18 or busted)): 25.7%

Now multiplied with my formerly calculated respective standoff probabilities:
0.0357 * 0.920 +
0.0097 * 0.774 +
0.0099 * 0.643 +
0.0105 * 0.483 +
0.0102 * 0.257 = 5.2%

Overall advance probabilityfor player 1 when doubling (Ken, please correct me): 59.0% + 5.2% = 64.2%.

Interesting. Seems like you shoulda doubled, Rick. :rolleyes:

Tournament probabilities get unwieldy in a hurry.
Click to expand...
Too true! Some surgeon with us? I would very much appreciate a cyborg-style coprocessor for my next tournament...
 
6 one way, half a dozen then the other!

I think having to do it all over again I would still stay on the Blackjack.

My main reason is "I would have felt like a dumb-ass had I given up the sure lead" for a maybe needed DD chance and lost.

You could have just stuck a fork in me...LOL!

O'well I had good luck on the regular tables anyway so I'll try to make it back for another shot at making the May finals.

Aces & Faces to all!
 
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