Week One, World Series of Blackjack 2

you can do it yourself

I don’t know what you mean by: “Did you analyze the % solely on the basis of potential busting, or did you also factor in the odds of Joe/Kevin/dealer still beating me even if I did hit to an 18? “
Odds must be inclusive of all given data. I (kinda) didn’t see the show so I will not include the deck composition, tells,etc..

(I said I “kinda” didn’t see it because I was busy playing Tunica Jack Binion World Poker Open and the show was on the monstrous main screen 20’ by 30’, no sound though. Quite a few players were glimpsing at it with some interest. However, two minutes before 7 p.m., just as Joe hit the second split trey, casino switched the screen to a basketball game. Damn!)

Okay, here is a part of step by step procedure, assuming infinite deck. I will just compare hitting to 18 starting with 13 and starting with 15.
You take first place if you win your hand, regardless what happens to your opponents. You also take first place if you push and Joe pushes or loses (his second hand).
You take second place if you push but Joe wins, and if you lose but Joe loses or pushes, as well. That’s all there is to it.

Hitting to 18 starting with 13 will get you 18 through 21 –10.3% of the times and bust about 58.9%.
Hitting to 18 starting with 15 will get you 18 through 21 – 8,9% of the times and bust about 64.4%.
Now you have to go through each dealer’s final outcome and multiply it by your and other players chances of winning/pushing/ losing to find out how much is contributed to the your final placing. (It sounds more complicated than it is.)

Dealer 17 – 13.43% times 10.3% x 4 (this is your 18, and 19, 20 and 21). You can’t push dealer’s 17 if you hit to 18
Total contribution to first place if dealer has 17 is 5.56%

For dealer’s 18 – 13.05% times 10.3% (this is your 19, 20 and 21). You push the dealer 10.3% x 13.05 x 8/13 (this is Joe’s having less than 19)
Total contribution to first place if the dealer has 18 is 4.88%.

Follow up with dealer having 19 through 21 and dealer bust and you will get the first place 33.01% of the times.

Now, do the same calculation but replace (hitting 13 to 18) 10.3% by 8.9% (hitting 15 to 18) and you will get the first place 28.46% of the times.

Similarly, you have to calculate chances for the second place.
Just as an example I will do one if dealer ends up with total 19.
12.52% [dealer’s 19] x (1- 10.3%x3) [this is you losing] x 8/13 [Joe losing or pushing] plus
12.52% [dealer’s 19] x 10.3% [this is you pushing x 5/13 [Joe winning] = 5.81%

When I calculated it exactly you I would have gotten second place 33.43% if you started hitting 13 to 18 and 34.83% starting with a hand of 15.

Hope it helps.
S. Yama

 

Rock on

You're really awesome for spending so much time on this. Thanks again.

This is one point i am unclear on; if i win my hand, and joe also wins BOTH his hands -- i take 2nd, not 1st. This is what i agonized over on the second hit (with the 15) -- that I could hit, get a 3 or 4, and STILL lose to Joe, accepting a lot of unnecessary risk for what would turn out to be the same result (2nd). Not being able to see his double card prevented me from knowing exactly what total i needed to hit to in order to have a shot at winning 1st.

I'm pretty shure my odds on taking 1st were much lower when faced with hitting the 15, simply because all the dealer had to do was bust or hit to a total that would beat/tie my total, while joe's doubled hand won. For instance, if i had hit to a 19, and the dealer pulled a 19 -- joe woulda lost the stiff hand, won the doubled hand (he had a 20), and cleared me by just enough to take 1st anyway. In fact, in hindsight the best thing of all was that i couldn't see joe's upside-down 10 -- it would've forced me into hitting again.

BTW, how was Tunica? Coming to Commerce for the LA Poker Classic?

-holly d.

 

correction

My bad, I should have read the write-up more carefully. I was under assumption that Joe lost his first split hand. In reality he stood on stiff – this makes only one but significant change in the numbers. Whenever dealer busts Joe gets the first place and Dave drops to second place even if he wins his bet. All numbers look like this:

Play:…………………........……..1ST Place……….2ND Place……1ST or 2ND.……3RD Place
Surrender………….......……...….30.3%……….….19.0%……...…49.3%……...…50.7%
Stand Stiff…………........………….0.0%………..…56.4%……...…56.4%…….……43.6%
Hit to 18 (start with 13)……...17.5%………..…33.4%……..….50.9%…...…….49.1%
Hit to 18 (start with 15)…...…15.1%……....…33.8%…..…….48.9%….……...51.1%

It doesn’t take much time at all if you are using spreadsheets. It is then a matter of writing the data correctly and dragging cells with right formulas.

I will try to write something about poker in Tunica, time allowing, later tonight.

S. Yama

 

Math Blitz

Ai yi yi, my head. Now i don't understand how much all the OTHER numbers have changed --

how can i have over a 60% chance of 1st or 2nd place with Surrender when 37% of the time, the dealer will bust and i will take 3rd? and dealer 17 or 18 will result in the same 3rd place? that's 2/5 (40%) of the possible other results (63% of the time the dealer hits to 17, 18, 19, 20, or 21), bringing my best-case scenario of advancing in 1st or 2nd thru surrender no better than (60% x 63%), or 37.8%. The key being, the only way i advance thru surrender is if the dealer hits to exactly 19, 20, or 21 due to Kevin's 19 and an assumed/actual 20 (ten in the hole) for Joe's double card (if he wins his double while losing the stiff hand, and I push or surrender, I still lose to him).
In fact, the only scenario that puts me in FIRST if i surrender is a dealer doing exactly what she did -- hitting to 21, something that will only happen (20% x 63%) or 12.6% of the time.

Likewise, I 'm pretty sure those other numbers for hitting are overly optimistic as well, especially for first place -- i 've gotta win for sure, and if i bust or hit to a smaller total, the chances are much much higher of slipping into 2nd or 3rd.

A sticky issue, for sure.

-holly d.

 

Correct numbers don't lie.

I noticed I gave 11.49% (dealer’s 21) to both your second and first place if you surrender and it belongs only to you taking first place. I corrected it in the table in my previous post. All other numbers stay the same. The number representing the 3rd place is whatever is left from 100% minus first and second place. No doubt I should have double-checked the table. Hey, but we did it together.

Since you still have objections to percentage of advancing here we go:
Forget what Joe had for his double for making playing decision this is inconsequential.

If you surrender you win if:
Dealer has 19 and Joe has 19 or less on his second hand. 12.5% x 8/13 = 7.7%
Dealer has 20 and Joe has 20 or less on his second hand. 12.1% x 8/13 = 11.2%
Dealer has 21 and Joe has anything on his second hand. 11.5% x 13/13 = 11.5%
7.7+11.2+11.5 = 30.35%

If you hit to 13 to 18 you end up with 18 through 21-- 10.3% each.
You win if:
Dealer has 17 and you have 18 thru 21 and if dealer has 17 and you have 17 and Joe has 17 or less.
13.4% x 4 x 10.3% + 13.4 x 10.3% x 6/13 = 5.56%
Dealer has 18 and you have 19 thru 21 and if dealer has 18 and you have 18 and Joe has 18 or less.
13.1% x 3 x 10.3% + 13.1 x 10.3% x 7/13 = 4.88%
Dealer has 19 and you have 20 and 21 and if dealer has 19 and you have 19 and Joe has 19 or less.
12.5% x 2 x 10.3% + 12.5 x 10.3% x 8/13 = 3.49%
Dealer has 20 and you have exactly 21 and if dealer has 20 and you have 20 and Joe has 20 or less.
12.1% x 20.6% + 12.1 x 10.3% x 12/13 = 2.4%
Dealer has 21 and you have exactly 21 no matter what Joe has.
11.5% x 10.3% = 1.19%
5.56+4.88+3.49+.2.4+1.19 = 17.52%

For hitting 15 to 18 replace 10.3% with 8.9% and you will get 15.1%
Exactly like in my table in the post above.

S. Yama

 

This one's gotta go down in the history books

Thanks for the astute analysis. Looking at these numbers makes me realize just how f#cked I was with this decision. Pretty much damned if I do, damned if I don't. Funny thing is, looks like I could've squeezed a few extra % of EV out of standing stiff, an option i never even considered... !

Can't wait til this week's episode, & more interesting & educational analysis --

-holly d.

 

$$

Dave,
Remind me what happened to the second place winners, and what were the prizes (all of them) and we will have more fun with your episode and the numbers.

S. Y.

 

I believe this is correct?

Yama, I think this is the correct prize break down in the first round.

1st = $10,000 and advances to the semi's
2nd = $ 5,000 and advances to a playoff with three others
3rd = $ 2,500
4th = $ 1,000
5th = $ 500

Playoffs: 4 per table (2nd place finishers)

1st = $ 5,000 and advances to the semi's
2nd = 0
3rd = 0
4th = 0

Round # 2 (Semi's) 5 players

1st = $10,000 and advances to the finals
2nd = $ 5,000 and advances to the finals
3rd = $ 1,500
4th = $ 1,000
5th = $ 500

Round # 3 (Finals) 4 players

1st = $250,000
2nd = $ 50,000
3rd = $ 10,000
4th = 0

 

Money!

Let’s put the dollar signs to Dave’s decisions.
Be advise that this process involves some simplification.

Firstly, we will assume that Dave is about average in skills against his possible future opponents. And I know at least two people who would ardently disagree with this statement and with each other’s opinion, to boot.
Secondly, average numbers don’t work perfectly but close enough to make it worthwhile looking at. The reason is that simple understanding of the weight of each place/prize in the scheme of prize structure will make an average in bj tournament skills player winning more than average. For example (another simplification), with five players if you add the first and the last prize and compare it to the third, or the second, third and fourth divided by three, you will see if “make it or brake it” or more cautious approach is advised.
Thirdly, total EV for better than average player being resultant of playing in more than one round gets further increased.
And fourthly, the better the player the bigger the spread in values we should attribute to each decision.

Ken in his post above worked out the initial table payouts equivalent to this:
1st = $44,600
2nd = $14,900
3rd = $2500
4th = $1000
5th = $500

I think that in Dave’s case we should include a premium for each and every television appearance. Just for the reason of strengthening his “Hollywood” image I would make it worth $5K and double it for additional “babe factor” value. So, let’s make it worth additional $10,000 for the second and the first place that would allow him charm the audience again. (Mathematically, first place should be worth more as it provides more (easier) chances to appear once again in the final.)

So, Dave, when he found himself in position of playing last hand of the round, having a 12 versus dealer 3, against two other contestants, …we know their bets and cards – was faced with three basic playing options (I skipped nonsensical double down, and in parenthesis is the EV without $10K “appearance value” added) and their Expected Value was as follow:

Stand:……………………………….........…..$15,100……( $9,500)
Surrender:……………………………….......$22,600……($17,600)
Hit: (intending to hit to 18): …….……$19,100……($14,000)

Once Dave had hit, two options remained:
Stand:……………………………….........…..$15,100……( $9,500)
Hit: (intending to hit to 18): …….……$18,200……($13,200)

S. Yama

 

you sly dog you...

Gotta have that Babe factor! And very true, a real consideration for me is the EV of additional TV exposure because it obviously has a real effect on my career (being tapped for 2 seasons of Celebrity Blackjack, as well as other television shows). Funny you bring this up, in my last analysis (on the 'Is it true?' thread) i was going to include this indeterminant "IT-FACTOR" but thought the only person who would understand the mathematical value of it was YOU!

Funny you also mention my skill level vs. potential future opponents -- since this game was actually the LAST qualifying one filmed, i knew going into it who my opponents would be for the Wild Card round. And at that table, I knew there was only one person I needed to worry about (can't reveal that info yet cuz this person's episode hasn't aired -- needless to say, Charlene didn't worry me) -- so i put my chances of advancing out of the wild card at a minimum of 40%, rather than the 'average' 25%. Once you see the lineup for the Wild Card game, i am sure you will agree with me. And that definately affected my decision to stand as well, rather than risk losing it right there.

interesting how poker players think a little bit differently about the mathematical totality of the situation, assigning all sorts of subjective -- but very real -- values to different criteria. kudos to you --

-holly d.