dvo125
dvo125
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February 20th, 2019 at 11:38:45 AM permalink
We're gearing up to go back Vegas which triggered reminiscence of our last trip. The highlight (or low light) of the trip was my buddy's meltdown at "The D" blackjack single deck. After a day of being thrashed by the house, he decided to watch us play blackjack. After the first shoe ended the house was down and my friend (we'll call him "G") reluctantly got in on the action.

Here's the scenario, and the first hand that was dealth:

Single Deck BJ, 3/2, at "The D"

The cards were dealt:

Dealer - 6, Facedown
"G" friend - 8,8
Me - 9,K
Other Friend, Q,K

"G" proceeded:
-Split 8's
-Dealt another 8
-Re-split, Dealt Another 8

"G" ended standing with the following 4 hands:
(1) 8,10 (2) 8,K (3) 8,2,9 (4) 8,5

Dealer flipped a 5 and hit for a 21.

"G" Proceeded by ordering a "quad" Jack and Coke. The rest of the night was a Blurr for him.

What are the odds on a single deck BJ table that friend "G" would get [4] splitting 8's then lose with 18/18/19/13 to a dealers 6?
OnceDear
OnceDear
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February 20th, 2019 at 12:01:08 PM permalink
Quote: dvo125

What are the odds on a single deck BJ table that friend "G" would get [4] splitting 8's then lose with 18/18/19/13 to a dealers 6?

It's a dead cert if you don't want it to happen.... Which, of course you wouldn't.
Those are the hands that make or break a session: Been there: Got the t-shirt. If it's not multiple re-splits, it's a mix of resplits and doubles that you just have to dip into your pocket for.
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
gordonm888
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gordonm888
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dvo125
February 20th, 2019 at 1:52:49 PM permalink
Quote: dvo125

What are the odds on a single deck BJ table that friend "G" would get [4] splitting 8's then lose with 18/18/19/13 to a dealers 6?



1. Given that your friend had (1) 8,10 (2) 8,K (3) 8,2,9 (4) 8,5 vs a dealer 6 in a single deck hand the odds of the dealer getting a 20 or 21 are

Dealer 20: 0.119
Dealer 21: 0.0998

so almost 22% chance of all 4 of those hands getting beat given a dealer 6.

If we take into account that your other friend and you held 9,K,Q,K in your two hands, the probability of the dealer outcomes was:

Dealer 17: 0.1388
Dealer 18: 0.1325
Dealer 19: 0.1391
Dealer 20: 0.1266
Dealer 21: 0.1035
Dealer Bust: 0.3595

or about 23% chance that all your friend's hands, which were 19 or lower, would lose..


2. Given an 8-8 pair vs 6 that is being split (with one deck), the odds of resplitting it twice to make four hands of 8 is roughly 0.8 %

When we take into account that that you and your other friend had 4 cards that were known to be 9,K,Q and K, means that the odds of the split 8-8 pair being re-split another 2 times was slightly higher: 1.3%.

3. However, we have not included the probability that none of the four re-split 8s would have made 20 or 21. I'm lazy and don't want to calculate that exactly, but hitting an 8 vs 6 does not often result in a 20 or 21. The only sequences that would do that are:

2, T -> 20
2,A ->21
3,9 -> 20
3,T ->21

Just making a rough guess (and ignoring all the other cards that one can see except the 4 8s, I would say that drawing to an 8 makes a 21 about 3.5% of the time and makes a 20 about 3.5% of the time. So the odds of having none of the 4 split 8's making a 21 was about 86.8% and the odds of not making a 20 or 21 in any of the four hands was about 75%.

4. So combining all this information, and accepting the short-cuts and approximations in section 3 above, if your friend starts with an 8-8 pair vs 6 the odds of resplitting it to quad 8s and then losing all four hands to a dealer 20 or 21 was roughly:

P = 0.008 * (0.22* 0.75 + 0.118* 0.0998 ) = 0.00141

or about 1 in 707.

5. Of course, this does not include the likelihood of being dealt an 88 pair vs a 6 in the first place, because we assumed that as the starting point. The odds of being dealt an 8-8 pair in single deck BJ is 0.00425, or 0.425%. An 8-8 vs 6 occurs with a probability of 0.00036, or 0.036%. So total probability of being dealt 8-8 vs 6 in single deck BJ and then re-splitting to quad 8s and losing all four hands to a dealer 20 or 21 was about 1 in 1.96 million.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
dvo125
dvo125
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February 20th, 2019 at 2:34:57 PM permalink
Hello gordonm888! Unbelievably impressive that you were able to compose this probability. Just sent this to the "guys" and we're getting a hell of a crack out of it. Love referencing that there was a nearly 3 time likely hood of being struck by lightning then my friend's experience.

Thanks again, appreciate the math and effort!

David
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