Poll

No votes (0%)
2 votes (40%)
1 vote (20%)
No votes (0%)
No votes (0%)
1 vote (20%)
1 vote (20%)
1 vote (20%)
No votes (0%)
1 vote (20%)

5 members have voted

terapined
terapined
Joined: Dec 1, 2012
  • Threads: 86
  • Posts: 5802
June 5th, 2018 at 9:17:16 AM permalink
Quote: PapaChubby

News on the expansion: Tampa Hard Rock Casino Expansion

Is the Asian food place to which you refer the noodle bar in the corner of the Asian gaming room? It was open when I was there a couple of weeks ago, and packed as usual. Maybe they're expanding, as they only had room for about 15 customers at the bar (and I could never find a seat).


Yes, thats the place I was referring to.
Spoke to another pit boss, i asked him if he knew any casinos with Saigon 5 and he did not know
"Everybody's bragging and drinking that wine, I can tell the Queen of Diamonds by the way she shines, Come to Daddy on an inside straight, I got no chance of losing this time" -Grateful Dead- "Loser"
gordonm888
Administrator
gordonm888
Joined: Feb 18, 2015
  • Threads: 50
  • Posts: 3282
June 5th, 2018 at 9:45:09 AM permalink
The hand: Joker-10-9-7-4

has 4 different ways to qualify, yielding point totals of 3,6,9 and 9! It might be easy for a player to not notice the highest scoring arrangement. So, there is an element of skill involved on a small number of hands that contain a joker.

Edit: The hand Joker-10-7-4-3 has 5 ways to qualify!
Last edited by: gordonm888 on Jun 5, 2018
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
charliepatrick
charliepatrick
Joined: Jun 17, 2011
  • Threads: 34
  • Posts: 2446
June 5th, 2018 at 12:53:47 PM permalink
?T743 : Interesting that the way to find the highest total is to add all 5 and see if you can get to a total of 10x. T743 add up to 24, so with ?=6 this can be achieved. Provided you can make 10 with two cards, that must be the maximum. Thus ?T4/73 or T73/?4 work (Joker=6). ?T7/43 or T73/?4 give a 7-hand (Joker=3), ?74/T3 gives a 3-card (Joker=9).
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1392
  • Posts: 23448
June 5th, 2018 at 12:56:37 PM permalink
Quote: terapined

No Saigon 5 card at the Seminole Tampa Hard Rock



Thank you for checking.

To be honest, I'm not surprised authentic Chinese food is a bit difficult to find in west Florida. Your next Vegas visit, I'll treat you to some good stuff.
It's not whether you win or lose; it's whether or not you had a good bet.
gordonm888
Administrator
gordonm888
Joined: Feb 18, 2015
  • Threads: 50
  • Posts: 3282
June 6th, 2018 at 10:25:11 AM permalink
I was wondering if the player will ever have a choice between arranging his hand so that it makes either 5 points or a lower point total 1-4. Because a 5 point hand has a 1:2 payout, I wondered whether optimal strategy would involve the player actually choosing the hand arrangement with the lower point total.

Quick summary: No. Choosing the arrangement of a hand with the highest point total is always correct.

I found 8 distinct hands in which it is possible to arrange the cards so that the hand is either 5 points or 1-4 points.

5 or 1 points: WT651
5 or 2 points: WT432, WT321, W8641, WT775, W8774, WT522 and W9322

For example, consider WT432:
5 points: WT4-32 EV= -0.03514
2 points: W43-T2 EV= -0.13524
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1392
  • Posts: 23448
June 6th, 2018 at 11:18:06 AM permalink
Quote: gordonm888

I was wondering if the player will ever have a choice between arranging his hand so that it makes either 5 points or a lower point total 1-4. Because a 5 point hand has a 1:2 payout, I wondered whether optimal strategy would involve the player actually choosing the hand arrangement with the lower point total.



Good question.

One one thing, I'm not even sure if the player is even allowed to set his hand. In this online game it is done automatically.

However, if you could, here is an estimate of the expected value of each total. This is based on independent decks between player and dealer.


Points EV
0 -28.70%
1 -22.70%
2 -19.12%
3 -6.71%
4 5.66%
5 -9.47%
6 31.49%
7 44.67%
8 59.20%
9 74.83%
10 91.41%
Does not qualify -68.65%


What this seems to show is that playing 3 or 4 would beat a 5 in expected value, but not 2 or less.
It's not whether you win or lose; it's whether or not you had a good bet.
gordonm888
Administrator
gordonm888
Joined: Feb 18, 2015
  • Threads: 50
  • Posts: 3282
June 6th, 2018 at 11:24:40 AM permalink
Quote: Wizard

Good question.

One one thing, I'm not even sure if the player is even allowed to set his hand. In this online game it is done automatically.

However, if you could, here is an estimate of the expected value of each total. This is based on independent decks between player and dealer.


Points EV
0 -28.70%
1 -22.70%
2 -19.12%
3 -6.71%
4 5.66%
5 -9.47%
6 31.49%
7 44.67%
8 59.20%
9 74.83%
10 91.41%
Does not qualify -68.65%


What this seems to show is that playing 3 or 4 would beat a 5 in expected value, but not 2 or less.



Yes, those are generic EVs not considering the five cards in your hand. However, all hands that can be arranged to make two or more different point totals have a Joker in them. Removing the Joker from the deck will change all of those values by a considerable amount. FYI:


]
Points EV with 53 cards EV -Joker removed
0 -28.70% -32.89%
1 -22.70% -26.58%
2 -19.12% -14.11%
3 -6.71% -0.01%
4 5.66% 13.08%
5 -9.47% -3.24%
6 31.49% 40.02%
7 44.67% 52.29%
8 59.20% 65.51%
9 74.83% 78.20%
10 91.61% 92.43%
Does not qualify -68.65% -66.44%


Not really sure what a zero point hand is, given that DNQ is a separate entry?

Anyway, your basic advice appears to be perfectly correct. A 5 point hand is worse than a 4 or 3 point hand.
Last edited by: gordonm888 on Jun 6, 2018
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
beachbumbabs
Administrator
beachbumbabs
Joined: May 21, 2013
  • Threads: 99
  • Posts: 14232
June 6th, 2018 at 1:40:05 PM permalink
If you got 5 ten-value cards, or 3 face cards and a 6-4, for example, all of those hands would both qualify and be worth 0. There would be a fair number of zero hands, I would think.
If the House lost every hand, they wouldn't deal the game.
charliepatrick
charliepatrick
Joined: Jun 17, 2011
  • Threads: 34
  • Posts: 2446
June 7th, 2018 at 3:25:33 AM permalink
I'm not sure how you can get a 0-point hand since you only subtract 10 if the total is over 10. A simple idea is If you add up all five cards and ignore the 10's then if you can qualify you are subtracting 10's so the last digit of the 2-card hand will be the same (or 10 if the digit is 0). Without a joker there is only one possible total, albeit you may be able to make it different ways (e.g. T7643 = T73/64 or T64/73; both total=30 so lo hand = 10).

With a joker you have a variety of totals which will be something like 32 35 38, 35 38 41, 29 32 35; thus the difference between them is 3 and if you can make a 5 then the other totals will include an 8 or 9 (as well as a 1 or 2).

Note According to the wizard write-up a picture card is worth 10 and all other cards are positive, thus two cards cannot add up to 0. The rule about exceeding ten says
Quote: https://wizardofodds.com/games/saigon-5-card/

the point value of the remaining two cards shall be the sum of the point value of the two cards, except if that total is greater than 10, then ten will be subtracted from the total.

Thus were people assuming two picture cards were considered as zero (e.g. TT866)?
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1392
  • Posts: 23448
June 7th, 2018 at 7:51:59 AM permalink
It's true you cannot get a zero point hand. I shouldn't have confused the topic by putting the line in my table. My value was simply equal to prob(dealer doesn't qualify)-prob(dealer qualifies).
It's not whether you win or lose; it's whether or not you had a good bet.

  • Jump to: