Odds of House to qualify?

Hi,


Can anyone tell me the odds of the house to qualify in Caribbean Stud Poker or similar? (to have at least AK)
wizardofvegas /games/caribbean-stud-poker/


I tried to search the answer, but couldn't find (sorry), so I will try my luck with a new thread.

I see the table shows 0.222501 in the "dealer doesn't qualify", so does that mean that there is ~78% for the dealer to qualify?

Regards,

I found that for the House to qualify with a pair or better there is a ~49.9% chance. What is the chance for the House to have an AK in the 5 cards?

Quote: Eufrazerijus

Hi,


Can anyone tell me the odds of the house to qualify in Caribbean Stud Poker or similar? (to have at least AK)
wizardofvegas /games/caribbean-stud-poker/


I tried to search the answer, but couldn't find (sorry), so I will try my luck with a new thread.

I see the table shows 0.222501 in the "dealer doesn't qualify", so does that mean that there is ~78% for the dealer to qualify?

Regards,

From the context of the table, I see that the Wizard's number 0.227385 at https://wizardofodds.com/games/caribbean-stud-poker/ for "Dealer doesn't qualify" is shorthand for "Dealer doesn't qualify after the player didn't fold."

Ignoring player's folding, I get a probability of 56.3187% that the dealer will qualify (at least AK with five cards.)


Here are the numbers I used to get 56.3187%:

The number of combination for the dealer's getting less than a pair is 1,302,540 (https://wizardofodds.com/games/poker/)

The number of ways for the dealer to get AK (including flushes, straights, and royals) with no pair is: 4 * 4 *combin(11,3) * 4^3 = 168,960.

The number of ways for the dealer to get a flush (including royals) containing AK is: 4 * combin(11,3) = 660.

The number of ways for the dealer to get a straight (including royals) containing AK is: 4^5 = 1,024.

The number of royals is 4.

So, the number of ways for the dealer to get just AK is: 168,960 - 660 - 1,024 + 4 = 167,280.

Then the number of ways for the dealer to get at least AK is: combin(52,5) - 1,302,540 + 167,280 = 1,463,700.

So, the probability of the dealer getting at least AK = 1,463,700/2,598,960 = 56.3187%

Wow, thank you so much! I really appreciate your answer. Any chance you could help with this then?:

When is it worth playing another player's "bad" hand (no pair)? As the ante for me would be free, I thought maybe it's worth calling the ante (e.g. win 10 or lose 10).

With the probability being 6.3187% above fifty-fifty, how much would I lower the odds if the visible house card is not A or K and players have 1, 2 or all 3 same cards as the visible house card? That seems harder to calculate :(

Many thanks,

If it's worth you playing then it's worth the original person playing.

You (or they) are placing £2 to either lose that £2 or win back the Ante £1 and its payout £1. Thus it has to be more likely than 50% the Dealer will not qualify to bet (it can be slightly lower if you have an AK-hand which might win). The original player has exactly the same decision as you, except it's possible you know something (e.g. you have AK and trips of the upcard, joke).

The chances of the Dealer making a hand, given you don't know their upcard, is as you described. However you have various additional pieces of information (i) the upcard (ii) your hand. Against an A or K you cannot just call on anything, the chances of the dealer qualifying are usually too high. However against a lower card having an A K and one of the upcards makes it a close call.

In essence, unless you realise the other person is folding too often, then by all means look at their cards to help your own close decisions, but otherwise ignore what they then do if they are playing correctly.

Quote: Eufrazerijus

Wow, thank you so much! I really appreciate your answer. Any chance you could help with this then?:

When is it worth playing another player's "bad" hand (no pair)? As the ante for me would be free, I thought maybe it's worth calling the ante (e.g. win 10 or lose 10).

With the probability being 6.3187% above fifty-fifty, how much would I lower the odds if the visible house card is not A or K and players have 1, 2 or all 3 same cards as the visible house card? That seems harder to calculate :(

Many thanks,

For a very simple example, suppose the dealer's up card is 7. Ignoring all other cards, I calculate the dealer's probability of getting AK or better is 0.546. But if you know a 7 is in a player's hand, the dealer's qualifying probability drops to 0.507.

But if you know the players have two 7's, the dealer has only a 0.465 chance of qualifying. And if the players have all three of the other 7s, the dealer has a 0.417 chance of qualifying.

So, it looks like your idea is good. If your friend wants to fold, then consider buying his hand if you know the players have at least two of the dealer's up card rank (but not when the up card is A or K.)

Of course these simple calculations ignore all the other cards besides the dealer's up card.

Wow, you did it again ChesterDog! I'd like to thank you with more than words!

By the way, are you doing this for fun?