June 30th, 2016 at 11:20:48 PM
permalink

Good Day, Sorry for the trouble but I want to know how does on Wizzard of Odds compute for the ante return table.

On the Pair Plus I already understand how the Pair Plus return table being computed. But for the Ante Return Table I'm a little bit stuck

Can someone enlighten me on how to compute for this.

Player wins with straight flush. which has a combination of 617,044.

Thank you.

On the Pair Plus I already understand how the Pair Plus return table being computed. But for the Ante Return Table I'm a little bit stuck

Can someone enlighten me on how to compute for this.

Player wins with straight flush. which has a combination of 617,044.

Thank you.

June 30th, 2016 at 11:50:23 PM
permalink

Quote:ChenLingGood Day, Sorry for the trouble but I want to know how does on Wizzard of Odds compute for the ante return table.

On the Pair Plus I already understand how the Pair Plus return table being computed. But for the Ante Return Table I'm a little bit stuck

Can someone enlighten me on how to compute for this.

Player wins with straight flush. which has a combination of 617,044.

Thank you.

Hello, and welcome!

I think you're referring to this page:

https://wizardofodds.com/games/three-card-poker/

And, scrolling down, to the ante return table.

The computation takes more into account than just all possible starting hands for a straight flush in a single hand. That hand still must be compared to all possible dealer hands, including a few to which it would lose, and a very few against which it would push.

The seven unit win refers to the ante bonus in which the player beats the dealer, whose hand qualifies. There are 617K+ combinations of two hands where that can happen. The player then receives 1 unit ante pay, 1 unit play bet pay, and 5 units ante bonus.

The next pay at 6 units would include two possible situations. Player has 3of a kind and dealer qualifies, Or player has straight flush and dealer does not qualify, so the play bet does not get paid, but the ante bonus does.

The rest of the table reflects similar values, especially whether the dealer does or does not qualify.

But I think maybe your question was the large numbers of each type pay. And that is caused by comparing each possible dealer hand from the remaining deck to a particular possible player hand.

You can't just figure the probabilities of a hand occurring in isolation (which does work for the pairs plus and other paytable bonuses), because the player must beat the dealer to win the ante and play bets.

I hope that makes sense to you.

If the House lost every hand, they wouldn't deal the game.

July 1st, 2016 at 1:25:01 AM
permalink

Yes, but what i am interested is how did you came up with the 617,044 combinations. using combination formula or permutation. If you can give me some insights.

July 1st, 2016 at 1:33:55 AM
permalink

^ Good explanation.

btw it's easy to check the total of all the outcomes is correct.

Number of player hands = 52 * 51 * 50 / 6 = 22100.

Now the deck has 49 cards so for each player hand there are 49 * 48 * 47 / 6 possible dealer hands = 18424.

If you multiply these together you get 407 170 400.

The easiest scenarios to check are SF ties or loses to SF.

(i) There are 48 ways for the player to get a SF. The dealer can only get a similar SF in the other three suits, so has 3 ways to tie. Hence the total is 48 * 3 = 144.

(ii) If the player has AKQ there is no way to lose to dealer. With KQJ there are only three AKQ's (as the player already has one of the suit's KQ), so only 3 ways for the player to lose. WIth QJT it's 6 ways, then JT9 four ways to get AKQ and 3 for KQJ QJT, so it's 10 ways. Continuing this gets 3+6+10+...+38+38 = 242 ways for the dealer to win. But this can occur for each suit the player has - so the total number is 4 * 242 = 968.

btw it's easy to check the total of all the outcomes is correct.

Number of player hands = 52 * 51 * 50 / 6 = 22100.

Now the deck has 49 cards so for each player hand there are 49 * 48 * 47 / 6 possible dealer hands = 18424.

If you multiply these together you get 407 170 400.

The easiest scenarios to check are SF ties or loses to SF.

(i) There are 48 ways for the player to get a SF. The dealer can only get a similar SF in the other three suits, so has 3 ways to tie. Hence the total is 48 * 3 = 144.

(ii) If the player has AKQ there is no way to lose to dealer. With KQJ there are only three AKQ's (as the player already has one of the suit's KQ), so only 3 ways for the player to lose. WIth QJT it's 6 ways, then JT9 four ways to get AKQ and 3 for KQJ QJT, so it's 10 ways. Continuing this gets 3+6+10+...+38+38 = 242 ways for the dealer to win. But this can occur for each suit the player has - so the total number is 4 * 242 = 968.

July 1st, 2016 at 2:06:20 AM
permalink

can you still continue to explain until we arrived to this number 617044. sorry for the trouble

July 1st, 2016 at 2:31:33 AM
permalink

I can tell you how you can get to it.

Take each SF in turn (as they have different results).

Work out how many hands can be formed from the remaining 49 cards that

(a) Form Trips

(b) Form a Pair

(c) Form a Straight (or SF)

(d) Form no hand (or might form a Flush)

Work out how many of (c) and (d) are SFs/flushes.

Work out how many of (d) do not qualify.

Subtract the SFs where the player loses or ties.

Add the rest up for each of the possible SFs (AKQ, KQJ...32A) giving a total of qualify and non-qualify; then multiply by 4 for each suit.

Take each SF in turn (as they have different results).

Work out how many hands can be formed from the remaining 49 cards that

(a) Form Trips

(b) Form a Pair

(c) Form a Straight (or SF)

(d) Form no hand (or might form a Flush)

Work out how many of (c) and (d) are SFs/flushes.

Work out how many of (d) do not qualify.

Subtract the SFs where the player loses or ties.

Add the rest up for each of the possible SFs (AKQ, KQJ...32A) giving a total of qualify and non-qualify; then multiply by 4 for each suit.