Here's a details.
- Flush in 4 cards pay 10 to 1
- Flush in 5 cards pay 20 to 1
- Flush in 6 cards pay 50 to 1
how to math probability and edge in this option?
Are flushes the only way to win this bet?
Without other payoffs, this sounds very expensive.
No other pays, and there is more information about this option.
Any Flush can be mixed between any results(4~6 cards)
So 'Flush in 4 cards' can be made after any results(used total cards 4~6)
It's probably easier to work out all the combinations of cards that can make up a round. Then you can either work through each of them working out how many of each would have different number of suited cards, or you could count the types of hands (e.g. 6 of a kind, 5-1, 4-2, 4-1-1 etc.) and work out the numbers for each hand type. One trick is usually for Baccarat one doesn't care about the individual ranks of picture cards and tens, as they all count as zero, so you could also treat them as all the same and assume each deck has, say, four "picture" Spades etc.Quote: Jay171717...how to math probability and edge in this option?...
Using infinite deck, there's a 140608/13^5 = a% chance of getting a 4-card hand, 112560/13^5 = b% chance of getting a 5-card hand and 118125/13^5 = c% chance of getting a 6-card hand. So the return is:Quote: ThatDonGuyI did a simulation with 1,000,000 8-deck shoes, and the house edge is about 15.45%
link to original post
(4/4^4 * a + 60/4^5 * b + 540/4^6 * c) * 11 +
(4/4^5 * b + 72/4^6 * c) * 21 +
4/4^6 * c * 51
=88.0009% or an edge of about 12%
I'd be very surprised if the infinite deck calculation could be so much different than 8-deck...normally it's within a few basis points.
So I believe someone's calculation is wrong. Could very well be mine since combinatorics are not my strongest point
Quote: Ace2Using infinite deck, there's a 140608/13^5 = a% chance of getting a 4-card hand, 112560/13^5 = b% chance of getting a 5-card hand and 118125/13^5 = c% chance of getting a 6-card hand. So the return is:Quote: ThatDonGuyI did a simulation with 1,000,000 8-deck shoes, and the house edge is about 15.45%
link to original post
(4/4^4 * a + 60/4^5 * b + 540/4^6 * c) * 11 +
(4/4^5 * b + 72/4^6 * c) * 21 +
4/4^6 * c * 51
=88.0009% or an edge of about 12%
I'd be very surprised if the infinite deck calculation could be so much different than 8-deck...normally it's within a few basis points.
So I believe someone's calculation is wrong. Could very well be mine since combinatorics are not my strongest point
link to original post
I did a simulation with an infinite deck, and got around 12% as well. That does seem strange.
Quote: ThatDonGuyQuote: Ace2Using infinite deck, there's a 140608/13^5 = a% chance of getting a 4-card hand, 112560/13^5 = b% chance of getting a 5-card hand and 118125/13^5 = c% chance of getting a 6-card hand. So the return is:Quote: ThatDonGuyI did a simulation with 1,000,000 8-deck shoes, and the house edge is about 15.45%
link to original post
(4/4^4 * a + 60/4^5 * b + 540/4^6 * c) * 11 +
(4/4^5 * b + 72/4^6 * c) * 21 +
4/4^6 * c * 51
=88.0009% or an edge of about 12%
I'd be very surprised if the infinite deck calculation could be so much different than 8-deck...normally it's within a few basis points.
So I believe someone's calculation is wrong. Could very well be mine since combinatorics are not my strongest point
link to original post
I did a simulation with an infinite deck, and got around 12% as well. That does seem strange.
link to original post
Really? Wouldn’t there be a significant greater chance of getting a flush on infinite deck than with an 8 deck shoe? And since each flush has a payout which is a multiple of the bet amount, the few percent difference in house edge makes sense? I think?
I’ve calculated the house edge of baccarat using infinite deck and it’s a negligible difference from 8-deck
Quote: Ace2In eight decks there are 104 cards of each suit. The effect of removing 4-6 cards shouldn’t be significant
I’ve calculated the house edge of baccarat using infinite deck and it’s a negligible difference from 8-deck
link to original post
Really? 99/411 = .2409
104/416= .2500
So you will get the 6 card flush 4% more often. And it pays 50 each time. Im not interested in doing the calculations for the 4 and 5 card flushes. But I stand by my guess that the difference between infinite deck and 8 deck with the pay table listed by the OP can be the from the 16% to the 12%.
Also, that's fine if removing 6 isn't a big deal (which as shown above it's not "not" a big deal), but what about removing 10, 20, or more?Quote: SOOPOOQuote: Ace2In eight decks there are 104 cards of each suit. The effect of removing 4-6 cards shouldn’t be significant
I’ve calculated the house edge of baccarat using infinite deck and it’s a negligible difference from 8-deck
link to original post
Really? 99/411 = .2409
104/416= .2500
So you will get the 6 card flush 4% more often. And it pays 50 each time. Im not interested in doing the calculations for the 4 and 5 card flushes. But I stand by my guess that the difference between infinite deck and 8 deck with the pay table listed by the OP can be the from the 16% to the 12%.
link to original post
Use a suit count and redo the math (if you can't sim) to find the breakeven/betting point, if it exists.
Prediction: Like most things baccarat, there's probably a special count and edge to be had, but you'll have to stand around and back count multiple games for hours to probably only get a few bets in. Thus, the max wager is important for 2 reasons... EV and the wild variance that will come with it.
Quote: Ace2In eight decks there are 104 cards of each suit. The effect of removing 4-6 cards shouldn’t be significant
I’ve calculated the house edge of baccarat using infinite deck and it’s a negligible difference from 8-deck
link to original post
True, but I did the following full-deal runs (not simulations) on this bet, and I get:
8-Deck Shoe HE = 15.228891 %
25-Deck Shoe HE = 13.035449 %
100-Deck Shoe HE = 12.258446 %
That said, it’s a sucker bet either way and I’d never make a bet with 12% or 15% edge. 0.75% is my limit