There is a promo here where if you hit 3 different flushes holding 2 cards of the same suit in your hand, you win some money. Then if you manage to do the fourth one as well, you win some extra money.
My question is: what are the odds of hitting 1 flush and then the other three remaining ones?
Lets say I hit flush of spades, what are the odds of making a second different flush (lets say hearts)?
And if I hit 2 flushes, what are the odds of making a third different one?
Thank you in advance
Quote: LinkerSplitHello again!
There is a promo here where if you hit 3 different flushes holding 2 cards of the same suit in your hand, you win some money. Then if you manage to do the fourth one as well, you win some extra money.
My question is: what are the odds of hitting 1 flush and then the other three remaining ones?
Lets say I hit flush of spades, what are the odds of making a second different flush (lets say hearts)?
And if I hit 2 flushes, what are the odds of making a third different one?
Thank you in advance
In what game? Over how many hands?
Quote: unJonIn what game? Over how many hands?
In texas holdem no limit, over 8 hours of game, avg 30 hands per hour
Quote: LinkerSplitIn texas holdem no limit, over 8 hours of game, avg 30 hands per hour
Depends on how you play? Are you a “any two sooted” type of person? You playing tight Skansky poker?
Quote: unJonDepends on how you play? Are you a “any two sooted” type of person? You playing tight Skansky poker?
Lets say I play any two suited cards I get over 8 hours
Quote: LinkerSplitLets say I play any two suited cards I get over 8 hours
You're most likely going to leave the poker room a loser.😀
Quote: MaxPenYou're most likely going to leave the poker room a loser.😀
I know :D , but the question is regarding the odds of hitting three different flushes in a session of 8 hours: does it get harder and harder after hitting the first, then the second and finally the third one?
If so, how hard does it getvin terms of %?
I believe it remains the same, but in math a belief is not enough so....
When starting with 2 suited cards you will make a flush by the river 6.4% of the time or 1 in 15.625.
So you are looking at a 1 in 250 hand event for a flush having to have 2 suited cards in your hand, if you stay to river every time there is opportunity to do so. 2nd flush will be 1 in 333 because you eliminated 1 of the available suits. Then the 3rd will be 1 in 500 because you eliminated 2 of the available suits.
Without having to have 2 suited cards in your hand you will see a 5 out 7 card flush about 3% of the time provided your in all dealt hands to the river.
But like others have pointed out there are going to be alot of variables in real play.
Hitting 3 DIFFERENT flushes with 2 cards in your hand in 8 hours is going to carry some long odds. In the millions:1 range. What is the payout for doing so?
Quote: MaxPen1 in 16 hands you will be dealt 2 suited cards.
When starting with 2 suited cards you will make a flush by the river 6.4% of the time or 1 in 15.625.
So you are looking at a 1 in 250 hand event for a flush having to have 2 suited cards in your hand, if you stay to river every time there is opportunity to do so. 2nd flush will be 1 in 333 because you eliminated 1 of the available suits. Then the 3rd will be 1 in 500 because you eliminated 2 of the available suits.
Without having to have 2 suited cards in your hand you will see a 5 out 7 card flush about 3% of the time provided your in all dealt hands to the river.
But like others have pointed out there are going to be alot of variables in real play.
Hitting 3 DIFFERENT flushes with 2 cards in your hand in 8 hours is going to carry some long odds. In the millions:1 range. What is the payout for doing so?
How’d you come up with 2 suited cards being 1-in-16? You do realize there are only 4 suites, right? I calculated the chance of getting 2 suited cards as 1-in-4.25, assuming a standard 52 card deck.
Now you need to figure out the chance of getting at least 3 of the same suit on the board. My phone battery is about to die so
0's are offsuit and 1's are suited, x's don't matter.... (reference for below) --
There are 50 cards remaining, 11 of which are suited. If the first 3 of the community cards are those, then flush. The same applies to any other specific 3 card grouping.
(11*10*9) / (50*49*48) =
990/117600
There are 8 such possible combinations......
00111
01011
01101
01110
10011
10101
10110
11001
11010
11100
(990*8)/117600 = 0.0673469 = 1-in-14.8484848484.......
The first flush is going to be:
1-in-(4.25*14.8484) = 63.1057 hands
Second:
1-in-(5.6666*14.8484) = 147.24564344
Third:
1-in-(8.6666*14.8484) = 275.93078688
Fourth:
1-in-(17.3333*14.8484) = 533.3025586
The chance of getting the second one is the chance of getting the first plus the second....getting the third is that number plus chance of getting the third....etc.
So, you should average 533.3025586 hands for every time you get all 4 flushes.
BTW my math might be wrong.