Odds of making three different flushes
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.
(532/533)^240
Chance of hitting it are 100% minus that, or:
1 - (532/533)^240 = 36.2820%
In 240 hands, you "should" be dealt 14 pairs of spades, 14 of hearts, 14 of clubs, and 14 of diamonds
For each pair, of the 2,118,760 possible boards, 462 have all five cards in that suit, 12,870 have four, and 122,265 have three, so 135,597 complete the flush.
The probability that none of the 14 pairs of spades make the flush is (1 - 135,597 / 2,118,760)14 = about 2/5
The estimated probability that you will get at least one "pair" of each suit that ends up as a flush is (3/5)4, or about 1/8 of the time.
Some quick simulation shows that, over a span of 240 deals, you get all four suits 1/8 of the time, and three of the four about 1/3 of the time.
I concur with 135,597 / 2,118,760 probability of completing a flush after being dealt 2 initial cards of same suit. So that times 4/17 is an overall probability of 1 in 66.4 of making any flush before any cards are dealt.
Then I did a simple Markov chain with 5 states: 0 to 4 flushes, with the conditions that you must get a flush (1 in 66.4) and it must be of a new suit to advance, except from state zero from which any flush will advance. So, for instance, if you already have 1 flush and then hit another, 3/4 chance you advance to 2, 1/4 chance you already had that suit and stay at 1. Obviously the final state of 4 flushes is an absorbing state.
After 240 iterations:
No flushes: 2.62%
One: 15.71 %
Two: 35.96 %
Three: 34.25 %
Four : 12.46 %
Sorry for the Markov chain. I see them as a last resort, almost like brute forcing a solution, when I can’t come up with a formulaic solution.
Incidentally, to answer OP’s question of “what’s the the chance of hitting 3 flushes if I already have 2”, need to know how many hands remaining.
Quote: Ace21 in 17 is the probability of being dealt a pair. The chance of being dealt 2 cards of the same suit is 4 in 17.
I concur with 135,597 / 2,118,760 probability of completing a flush after being dealt 2 initial cards of same suit. So that times 4/17 is an overall probability of 1 in 66.4 of making any flush before any cards are dealt.
Then I did a simple Markov chain with 5 states: 0 to 4 flushes, with the conditions that you must get a flush (1 in 66.4) and it must be of a new suit to advance, except from state zero from which any flush will advance. So, for instance, if you already have 1 flush and then hit another, 3/4 chance you advance to 2, 1/4 chance you already had that suit and stay at 1.
After 240 iterations:
No flushes: 2.62%
One: 15.71 %
Two: 35.96 %
Three: 34.25 %
Four : 12.46 %
Incidentally, to answer OP’s question of “what’s the the chance of hitting 3 flushes if I already have 2”, need to know how many hands remaining.
Can you show where my math went awry?
How can you have a 1 in 533 chance (or any chance at all) of hitting 4 flushes “on any given hand?” Impossible to hit 4 flushes in less than 4 hands.Quote: RSCan you show where my math went awry?
Fellow Math Geeks:
What would be the odds of hitting all 4 flushes in the first 4 hands played? Free t-shirt to first correct formulaic answer.
Quote: Ace2How can you have a 1 in 533 chance (or any chance at all) of hitting 4 flushes “on any given hand?” Impossible to hit 4 flushes in less than 4 hands.
Fellow Math Geeks:
What would be the odds of hitting all 4 flushes in the first 4 hands played? Free t-shirt to first correct formulaic answer.
I would say that it's not possible (obviously). But I would think logically you could treat it as 1 event that has a 1-in-533 chance of occurring. Although looking over the math, it doesn't seem like you would average the 4-flush cycle in 533 hands.
Using a different example that can be easier to follow -- the chances of rolling a 12 with 2 dice is 1-in-36. Rolling a 7 with 2 dice is 1-in-6, and rolling a 7 two times in a row is 1-in-36.
After X rolls, you'd expect to hit 36/X 12's and 36/X double-7's, right?. After X rolls, your probability of hitting at least one 12 is 1 - (35/36)^X.....wouldn't it be the same probability for rolling double-7's? The only thing I could think of why it might be slightly off, is that with double-7's, you need 2 rolls and not 1, so perhaps it should be X+1 in the case of the double-7's.
The average number of rolls to hit a 12 is 36. The average number of rolls to hit 2 consecutive 7s is 42. But even knowing that, you can’t always use the averages in the same way because the first scenario is binary: you can win on any roll. But in scenario 2 you can only win if you’re already in a state and then roll another 7. It seems like a minor difference but drastically changes how you can use some of their numbers.
Quote: Ace2How can you have a 1 in 533 chance (or any chance at all) of hitting 4 flushes “on any given hand?” Impossible to hit 4 flushes in less than 4 hands.
Fellow Math Geeks:
What would be the odds of hitting all 4 flushes in the first 4 hands played? Free t-shirt to first correct formulaic answer.
First hand: 4/17 (probability of getting two hole cards of the same suit) x 135,597 / 2,118,760 (probability that the board fills the flush)
Second hand: 3/17 (two hole cards in the same suit, in one of the three remaining suits) x 135,597 / 2,118,760
Third hand: 2/17 x 135,597 / 2,118,760
Fourth hand: 1/17 x 135,597 / 2,118,760
Solution: (24 x 135,5974) / (17 x 2,118,760)4 = about 1 / 207,448,921
Quote: unJonSo we are ignoring the contribution of flopping a flush draw on the flop with unsuited hole cards?
I took it from the OP that suited hand was a condition of the promotion. Perhaps it's not?
Quote: beachbumbabsI took it from the OP that suited hand was a condition of the promotion. Perhaps it's not?
Withdrawn! Just reread the OP and agree.
