The odds of being dealt a 4 of a kind in your first 4 cards of any draw or stud game is ~1/21,000. Odds of a 4 of a kind in your first 5 cards is ~1/4,000.
Then you missed the page title as well.Originally posted by: mobobuff
That's 5-card. OP is talking Hold 'Em.
Edit: I guess the page title could be misleading/wrong.
Originally posted by: TuxDave
The math in this thread makes me
Yeah, it would make me cry too.....if I were in pre-school!
True answer to the initial question, odds of 2 four of a kinds in a row in texas hold-em is 1 in 481,867. Trust me.
(this is assuming the 4 of a kind is using one or both cards in your hands, does not include the chances a 4 of a kind is on the board)
(this is assuming the 4 of a kind is using one or both cards in your hands, does not include the chances a 4 of a kind is on the board)
mobobuff
Lifer
- Apr 5, 2004
- 11,099
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Originally posted by: manly
Then you missed the page title as well.Originally posted by: mobobuff
That's 5-card. OP is talking Hold 'Em.
Edit: I guess the page title could be misleading/wrong.
Yeah, I think that first table is there just for straight 5-card odds, because most people will be looking for those first. Notice the header rule line that divides the top table from the rest, and the tables below are defined as "Hold 'Em" hands. Also, those odds in the first table are pretty low (bad) compared to my experience playing Hold 'Em, but they appear correct for 5-card.
It's based off a previous post, now including the percentages of being dealt pair hold cards both time, no pair both time, or one pair and one not a pair.
This page confirms you're right but I don't think the Texas Hold'em odds for the monsters (quads or better) would be much higher than the 5-card odds.Originally posted by: mobobuff
Originally posted by: manly
Then you missed the page title as well.Originally posted by: mobobuff
That's 5-card. OP is talking Hold 'Em.
Edit: I guess the page title could be misleading/wrong.
Yeah, I think that first table is there just for straight 5-card odds, because most people will be looking for those first. Notice the header rule line that divides the top table from the rest, and the tables below are defined as "Hold 'Em" hands. Also, those odds in the first table are pretty low (bad) compared to my experience playing Hold 'Em, but they appear correct for 5-card.
- Dec 29, 2002
- 7,402
- 0
- 71
What have I done...it's a hurricane of math!
I think cubby may be the closest so far, I don't know about his/her math, but the player had a pocket pair on the first hand, and not on the second. 1 in 12,000 sounds about right. But, then again, this is online poker and not real life poker.
I think cubby may be the closest so far, I don't know about his/her math, but the player had a pocket pair on the first hand, and not on the second. 1 in 12,000 sounds about right. But, then again, this is online poker and not real life poker.
DrPizza
Administrator Elite Member Goat Whisperer
Originally posted by: RapidSnail
Didn't read the thread:
P = (4/52) * (3/51) * (2/50) * (1/49)
P = (24/6 497 400)
P = 1/270 725
Dependence(P^2) = (1/270 725)^2
Dependence(P^2) = 1/73 292 025 625
?
Edit: I may have misapplied dependence. Dependence is whether or not one probability affects the probability of another outcome, if I recall correctly. My reasoning is that if the first four-of-a-kind doesn't occur, then the second four-of-a-kind can't occur with regards to the criteria. Or does the criteria not affect the outcome?
Edit 2: I guess my answer only works if after each card is delt, the next is delivered out of random (no linear progression through the stack). I'm not sure how to make it work for traditional poker though.
Not quite the way to work out the probability. The problem is that there are 5 cards in the hand, not 4. Also, you're working more toward the probability of having 4 of a particular rank. Thus, after your first card, which could be *any* card from the deck (you're not limited to just 4), the probability of the second card matching the 1st card is 3/51, then 2/50, and 1/49. But the probability of the 1st card is 52/52. Also note: the probability of the 5th card not matching the first 4 is 48/48.
Unfortunately, this still doesn't lead to the correct probability because no one said you had to be dealt your 4 of a kind on the first 4 cards. The 5th card could have been the 1st, 2nd, 3rd, 4th, or 5th card. Thus, there are 5 different ways to be dealt that hand.
5 * 52/52 * 3/51 * 2/50 * 1/49 * 48/48 = 74880/311,875,200
Usually, people working with probabilities wouldn't prefer to do it this way - it's too easy to make a logical mistake.
The better way to do it is the number of possible 4 of a kind hands divided by the total number of possible hands.
The total number of possible hands is the number of combinations that can be made from 5 cards. 52C5 (or whatever notation you want for combinations) This is 2,598,960 hands
(or 52*51*50*49*48 divided by the number of arrangements of 5 cards: 5*4*3*2*1)
The total number of 5 card hands with 4 of a kind: Choose one rank out of the 13 (i.e. you could have four A's, four 2's, etc.) There are 13 different choices, times 48 other cards remaining. Four 2's and the 3 of clubs, Four 2's and the 3 of spades, Four 2's and the 3 of diamonds, four 2's and the 3 of hearts; four 2's and the 4 of... i.e. 13*48 possible hands.
So, your probability is 13*48 = 624 / 2,598,960
There are quite a few ways to determine the number of 4 of a kind hands.
Since there's 4 of a kind, they'll be of two different ranks. One of the ranks has 4 out of 4 of that rank, the other rank has 1 out of 4.
Choose 2 ranks, choose 4 out of 4 of that rank, choose 1 out of 4 of the other rank (and the tricky part, times 2 because the 4 out of 4 can be either of the two ranks that you select):
13C2 * 4C4 * 4C1 *2 = 624
If you think card distribution is messed up online, wouldn't it be foolish of you to continue to play?
Let's not turn this into an online poker argument, the sheer number of hands you see playing online guarantees you'll see some crazy stuff.
That said I've lost to quads twice in less than an hour playing live, so it happens there too.
Viper GTS
That said I've lost to quads twice in less than an hour playing live, so it happens there too.
Viper GTS
DrPizza
Administrator Elite Member Goat Whisperer
Oh, and in Texas Hold'em, essentially it's a 7 card hand.
Thus the probability of 4 of a kind would be:
13*48*47*46 divided by the total possible number of 7 card hands.
(13*48*47*46) divided by (52*51*50*49*48*47*46)/(7*6*5*4*3*2*1)
1,349,088 out of 133,784,560 = 0.010084033613445
On the next hand, the probability would be the same, therefore that probability times itself. 0.000101687733917096
Or 1 in 9,834
Of course, it's somewhat implied in the problem that the 4 of a kind were unique to that one player, i.e. he didn't get the 4 of a kind from the 3 cards on the flop plus the turn or river. At least 1 of the cards in his hand was the 4th of the 4 of a kind.
The above probability was a pita using the calculator on Vista. I typed it wrong at least 4 times. I'm not doing any more calculating!
Hint: the total number of 7 card hands is the same. However, restrict it so that at least 1 of the two cards dealt to the person is one of the cards from the 4 of a kind. i.e. figure out the proportion of those 4 of a kind hands where 1 or 2 of the 4 are dealt to the individual. If it's half of those hands, simply divide by 2. I'll leave it up to the rest of you to deal with this similarly trivial probability. I've done my share.
Thus the probability of 4 of a kind would be:
13*48*47*46 divided by the total possible number of 7 card hands.
(13*48*47*46) divided by (52*51*50*49*48*47*46)/(7*6*5*4*3*2*1)
1,349,088 out of 133,784,560 = 0.010084033613445
On the next hand, the probability would be the same, therefore that probability times itself. 0.000101687733917096
Or 1 in 9,834
Of course, it's somewhat implied in the problem that the 4 of a kind were unique to that one player, i.e. he didn't get the 4 of a kind from the 3 cards on the flop plus the turn or river. At least 1 of the cards in his hand was the 4th of the 4 of a kind.
The above probability was a pita using the calculator on Vista. I typed it wrong at least 4 times. I'm not doing any more calculating!
Hint: the total number of 7 card hands is the same. However, restrict it so that at least 1 of the two cards dealt to the person is one of the cards from the 4 of a kind. i.e. figure out the proportion of those 4 of a kind hands where 1 or 2 of the 4 are dealt to the individual. If it's half of those hands, simply divide by 2. I'll leave it up to the rest of you to deal with this similarly trivial probability.
I've done my share.
Throckmorton
Lifer
- Aug 23, 2007
- 16,829
- 3
- 0
I don't play poker at all... Isn't "four of a kind" 4 of the same suit?
That means the odds of 4 of a kind is:
((((((1 / 4) * 12) / 51) * 11) / 50) * 10) / 49. = 0.00264105642
I just looked it up, and it's 4 of the same rank....
Which means it would be:
4/52 * 3/51 * 2/50 * 1/49 = 3.69378521 × 10^-6 = 0.00137174211
So the odds are better than than 1 in 1000 to get 4 of a kind.
0.00137174211 ^ 4 = 3.54070616 × 10^-12
That means the odds of 4 of a kind is:
((((((1 / 4) * 12) / 51) * 11) / 50) * 10) / 49. = 0.00264105642
I just looked it up, and it's 4 of the same rank....
Which means it would be:
4/52 * 3/51 * 2/50 * 1/49 = 3.69378521 × 10^-6 = 0.00137174211
So the odds are better than than 1 in 1000 to get 4 of a kind.
0.00137174211 ^ 4 = 3.54070616 × 10^-12
mobobuff
Lifer
- Apr 5, 2004
- 11,099
- 1
- 81
Originally posted by: DrPizza
Oh, and in Texas Hold'em, essentially it's a 7 card hand.
Thus the probability of 4 of a kind would be:
13*48*47*46 divided by the total possible number of 7 card hands.
(13*48*47*46) divided by (52*51*50*49*48*47*46)/(7*6*5*4*3*2*1)
1,349,088 out of 133,784,560 = 0.010084033613445
On the next hand, the probability would be the same, therefore that probability times itself. 0.000101687733917096
Or 1 in 9,834
Of course, it's somewhat implied in the problem that the 4 of a kind were unique to that one player, i.e. he didn't get the 4 of a kind from the 3 cards on the flop plus the turn or river. At least 1 of the cards in his hand was the 4th of the 4 of a kind.
The above probability was a pita using the calculator on Vista. I typed it wrong at least 4 times. I'm not doing any more calculating!
Hint: the total number of 7 card hands is the same. However, restrict it so that at least 1 of the two cards dealt to the person is one of the cards from the 4 of a kind. i.e. figure out the proportion of those 4 of a kind hands where 1 or 2 of the 4 are dealt to the individual. If it's half of those hands, simply divide by 2. I'll leave it up to the rest of you to deal with this similarly trivial probability.
That blows my mind, obviously I'm not questioning you math, but I haven't played more than 1,000 hands on Bodog Hold 'Em and I've had at least 5 4OAKs that I can remember. That would mean my experience is roughly 1 in 200.
DrPizza
Administrator Elite Member Goat Whisperer
Originally posted by: mobobuff
Originally posted by: DrPizza
Oh, and in Texas Hold'em, essentially it's a 7 card hand.
Thus the probability of 4 of a kind would be:
13*48*47*46 divided by the total possible number of 7 card hands.
(13*48*47*46) divided by (52*51*50*49*48*47*46)/(7*6*5*4*3*2*1)
1,349,088 out of 133,784,560 = 0.010084033613445
On the next hand, the probability would be the same, therefore that probability times itself. 0.000101687733917096
Or 1 in 9,834
Of course, it's somewhat implied in the problem that the 4 of a kind were unique to that one player, i.e. he didn't get the 4 of a kind from the 3 cards on the flop plus the turn or river. At least 1 of the cards in his hand was the 4th of the 4 of a kind.
The above probability was a pita using the calculator on Vista. I typed it wrong at least 4 times. I'm not doing any more calculating!
Hint: the total number of 7 card hands is the same. However, restrict it so that at least 1 of the two cards dealt to the person is one of the cards from the 4 of a kind. i.e. figure out the proportion of those 4 of a kind hands where 1 or 2 of the 4 are dealt to the individual. If it's half of those hands, simply divide by 2. I'll leave it up to the rest of you to deal with this similarly trivial probability.
That blows my mind, obviously I'm not questioning you math, but I haven't played more than 1,000 hands on Bodog Hold 'Em and I've had at least 5 4OAKs that I can remember. That would mean my experience is roughly 1 in 200.
Look at my post closer: the probability of 2 in a row is 1 in 9834.
The probability of getting 4 of a kind is around 0.01 or roughly 1 in 100. (assuming you stay in) Hmmm... that almost seems a little high. Maybe someone should check my calculation.
- Dec 29, 2002
- 7,402
- 0
- 71
Well, It wouldn't necessarily be an advantage or disadvantage. I'm either the one getting a straight flush/four a kind, or I'm the one on the other end getting shafted. I only play with play money anyways though, because of the card distribution. Do any online poker services feature live dealer/decks? That would be a little more trustworthy. Still I probably wouldn't play with real money online, that just seems ridiculous to me.
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