Monday i got quads over quads and tuesday set over set with the exact same hand.





I didnt get 66 in between so it was in a row. What are the odds of quads over quads and shortly after set over set.
The funny thing is that yesterday i folded 66 pre because of a huge raise. Flop was QT3r, both players check. Turn 6 and 1 player had TT. So it would have happened again. So i would have hit the "dream" TURN again lol.

What are the odds of this happening?

Is it possible to calculate that? Sorry but im a math fish

50%
it happens or it doesn't

funny this happened to me online last week, i was saying the same thing in the chatbox "what are the odds!"

There are many spots where you could start this calculation with varying degrees of difficulty depending on what you want.

Do you want to include the probability that you and your opponent are each dealt pairs?

Do you want to just assume you already have pairs and only are concerned about postflop?

Do you want to assume you both see all 5 cards or are there restrictions like you both have to have a set on the flop or turn?

Do you want to know the odds apecifically of hero losing or just any set over set or quads over quads scenario? For example if hero holds the pair AA he can never be beaten in such a scenario but if hero holds 22 he is slightly more likely to be beaten in such a scenario. Also hero can win by making quads or straight flush when he has a lower set and villain doesn't improve.

Also whether you play a particular pair at a certain position and how many players are at your table can change things slightly as well.

As a rough estimate just assume both you and your opponent have a pair and you always loose the showdown and see all 5 cards.

For you and your opponent to have 2 hole cards and both make quads that means 4 out of the 5 board cards are a fixed at the 2 ranks you need so that means the 5th board card can be any of the 44 remaining board cards so there are 44 boards that make you quads over quads by the river. The total number of 5 card boards is 48C5 where C means the combination or choose function.

48!/(48-5!)*5! = 48!/(43!)*5! = 48*47*46*45*44/120 = 1,712,304

So 44/1,712,304 * 100= .002%

Now the probability you have set over set (or better) means that 2 of the board cards are fixed for you and your opponent's rank. I am going to include fullhouses but exclude 4 of a kind for you (meaning you can never win). That means there are 3 remaining board cards to be chosen. Meaning there are 46 unexposed cards. Of those 46 we never want to be able to draw quads so there are only 45 cards in the deck that we want to count for this scenario.

(45c3)/(48c5) = (45*44*43/6)/1,712,304 = 14,190/1,712,304 *100 = .828%

So I would say roughly there is a .002*.828 = .0016% chance of this happening.

Assuming my math is correct.

Someone will probably come in and blow the doors off this post any way but I was bored and wanted to at least practice this problem because it had been awhile.

Rigged RNGs man. That’s why I don’t play live.

+1. Haha

Thx man for taking the time!

So you mean there is a .0016% chance to have quads over quads and shortly after set over set or do i get this wrong?

Is it possible to calculate the probability for it to happen 2 times in a row with the exact same hand like it happened to me?
And what about 3 times in a row like it almost happened? Or is this not possible?

In reality the number should be much higher than .0016%, depending on how many hands you've played in your lifetime and so on. Let's say it's 1%.

Then we can say it's also 1% chance that it happens with 55, 1% that it happens with 77, 1% that it happens with 88...

Then there's 1% chance that we get straight flush vs straight flush two times, 1% chence that we get set vs set 5 times in a row and lose, 1% that we lose straight vs a pair 2 times in a row...

If we add that all up we get a number like 150%, meaning it's completely normal for things like this to happen every once in a while. It would be much weirder if you played ~1 mil hands and never had something like this happening.

Thanks for the follow up I was brushing up on the math of this to make sure my follow up post eas accurate.

Correct. This is a rough estimate and assumes you always loose, or put it another way you always have a pair lower than your opponent and you never hit a straightflush or quads to win by the river.

Not sure exactly what you're asking here. Becaue for the final probability I provided you would have to have both scenarios happen to you. It also gets more complex as you'd have to start talking about the individual probabilites of holding say 66 vs AA. Then you'd also have to consider what game type you're playing because maybe 22 doesn't get played very often UTG especially the more players behind but always gets played on the button etc.

I mean the odds of quads over quads and set over set both times with with a specific hand like 66 vs 99 and it happening in a row. Or quads over quads and 2 times in a row set over set hitting always the quads or the set ott.

As far as I can tell that is what I provided you give or take a few edge cases (straight flushes for example) and ignoring the number of higher pairs that could actually beat you.