All 3 people got dealt the same poker hand
Wednesday, 12 Mar 2025
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You calculated the chance of exactly this hand. The chance of having 3 matching hands in random order (high card first or low card first for the second and third hand) is actually much higher.
I need /r/theydidthemath to run checks on this
Nevermind, my maths does not check out. Turns out it’s more like 1 in 40,000 as calculated by u/eloel- in [this post](https://www.reddit.com/r/theydidthemath/s/bvE4RB85IU)
Aces and eights, the dead man’s hand!
My question is who ended up actually winning?
They should all get tattoos commemorating the event.
There are, for the purposes of our calculations, three poker hands: pocket pair, distinct cards suited, and distinct cards off-suit.
The first player is (4 choose 2)•13/(52 choose 2)=1/17 to have a pocket pair. If they do, it’s impossible for *both* other players to get the same pocket pair.
The odds of the first player getting two cards of the same suit is (13 choose 2)•4/(52 choose 2)=4/17. Assuming this, the odds of the next player getting the same two ranks in another suit is 3/(50 choose 2)=3/1225, and then assuming *that* the odds of the third player getting the same two ranks in one of the last two suits is 2/(48 choose 2)=1/564. All together, the odds of three players getting the same hand of this type is thus 1/978775
Lastly, we thus get a 12/17 chance of the first player being two distinct cards, off-suit. The odds for the third player will depend on how much suit overlap there is between the first two, so we’ll break down into three cases:
The second player is 1/(50 choose 2) to get the same two cards in the same two suits (so e.g. first player 8H, AC, second player 8C, AH). In this case, the third player is 2/(48 choose 2) to get the same two ranks in distinct suits.
The second player is 4/(50 choose 2) to have one suit overlap with the first player, and if this happens, the third player is 3/(48 choose 2) to get the same hand.
Lastly, the second player is 2/(50 choose 2) to get the same cards with no suit overlap, and if this happens, the third player is 4/(48 choose 2) to match them.
Putting these together, we get (1•2+4•3+2•4)/((50 choose 2)•(48 choose 2))=22/(25•49•24•47); multiplying by the odds for the first player, we get 11/978775.
Thus, the odds of three of 3 players getting the *same* hand are 12/978775, or roughly 1 in 81,565.
This math is wrong. Assumes they all get this specific hand, not the same predetermined hand. Just wrong
Great shuffling, dealer.
Aren’t these technically different hands because of suit?
Who wants to play those eights and aces?
Everyone was dealt deadman’s hands. I would quit playing and go to bed lol
Add in the odds that you get to see all 3 had the same cards.
Assuming all you care about are the chances the hands are all the same and not what the specific hand is, the odds are about 0.0024%.
For the first hand, all we care about is that it isn’t doubles, so 52/52 * 47/51.
For the second, the first card can match either and the second must match the other, so 6/50 * 3/49.
For the third, it’s the same but with lower odds, so 4/48 * 2/47.
Multiply those all together, and you get about 0.000024, so more like one in 40,000 than one in a billion.
If you want it to specifically be Aces and Eights, just replace that first section with 4/52 * 4/51. In that case you get 0.000000145, so around one in seven million.
This is wrong, but also, what the fuck is 39 in a billion.
One of my favorite stats about cards, there are more ways to shuffle a 52 card deck than atoms in/on earth
Dead man’s hand combined.
you just calculated the probability of having three specific hands
Mildly interesting how shit this person is at math.
RIP
this is all of their lifetime luck used up at once
They’re all dead men.
Player 1 draws card 1. Any card is valid.
Player 1 draws card 2. Card two must not match rank with card 1. Odds: 48/51
Player 2 draws card 3. Card three must match rank with either card 1 *or* card 2. Odds: 6/50
Player 2 draws card 4. Card four must match rank with player 1’s other card. Odds: 3/49
Player 3 draws card 5. Card five must match rank with either card 1 *or* card 2. Odds: 4/48
Player 3 draws card 6. Card six must match rank with player 1’s other card. Odds: 2/47
Multiply these together. Overall probability of this type of occurrence is 1 in ~40,782
It’s worth noting that our brains are hard wired to see patterns, and there are potentially thousands of patterns that you are likely to notice that may occur in a given shuffle. Seeing a pattern that seems notable after the fact from a particular deal, with no specific notion in advance about what that pattern should be – that’s probably more like 1 in 100 or even better.
r/untrustworthypoptarts
Correct odds: 0.00002452044 = about 1 in 40000
copy paste from: https://www.reddit.com/r/theydidthemath/comments/1i7p6mq/request_all_3_people_got_dealt_the_same_poker/?sort=top
But, it was just as likely as any other possible combination.
Fun fact: the number of combinations you can make with one deck of 52 cards is so unbelievably big, it’s bigger than the amount of atoms in the entire universe! Bigger than all the grains of sand that can be found in the entire universe. Bigger then everything!!!!
I’m not good with math but it seems like the bottom part should be 49x48x47x46. Two of the cards dealt don’t matter because they can be any two cards, it’s the next two pairs that need to be matched to it. The math you did is for dealing specifically an 8 and an Ace to three people.
The left side of your math is right, because the aces it doesn’t matter which suit. The denominator on the right is also correct, the numerator on the right is wrong as it should be 3x2x1 as you also have to figure that the suit of the 8 does not match the ace.
Really 9.8 in a billion.
what was the action like? did someone try to push all in pre flop?
One more hand left in the deck..
That’s cool but I’d rather win the lottery with those odds
Decompress the main shuttle bay. The explosive reaction may blow us out of the way.
r/balatro
One time I was playing poker with 5 or 6 people. We were playing different games after playing a long game of Texas hold em. We were playing 5 card draw, my brother and BIL both threw down a Royal Flush. They were about to get in a fight because we had to look up which suit wins. They ended up splitting the pot, but was a crazy scenario.
Not even fking close
It is not possible that more than one player gets delt the same hand as another.
I played in a poker tournament where all four people all in had Ace King (off suit)
It was chopped.
So, who won?
I’m going to assume that a new deck accidentally faro shuffled was involved.
I’ve had that happen with Ace 7, three of the four people playing were dealt that hand
The chances of three people having the same hand, but not caring what hand that is:
First draw two random cards to the first player.
The first card to player 2 can be any of the 3 aces or 3 8s, so 6/50.
The second card to player 2 can be any of the three of the cards they don’t already have, so 3/49
The first card to player 3 can be either of the aces or 8s, so 4/48
The second card to player 3 can be either of the 2 remaining cards they don’t already have, so 2/47
Probability = (6*4*3*2)/(50*49*48*47)
1 in 38,000
Edit: Actually, you need the first player to not get two of the same card or you can’t all get the same hand. So multiply by 12/13 to correct for that. It’s about 1 in 40,000
Have you checked if you are in a dream?
You need to wake up
Please wake up