[scatcher]

I lack the mathematics necessary to calculate the odds of a single person getting broadway back to back three hands in a row.

Anybody able to figure it out?

[tikipirate]

Yep. Without replacement, ie shuffling the cards back together and re-dealing, the odds would be (1/254.8 ) * (1/254.8 ) * (1/254.8 ). But since you most likely did shuffle and replaced the dealt cards back in the deck, it's an independent trial each time.

This is little consolation if the same guy keeps pulling broadways out of thin air to take your chips, but all things being even, some strange things can happen when playing cards.

One of my fav sites for this kind of stuff
http://wizardofodds.com/

enjoy

[scatcher]

While I would agree that every single deal must be treated as statistically isolated for judging the possibility of broadway arriving in that hand, when it's dictated that it must happen 3 times in three consecutive shuffles I believe the odds do have to be adjusted.

Changing the focus from a single hand to a set of three makes it a fundamentally different question.

PartyPoker sent me a snail-mail letter to say they had deposited $25US in my account that was mine to play with and keep, the caveat being I had to amass 75 Party Poker points before i could withdraw any. I was doing alright - got it up to about $90 - and then I ran into 3 consecutive broadways from the same guy. I had two pair, trips, and a baby straight against him.

[tikipirate]

Actually the only time that (1/254.8 ) * (1/254.8 ) * (1/254.8 ) is a valid expression is when
1) Three broadway striaght arrive in three consecutive hands
2) This there is no replacement, and without shuffling. I guess the best way to do this would be to have three identical decks, shuffle each, deal one hand, switch decks, deal another, switch, deal the last.

The bottem line is that, as much as we'd like it to, probability dosen't work really good in hand to hand situations; it works very good "inside" a hand; what the probability is of red/black card coming on the river given the other four cards; the probability we have of getting the one card we need to make our gutshot straight, etc. But comparing independent random events just dosen't work so well.

[jojobinks]

Originally Posted by tikipirate

The bottem line is that, as much as we'd like it to, probability dosen't work really good in hand to hand situations; it works very good "inside" a hand; what the probability is of red/black card coming on the river given the other four cards; the probability we have of getting the one card we need to make our gutshot straight, etc. But comparing independent random events just dosen't work so well.

that's nonsense. it's series probability, which is pretty basic probability. clearly the chance of hitting heads 3x in a row is (1/2)^3, right? getting the same hand 3x in a row is the same. take the probability for the single event, and multiply by the other events.

different decks don't matter. nothing else matters. as long as the deck is well shuffled enough to be random, this works.

[scatcher]

That fits very nicely with how I thought it ought to work.

Statistically improbable, but of course the poker Gods laugh in our faces whenever somebody says that.

[tikipirate]

I guess what I meant was these probabilites arn't useful or meaningful. You're not going to change anything because of an event happening. If you see two quads back to back, then pick up any pocket pair the very next hand, are you going to reason that the odds of you hitting quads this hand are any different than they would normally be? Each trial (hand) is independent, one hand (ideally) has no bering on the next.

It looks like the post got deleted but where ever 1/254.8 came from, I think that is for 5 card, and it's for any straight. Due to the community cards, you'd have to use the 7-card stud probabilities; in the end this make it slightly more likely to happen.