There are a lot of people on this board smarter than me, so I’d like you help. Assume you are playing video poker. You are playing the machine where whatever cards you choose to keep propagate across 10 hands. You are dealt 3 aces. You choose them and so now all ten hands have three aces. So, in any one hand, the odds of getting the 4th ace are 2 in 47. How does having 10 hands improve the odds? What is the formula? Thanks
First hand odds on the turn: 1/47 Second hand odds on the turn: 1/46 ... Tenth hand odds on the turn: 1/38 First hand odds on the river: 1/37 ... Tenth hand odds on the river: 1/28 The sum of all of these percentages is 54.65%.
Since it’s poker, I get to draw 2 cards to each hand. The same remaining 47 cards exist in each of the 10 hands, so each of the 10 hands gets two pulls. So each are 1 in 47 and then 1 in 46.
I was thinking along those lines, but then the situation exists where once you have 24 decks, the odds pass 100% and that can’t be true. Do the odds stay .043 regardless and there is just the illusion that your odds would be better?
You're going to have to explain this more clearly. 24 decks? From what I think I understand, you get dealt 10 hands and pick whatever cards you want from one hand, which are replicated across the 9 other hands. Then each hand has a fresh deck, minus the cards you discarded from the one hand. Right?
Let’s simplify it down to two hands. I get dealt 5 cards. Three aces, a king and a queen. So, 47 cards are left for hand one and three of the aces, a king and a queen are not among the 47. I keep the three aces. The three aces are now in play for both hands. Each hand has 47 cards left. The same 47 cards, but they are arranged differently. Hand one has a .043 chance of drawing the 4th ace as does hand 2.
I still don’t understand the question. What do you mean by “the 3 aces are now in play for both hands?” And what do you mean “each hand has 47 cards left?” Are these two separate decks?
Video poker has some games where there is an option to play X number of hands in any one bet. You are dealt one hand of 5 cards from a 52 card deck. So...that first hand has 47 cards left. In my example, I keep 3 aces and throw away the other 2 cards. All the hands now have 3 aces and a “draw” deck of 47 remaining cards as the original 5 dealt cards are no longer in any deck. So, in each of the hands’ remaining decks, I get two cards. I have a 1 in 47 chance of getting the 4th ace on the first draw and a 1 in 46 chance of getting the 4th ace (assuming it wasn’t the first drawn card) on the second draw. So, that happens for hand 1. The it moves to hand 2 and, again, there is the 1/47 and 1/46 draw scenario from its remaining 47 cards. This repeats itself for every hand I bet.
I found a formula that might apply 1 - (( 1 - Y ) ^ X) Y = odds (around .043 in my example) X = number of attempts
Sorry @bobrek I don’t know enough about online poker or what you’re asking to answer this for you. And I’m not fully understanding the question even after your explanation. But in general in probability theory you have to consider the sample space as a first step when coming up with something of this sort. I think with what you’re trying to do a simple combination count would be sufficient for that. Then you’d have to either calculate a cumulative probability (simply adding probability up) or alternatively calculate some sort of odds ratio. I can tell you that “odds” in the sense that you posted above isn’t what an odds ratio is in probability. So that might be a good place to start.
Assuming each hand has its own deck of 47 cards with 1 ace: There is a 95.7% (46/47*45/46*100%) chance an ace doesn't come up in each hand. There is a 64.7% (95.7%^10) chance that an ace doesn't come up in all ten hands. There is a 35.3% (100%-64.7%) chance that an ace comes up in at least one hand out of ten.
if I remember correctly from biostat this sounds right (that was awhile ago but the numbers def work w/ formula), and to add @Joe Joe has the right numbers here as well. So... Are you planning on gambling?
Hahah, yeah I understand, if I am really focusing on the odds, have to not drink as much... If I drink a lot at the tables odds don't exist... so the 5 cent slots and free drinks sound good
