Quick one here.
5 card draw. You’re dealt three of a kind. Obviously your odds of winning are already pretty good. What’s the better draw? One card to increase your odds of drawing a card that pairs up your loner to get the full house, or two which increases the odds of pulling 4 of a kind. I would think draw one is better. Also, any change in the odds on whether you drop your high loner or your low loner? Thanks ahead of time.
I’m going to have a tough time backing this up, but I’ve read that odds-wise your best chances of improving the hand are by drawing two.
A quick cut at this is that if you draw a single card, there are 4 that can really help you (1 that can give you the 4 of a kind, and 3 that can pair up the off card and give you a full house). If you draw two cards, you now have two chances to pull the single card that will give you 4 of a kind, plus lots more ways to get a pair in the two you draw to get the full house. No actual math there, but it should be easy to see there are more different ways to improve when drawing two cards.
Of course I’m ignoring the cases where you just end up with a higher kicker, as they don’t matter. You can’t tie on a 3-of-a-kind (unless of course you’re playing with wild cards, which I’ve also ignored).
Now in practical terms of course it may be a better idea to only draw a single card anyway. If I’m up against only a few opponents and I’m confident my 3-of-a-kind are going to hold up without improvement, I sometimes draw a single card to make it look like I’m drawing to a straight or flush with the hopes they might bet into me thinking I didn’t hit. Whether this is a good move or not depends on a lot of things, but it’s worth thinking about.
Oh, one more thing. If you do decide to draw a single card, it doesn’t matter if you discard the low card or the high one. In both cases your odds of pairing the card you kept (or hitting the 4-of-a-kind) are exactly the same. Since the rank of the full house (if you hit it) is determined by the 3-of-a-kind portion, and you can’t tie on that, the rank of the pair you filled with doesn’t matter.
According to my fiendish calculations, your probability of improving your hand if you draw one card is 4/47, because there are 4 cards that can help you out of the 47 that might be drawn.
If you draw two cards, your probability of improving your hand is 5/47. Here is my work:
The probability that the first card drawn will give you 4 of a kind is 1/47. If that happens, the second card drawn doesn’t matter; it can’t improve your hand.
If the first card doesn’t improve your hand, then there is a 4/46 probability that the second card will. This is because there are now 46 unknown cards, one of which will match your 3 of a kind, and three of which will match the first draw.
So, your probability of helping your hand by drawing two cards is equal to the probability of the first card helping, added to the contingent probability that the first card won’t help but the second one will. In other words,
1/47 + 46/47 * 4/46 = 5/47.
So you have a significantly better chance of improving your hand when you draw two.
I think you both have flawed math (but am no where skilled enough in math/statistics to back that up). Somewhere in the equation has to be the odds of pulling a pair on a two-card draw. Doesn’t it?
If you keep four (the three-of-a-kind and a lone card) and draw only one, you can draw one of four cards that can improve your hand, one that matches the three-of-a-kind and one of three that match the lone card (4/47 chance of improvement).
If you keep three and throw out two, the first of two cards you recieve can:
For case 2, the second drawn card can either match the three-of-a-kind (1/46) or the first drawn card (2/46). For case 3, the second drawn card can either match the three-of-a-kind (1/46) or the first drawn card (3/46). Overall, your chance of improving you hand by drawing two is:
(1/47) + (6/47)(3/46) + (40/47)(4/46) = ~(4.87/47). So you’re slightly better off drawing two.
And in the the case where you only draw one, it doesn’t matter which lone card you keep; there are four cards remaining in the deck that will improve your hand either way (one matching the three-of-a-kind, and three matching the loner you keep).
That’s true, you do have to account for that. I sorta-kinda addressed it by saying you had “lots more ways to get a pair in the two you draw”, but mostly dodged by not doing any actual math. Important to catch though as this is part of the reason drawing two is statistically better than drawing one.
Of course, we’re neglecting the intangibles. Sometimes how you draw is dependent on what you’re trying to sell to your opponents, not to mention how your opponents have drawn.
If you have trip aces, and you’re the last to draw, and your opponents all draw three – draw two; if you draw only one, they’ll fold like a road map.
On the other hand, if you have trip 9’s and and your the first to draw, you might as well represent four of a kind, and draw one.
I have that in my calculations, Gazoo. Here it is (with emphasis added):
… Caldazar’s answer is more complete than mine. So ignore mine and read his 
It seems easier to use all 47 other cards to figure out this problem, even though in a real game (obviously) less cards would be available. Let’s assume you have A A A 2 5, just to make it easier to talk about.
If you ditch the 2, then there are only 4 cards that can help you: another A, or any of the three 5’s left. 4/47= ~.0851
If you ditch two cards, then there are 47*46/2= 1081 different 2-card combos that you could recieve.
If you get a pair, that would improve your hand. There are three 2’s still left. There are 32/2= 3 ways to get a pair of 2’s off the draw. There are three 5’s still left. There are 32/2= 3 ways to get a pair of 5’s off the draw. There are 4 of these ten denomiations left (3, 4, 6, 7, 8, 9, 10, J, Q, K). There are 43/2= 6 ways to get a pair for each denomination. 610= 60. So, there are 3 + 3 + 60 = 66 ways to get a pair off the draw.
Drawing an “A+any other card” would also improve your hand. There are 46 ways to get an “A=any other card”.
So, we have 66 + 46= 112 hands that will improve our trip aces. 112/1081= ~.1036= ~4.87/47
So I guess I get the same thing as Caldazar… Of course, that’s assuming all 47 other cards are in play. If you’re playing 7-stud, you’re gonna “see” other cards, and your odds will be less.
jackbatty has the right answer, strategically speaking. There are many times when you’ll want to draw one card when the ‘odds’ suggest drawing two, and vice versa. Those strategic choices will far outweigh the the odds improvement from the mathematically corrent strategy, if your goal is to take the money.
I think it’s easiest to think of it this way: if you draw two, look at the first card you draw. It might (1/47) give you a four of a kind. If not, the situation is that you have three of a kind and a single, exactly as if you just drew one.
So your first draw is a ‘free’ shot at getting your four of a kind.