In Omaha Hi/Lo, what are the odds of being dealt a starting hand with an Ace and a 2 in it? If you are unfamiliar with Omaha, there are 4 cards in a starting hand. I would also appreciate the mathematically explanation of how to figure this out.
2006-08-21 15:29:36 · 3 answers · asked by mysticman44 7 in Games & Recreation ➔ Card Games
No, because the chances of getting an Ace and a 2 when 2 cards are dealt is 1.2%, because you have an 8 in 52 chance of the first card being an Ace or 2, and a 4 in 51 of the second card being the other. 8/52 * 4/51 equals 1.2%. Since there are 4 cards dealt, the odds have to be higher.
2006-08-21 16:50:54 · update #1
Okay, it's been a few years since I was last in a math class, but this is doable...
There are 4 aces in a deck of 52 -- 1/13 odds of being dealt an ace in one turn. The odds of getting a deuce in a single turn are the same. However, the odds of getting them consecutively in two turns requires those odd to be multiplied -- 13x13 = 169 or 1/169.
Now, since we are dealt four cards, and not just two, our chances of getting the desired cards are doubled to 2/169 or roughly 1.2% of the time.
But consider this... each player has those odd of 1.2%. So if there are 8 players at the table, the odds are approaching 1 in 10 that SOMEONE at the table will have an ace/deuce.
This all assumes that my rusty math is good. Anyone confirm this?
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Your added detail could be correct -- I'll link this question to the education category, and we'll try to get some math teachers up in this place. If your supposition is correct, the remainder of the problem would still be true -- four cards dealt would double the odds, in this case to 2.4%.
2006-08-21 16:40:53 · answer #1 · answered by Unknown User 3 · 1⤊ 0⤋