There are 6 different 2-card combinations that WILL contain a pair of Aces — Comb(4, 2):

* |
♦ |
» + |
* | ||||

L |
V |
* |
L |
Y |

The number of 2-card combinations that WILL NOT contain a pair of Aces is Total Possibilities (1,326) — WILLs (6) = 1,320 WILLNOTs.

The odds against being dealt Aces in the hole in Hold'em are:

WILLNOTs (1,320) : WILLs (6) 1,320 : 6 Reduce 1,320/6 : 6 / 6 220 : 1

Another way is to determine the probability of being dealt Aces in the hole, by multiplying the probability of being dealt the first Ace by the probability of being dealt the second Ace:

This probability, expressed as a percentage (.0045249 * 100), is less than one half of one percent or .45249%.

As a check, we can convert the probability expressed as a decimal ratio to 1 (.0045249) back to odds:

WILLs^^— = Probability as decimal ratio to 1 Total Possibilities

Odds = WILLNOTs : WILLs

Total Possibilities = 1

- WILLs = .0045249 WILLNOTs = .9954751

WILLNOTs : WILLs .9954751 : .0045249 Reduce

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