In Texas Holdem, what at the odds of 3 aces on the flop, then the 4th ace drawn on the first river card? I played Texas Holdem with some friends last night. On one hand, the dealer burned a card, drew 3 aces on the flop, then burned a card and drew the 4th ace on the first river card. My question is, what are the odds of happening? It felt like witnessing a miracle.
@doubled answered this to apply to poker which is appreciated. I wonder now what we might expect if we ignored the poker aspect and did include the burn cards from a completely randomized deck:

*

*Non ace

*Ace

*Ace

*Ace

*Non ace

*Ace

I would think we'd have the following?
48/52 * 4/51 * 3/50 * 2/49 * 47/48 * 1/47

 A: This is the same as the probability of successively drawing any 4 specific cards in a row (as @whuber points out, seeing 4 Aces is just as likely as seeing any 4 specific cards, and what makes Aces special is that we as poker players put weight to that specific realization of cards). We have 4 aces to draw from for the first card, 3 for the second, 2 for the third, and 1 for the fourth. Each time, there is one less card in the deck left, and conditional on adjusting for the number of cards left, each probability is independent, and so the probability is simply:
$$P(\text{drawing 4 aces in a row}) = (4/52)*(3/51)*(2/50)*(1/49) \approx .00037\%$$
I multiplied the answer by 100 to get a percentage. Note that this assumes that the deck was truly randomly shuffled, which in friendly play is almost never truly the case (indeed, this may be one reason why we often 'burn' a card, and this randomness assumption is also why the burning  of a card has no effect on the probability), but regardless, a crazy and fun event to happen :).
