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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:

  1. Non ace
  2. Ace
  3. Ace
  4. Ace
  5. Non ace
  6. Ace

I would think we'd have the following?

48/52 * 4/51 * 3/50 * 2/49 * 47/48 * 1/47
Andrew
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    You will likely get better answers if you could rephrase your question without the poker-specific terminology. – Stephan Kolassa Jun 15 '20 at 15:19
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    You witness a miracle in every game. For instance, last night I saw a game in which the flop and the first river cards comprised the two of spades, the three of spades, the seven of diamonds, and the Jack of hearts. What were the chances of that?! – whuber Jun 15 '20 at 17:10
  • Re the edit: you seem to be supposing all six events are independent, but they obviously are not: events 2,3,4 and 6 imply 1 and 5 must occur. – whuber Jun 15 '20 at 19:24
  • @whuber surely you can understand why seeing 4 aces in a row is more exciting than a 2, 3, 7 and Jack. Exciting enough to delve in to the probability of that happening. I'm here to learn. You seem to be here to reply with snarky comments. I'm asking what the probability is of the entire scenario happening and offering my attempt at a solution. Feel free to correct me. – Andrew Jun 15 '20 at 19:36
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    My comments are intended to be instructive, not snarky, so I'm sorry that you are reading them that way. A deeper issue concerns the meaning of "entire scenario:" would that be observing four aces exactly as you initially described? Would it mean observing four aces by the end of the game? Observing four aces during one of the night's games? Observing four of any kind in any of those situations? Arguably, the most pertinent characterization is "what's the chance of observing something during a night of poker that looks unusual?" The answer to that is almost 100%. – whuber Jun 15 '20 at 20:51
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    For more about this kind of question, please see https://stats.meta.stackexchange.com/a/3207/919. (An old meta-joke is that the chance of not observing something unusual is so small that this, too, must be considered "unusual" and therefore you will always observe something unusual!) – whuber Jun 15 '20 at 20:52
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    Hey @Andrew. I just noticed your updated post asking for further questions--it's awesome that you're interested in these probability questions--typically this should be a separate post that fully explains the question and what you're interested in. For example, does the order matter? Similarly, you mention 'ignore the poker aspect,' and it's not fully clear what you mean by that, or whether there's some misunderstanding there that we can all better understand. Typically, these kind of reasons suggest a separate post to really isolate your question, and... – doubled Jun 15 '20 at 21:46
  • ...also gives you an opportunity to fully show your work and how you've thought about it. – doubled Jun 15 '20 at 21:47
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    Also, to respond to your comment about @whuber's comment, I think such comments are indeed thought-provoking, and more generally, maybe I'm too optimisitic, but I think one of the virtues of sites like this is one is that typically, people genuinely want to help by providing new ways to approach a problem. When I first learned about probabilities, I think a comment like that one would have really helped me understand what we really mean by 'low probability events.' 4 Aces is awesome, but so is any 4 of a kind, and similarly seeing A-K-Q-J-T suited, or indeed any straight flush, would be cool:) – doubled Jun 15 '20 at 21:51
  • There's no "first" river card. Did you mean the turn card? – xehpuk Jun 16 '20 at 00:02
  • @xehpuk yes, I meant turn card. Surprised I made it this far thinking they're both river cards. – Andrew Jun 16 '20 at 17:05
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    @whuber apologies, tone is easily lost in text. Appreciate the input! – Andrew Jun 16 '20 at 17:08

1 Answers1

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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 :).

doubled
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