So this is a Bayesian question in words first and then I'll try to put a little mathematical meat on it. Admittedly, this will eventually be about teleological reasoning, but I would like you to just consider the thought experiment as it is posed.

Suppose you are seated at a poker table for the very first time and, for your very first hand of poker, you are dealt a Royal Flush. What are you gonna think? Might you suspect that the deck was stacked or not?

We could amplify the question and ask to consider what you might think if, instead of just one hand, for your very first and second hand of poker, you get for both hands a Royal Flush. Would you think that you're just very lucky or would you suspect that the dealer is stacking the deck?

In both cases, the alternative is accepting very long odds on the outcome.

Let event A be that the dealer (who might be beneficent) stacks the deck.

Let event B be that you receive a Royal Flush on your very first hand of poker, dealt by this dealer.

¬A means "not A" or that A did not occur. It's the logical complement of event A.

P(A) is the probability of event A.

P(B) is the probability of event B.

P(A|B) is the probability of event A occurring, given that B occurred.

We know that:

P(A) + P(¬A) = 1 (and the same for B.)

P(B|A)P(A) + P(BA)P(¬A) = P(B)

and Bayes' Rule:

P(AB) = P(A|B) P(B) = P(B|A) P(A)

We want to solve for P(A|B) in terms of everything else. When I solve for it, I get this expression:

P(A|B) = _________________________________

          1 + (1/P(A) - 1) P(B|¬A)/P(B|A)

Now P(BA) is the probability of being dealt a Royal Flush when the deck is not stacked and that is well known to be a very small probability: 4 × (47! 5!)/52! = 1/649740

Now, with the assumption that the dealer is beneficent (he or she likes you), so that if the dealer would stack the deck, they would stack it to favor you. So P(B|A) would be the likelihood of a Royal Flush if they stack the deck and is much larger than P(BA).

So the question is if the right-hand term in the denominator is closer to 0 (much less than 1) or closer to infinity (much greater than 1). If that term equals one, it's even a posteri odds that the deck was stacked. If you insist that P(A) is zero (or astronomically small) in the first place, that it is essentially impossible for the dealer to stack the deck, then of course it is still impossible if you get a Royal Flush. 1/P(A) is astronomically large and even when multiplied by a small number, it's still large, the denominator is large and you get something near zero as a result.

But what if I do not grant that? What if we start with an assumption that the odds are 50 to 1 that the deck is randomly shuffled? Or 100 to 1? Or 1000 to 1 (in favor of the deck being randomly shuffled)? Then 1/P(A) is not astronomically large and the smallness of P(BA)/P(B|A) is salient.

So what do you think? You are seated at a poker table for the first time and, for your very first hand of poker, you are dealt a Royal Flush. Would you suspect that the deck was stacked or not?

  • 1
    Is there a question here beyond "what do you guys think?"? This is not a discussion forum.
    – Conifold
    Apr 11, 2020 at 6:12
  • yes. read the question. Apr 11, 2020 at 6:12
  • The question is plainly stated at the beginning: "Suppose you are seated at a poker table for the very first time and, for your very first hand of poker, you are dealt a Royal Flush. What are you gonna think? Might you suspect that the deck was stacked or not?" Apr 11, 2020 at 6:13
  • Yes, there is also "Might you suspect that something nefarious might be happening?" All of which are variations on the last one, and off-topic here. You mentioned something about "teleological reasoning", maybe something there is not of the "what do you think" sort?
    – Conifold
    Apr 11, 2020 at 6:16
  • 1
    If you want to use latex in a question or answer, go to a site like this, where you can create and render your latex, download it as an image, and post the image to the site (this site only allows you to download images in CodeCogs mode). it's not optimal, but it's better than nothing. Apr 11, 2020 at 14:19

5 Answers 5


The short answer is that you can't tell, because you are working from a single data point. For all you know, the dealer has dealt hundreds of thousands of hands and you just happened along at a propitious moment. Royal flushes are rare, but they do happen, even with properly shuffled decks. Winning the lottery ten times in a row is different because you have ten data points. Each win is unlikely and the combination of all of them is extremely so.

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    Thank you for putting it into an answer. First thing I would say is that you can put a compound event (or a set of events, whether they are sequential in time or not) into a box and call that a single event. It would be like me asking "Imagine you are seated at a poker table for the first time and, for your very first 10 hands of poker, you are dealt a Royal Flush each hand. Would you suspect that the deck was stacked or not?" Now we're back to a "single data point". It's a quantitative difference (even longer odds), not a qualitative difference. So the question remains. Apr 12, 2020 at 5:12
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    True, but your example has a probability of roughly one in a million. That is not so small given the number of poker hands that are dealt in the world. If the example were changed to an extremely improbable event, such as playing bridge and all four players being dealt 13 cards of the same suit, then I would definitely conclude the dealer was cheating.
    – Bumble
    Apr 12, 2020 at 6:21
  • thank you for answering the question. so if the odds are 1/649740 , you would think that you were simply "lucky" and that the deck was not stacked, but if the odds were significantly smaller, like (1/649740)^10 , those odds would be so small that you would suspect that the deck was stacked. do i represent your position correctly? Apr 12, 2020 at 16:32
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    The probability has to be taken together with the base rate of incidence. Millions of poker hands are played every day, so a 1/649740 rate of incidence is fairly high in context. Ten royal flushes in a row on consecutive hands to the same player is so improbable even given the base rate, that we could safely say the dealer is cheating.
    – Bumble
    Apr 13, 2020 at 6:46
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    If you are going to reason in a Bayesian fashion then one of the applicable principles is the principle of total evidence, i.e. you must conditionalise on all of your available evidence. I cannot pick and choose among the evidence and ignore what I know or at least reasonably believe to be true. Examination questions and puzzles that ask you to use only the information that is present in the question are artificial and unsatisfactory.
    – Bumble
    Apr 14, 2020 at 8:43

TLDR: It depends on the prior probabilty you assign to the proposition that the dealer is stacking the deck in your favor, and to the conditional probability of being dealt a royal flush given that the dealer is stacking the deck in your favor.

It sounds to me like you've hit upon the "problem" of the priors in Bayesian epistemology. Bayes' theorem tells us how to update our beliefs in light of new evidence, but it doesn't tell us how to assign the initial ("prior") probabilities that get updated. This is analogous to propositional logic, where the rules of logic allow us to infer propositions from other propositions, but it doesn't tell us how we should assign truth values to atomic propositions. Whether or not you conclude that the dealer is stacking the deck in your favor will depend crucially on the prior probability you assign to the proposition "the dealer is stacking the deck". For a known cheater who has been caught doing such things before, we might assign a high probability to the proposition.

I, personally, would assign a very low prior probability to the proposition that the dealer is stacking the deck in my favor; I don't personally know any dealers, nor do I know anyone who knows any dealers, and even if I did I have no reason to think any of them would do such a thing (wouldn't they consult with me first and try to get a cut?). For the sake of argument, let's say that I assign the probability of a particular dealer stacking the deck in my favor a value of one in a billion (i.e. not impossible, just very unlikely). Let's assume for the sake of argument that this dealer can shuffle perfectly, so that he could guarantee I would always get a royal flush. So, P(royal_flush|deck_stacking) = 1. Next let's compute the probability of P(deck_stacking|royal_flush) using Bayes' theorem and the information above:

P(deck_stacking|royal_flush) = [P(royal_flush|deck_stacking) * P(deck_stacking)] / P(royal_flush).

Let's calculate the value of the denominator using the law of total probability:

P(royal_flush) = (1/649740 * 0.999999999) + (1 * 0.000000001) = ‭0.00000154007717

(It's approximately the same as getting a royal flush without deck stacking, since the likelihood of deck stacking is so low, according to my priors.)

Okay, so going back to Bayes' theorem:

P(deck_stacking|royal_flush) = [P(royal_flush|deck_stacking) * P(deck_stacking)] / P(royal_flush)
                             = 0.000000001 / 0.00000154007717
                             = 0.0006493181118

Still, a very low number. So no, given the value above, I would not conclude that the dealer was stacking the deck in my favor.

What about two royal flushes in a row? Now things change considerably, since two royal flushes in a row have a much lower probability:

P(deck_stacking|two_royal_flushes) = [P(two_royal_flushes|deck_stacking) * P(deck_stacking)] / P(two_royal_flushes).

The denominator is:

P(two_royal_flushes) = (1/422,162,067,600‬ * 0.999999999) + (1 * 0.000000001) = ‭0.0000000010023688

The deck stacking term, the (1 * 0.000000001) "weighs" a lot more here, since two royal flushes occurring by chance are so much more unlikely. Plugging this value into Bayes' theorem gives us:

P(deck_stacking|two_royal_flushes) = [P(two_royal_flushes|deck_stacking) * P(deck_stacking)] / P(two_royal_flushes)
                                   = 0.000000001 / 0.0000000010023688
                                   = 0.998

Now it seems rational to conclude that the dealer is stacking the deck in your favor. The effect is compounded the more royal flushes you get dealt.

Again, the above analysis is all very sensitive to the prior probability you assign to the dealer stacking the deck in your favor. But for most of us, I imagine this probability would be quite low, and unless you increase it several orders of magnitude over one-in-a-billion, luck is still more likely than the dealer stacking the deck for a single royal flush.

(I'm a visual thinker, so I find this is all easier to understand intuitively if you draw some "Bayes' bars", that represent your sample space. You segment it into two regions representing the prior probabilities, and shade in a region of each of these two segments representing the conditional probability of (some number of) royal flushes representing the conditional probabilities. Finally, you compare the sizes of the shaded regions. I may upload some visuals later.)

  • this is interesting (but i haven't read through all). i also would assign an extremely low a priori likelihood that the dealer is stacking the deck. so i would say P(A) is small, but not astronomically small. not that i know any dealers nor can imagine how to get away with it, but i accept that i am no expert and that some skilled card shark can do things that i might not see. so 0<P(A)<<1. but even so, since i already know how unlikely it is to get such a hand for any specific hand (and i specified it as your very first hand), my choice is between the unlikely P(A) and the unlikely P(B|¬A). Apr 15, 2020 at 1:49
  • @robertbristow-johnson Yes, it comes down to just how small the prior P(A) actually is. Just to anticipate where you might be going with this (given that you mentioned teleological reasoning, and given your recent comment on Bumble's answer) I think you'll really like John Hawthorne's discussions here, here, and here on Bayesian teleological arguments. I'm in complete agreement with what he says in these talks. Apr 15, 2020 at 17:40
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    Whether or not one thinks fine-tuning is a good argument will partially depend on the prior probability one gives theism. Like our dealer-deck-stacking hypothesis, B is evidence for our hypothesis, but whether P(A|B) high enough that we should believe it will depend on how serious we took that hypothesis in the first place. If we were on the fence about theism/deck stacking, fine-tuning/royal flush is awesome evidence for theism/deck stacking. If we think theism/deck stacking is astronomically improbable in the first place, fine-tuning/royal flush might not be enough to tip the scales. Apr 15, 2020 at 17:41
  • The question really is if P(A|B) is as much as 1/2 (so stacking is more likely than random if you got a royal flush), but that depends on if (1/P(A)-1) P(B|¬A)/P(B|A) is greater than 1 or less than 1. If you make P(A) to be virtually zero, then that quantity is large even if P(B|¬A) << P(B|A). But that's starting out with the prejudice that the dealer just virtually cannot be stacking the deck, so getting a royal flush is not gonna change your mind. I wanted to pin you down on what you thought P(A) and P(B|A) are, since those are the unknowns. Apr 21, 2020 at 3:19
  • //" If we were on the fence about theism/deck stacking, fine-tuning/royal flush is awesome evidence for theism/deck stacking. If we think theism/deck stacking is astronomically improbable in the first place, fine-tuning/royal flush might not be enough to tip the scales. "// ------ I agree with this. It comes from what values you might assign to P(A) and P(B|A). Apr 21, 2020 at 3:26

I am not sure if I am answering your question correctly, partially because I am unsure as to what you are trying to achieve with your question, but looking into Bayesian statistics I found the following link:


There are various loopholes in Bayesian inference: 1. In most real-world examples, obtaining the prior probability is often hard. 2. When both prior and likelihood become too complicated, MCMC (Markov chain Monte Carlo) sampling is used. This is extremely slow in real instances. 3. In order to quantify the prior knowledge, the user can influence the result, unknowingly or knowingly.

..It sounds like the best you could do is compare a normal distribution of the odds the dealer is cheating with a a normal distribution chart for the odds of getting royal flushes, but this would not give you certainty either way. you should google the Markov chain Monte Carlo which might lead you towards what you are wondering.

With the information you presented, we can calculate the odds of getting two royal flushes in a row and we can not calculate the odds of the dealer cheating. I agree with bumble that you could not know if the deck was truly stacked as their is not enough information.

  • Well, it looks like i have to make a decision on the bounty. The question did not provide you with additional experiments for you to get an idea of what is likely. Just one experiment where you know that the game is poker and that the odds of getting a royal flush from a randomly-shuffled deck is 1/649740 . You can assume that P(A) is small (you wouldn't expect, from the outset, that stacking the deck is common) but not zero (because you know it's possible and that there is this notion of stacking the deck, so it's not unheard of). Would you consider P(A) to be so small that P(A|B) < 1/2 ? Apr 21, 2020 at 3:12

I think there might be a natural way to progress by stepping outside of the Bayesian formalism for a bit. The question I always want to start from in any problem where we think Probabilistic reasoning might be applicable is to ask “What is our state space?”

Poker has an obvious state space as a finite game. The 52 cards in a deck can be stacked in (Factorial(52)) ways, but we’re interested in the 5 cards you are specifically given and can collapse the order of those 5 cards, so it’s just 52C5. The configuration B that you are interested in is that the 5 cards you have been dealt match one of 4 patterns - the TJQKA of a matching suit in any of the 4 suits. This is a very small fraction of the poker state space, and so we say that it has a low probability in a random game.

Clearly, however, the state space of the poker game is not itself the deciding factor in your scenario, because now we introduce a Dealer. In your game you’ve been “dealt” a Dealer, who has dealt you 5 cards in some configuration. Depending on the properties of the dealer, you may have been dealt different configurations - some dealers might only deal busted flushes, others might consistently deal a mix of mid-value hands, and others might be completely fair.

So, what were the odds, given only one piece of information, that we were dealt this particular dealer (or one satisfying some more general property)? The answer to that question requires that we enumerate the state space of dealers, and look at what the information we’ve received about this hand tells us about what possible dealers we might have been dealt.

What is a dealer? A hugely simplified model would say that a dealer in abstract is a one-time function from the set of 52 cards to a 5-card sample set. In this view it should be clear that there are exactly as many dealers as there are 5-card permutations, and so the probability that you would get this hand from this dealer is always 1. However, this only allows for dealers that would then continue to deal you this exact hand. So this is maybe not so interesting as a model.

It seems like we could propose that dealers themselves model a local probability distribution. Each dealer deals each possible hand with some probability. For example, a “Fair” dealer assigns each hand the same probability, a “Constant” dealer gives exactly one hand with probability 1, a “Weighted” dealer gives some hands more often than others.

This is a wholly mathematical model, so there seems to be a basis for some discussion of principles independently of any given prior configuration! Instead of asking “what should our starting Bayes weight values be”, we say “we know we’re somewhere in this particular state space, and want to talk abstractly about how what conclusions we can draw given particular events”.

Unlike the simple Poker model, though, progressing an answer to your question is tricky given that this model is uncountable. Each Dealer is a product of 2.6 million dimensions, and our measure on each of those dimensions is the open (0,1) interval. We have continuum many such possible dealers, so even acknowledging that the probabilities of all of the possible poker hands sum to unity, trying to talk in such absolute generality about this will need some powerful analysis.


So, let’s take this core idea and apply it to a simpler game. Cho-Han dice rolling has a similar concept but we’re only interested in odds/evens. Each dealer has some weighting of evens and odds, and the probabilities of these weightings sum to unity.

We still have a similar problem but the analysis involved is much more tractable; each dealer can be modelled as a single (0,1) value representing the odds of getting an Even number. Let’s call this P(i)(E) - representing the chance that dealer i rolls Even.

Again, there are continuum many possible dealers. However, when we roll the dice and the outcome comes up Even, we start to know something about where in the probability distribution we are. It is more likely that we are in a world where our dealer skews Even than it is that they skew Odd. It is impossible that we might have gotten this result from a dealer that only throws Odd, for example.

Graph of probability of an outcome vs the distribution space of a weighted Cho-Han thrower

For a very loose thought, if we to line up all of our P(i)(E) probability functions in a graph like this, we’d end up with a 1x1 square of possibilities. If each of these possibilities was in some general sense “equally random” but we know that the result of the throw was Even, then we can state some general results about the likely chance our random point fell within a given weighted throw. For example, it looks like the area under this graph is halved at 1/root(2), rather than (as intuitions sometimes suggest) 0.5.

(This basically is a much simplified sketch what Bayesian Learning does in each update, and taking integrals is what we do when we want to make predictions or classifications using our learning to date)

Now this isn’t the same as saying the thrower is confirmed biased. In fact our model pretty much assumes a probability of zero that any given thrower has a bang-on 50:50 odds/evens chance (though they can be arbitrarily close). The point, rather, is simply that we can start to assign numbers to just how much bias they have shown given the outcomes they have demonstrated thus far.

And if, as we often want to do in Bayesian learning, we are interested in using past data as a measure of prediction and a suggestion for future action, this bias measure gives us a first glimpse at what we might expect to happen next.


So, starting to push back to the point at hand, what if instead we talked about a d3, and asked about a thrown result of 3? Well, we’re adding an extra dimension because we’re adding another free variable here (each dealer can now vary their biases for a 1 or 2 in addition to the bias for a 3), so now we’re talking about volumes rather than areas when we take integrals.

This actually suggests we’re more likely to be weighted towards 3 (being non-committal about the other two values) than the Heads case, since 2^(-1/3) is bigger than 2^(-1/2). And that makes sense, because getting a 3 on a d3 is in a sense more surprising than getting Evens on a dice roll.

I really do not have the grasp of analysis needed to translate this to the Poker case. However, there absolutely seems to be some foundation to the idea that the more apparently unlikely a scenario in an unfamiliar game we encounter is, the more we ought to consider it likely that the dealer is weighted in favour of that outcome. And this is a property we ought to assume holds regardless of our choice of priors.

Any conclusion we might draw is itself only probabilistic. And we haven’t considered more complex models of dealership - the Markov Chain model Noah suggests allows not just for a set probability for each state but also the ability to vary distributions over time, such that our dealers could switch up their strategies. This analysis gets very mathematically interesting in even higher dimensions.

But with only one piece of evidence about how a dealer is weighted, and assuming all possibilities of dealership were independently likely to begin with in this one-time game, it’s more likely to be similar than not.


While in the end you can always get just about any answer you want based on how you set the prior probabilities, as you point out; and while you can certainly consider a sequence of multiple events to be a compound event, as you point out; I feel it is important to point out the exponential effect of multiple observations in Bayesian reasoning. Suppose you receive n royal flushes in a row denoted by B_1, B_2, ..., B_n. Then P(A|B_1..B_n) is proportional to P(A)P(B_1|A)...P(B_n|A) and P(¬A|B_1..B_n) is proportional to P(¬A)P(B_1|¬A)...P(B_n|¬A). Thus every time that you observe another royal flush the ratio between these two probabilities changes by another factor P(B|A)/P(B|¬A) (which in this case is very large), so the total effect on the ratio is (P(B|A)/P(B|¬A))^n which is huge. Therefore it's not hard to argue that any reasonable prior would quickly be overwhelmed by this repeated evidence, so that for any reasonable prior you'd quickly suspect the deck is being stacked (which aligns with intuition). (In another response Adam Sharpe worked this out in detail for two royal flushes illustrating the point!)

  • well, i included the amplified question simply to make a point to those who glibly say they would just feel lucky and not suspect anything amiss. but i think it really just squares P(B|¬A) to make it even smaller. it's a quantitative issue, not really a qualitative difference in the nature of the question. Apr 15, 2020 at 1:54
  • That's true for two observations. However in the limit as n gets large one might start to consider this something like a qualitative difference, along the lines of en.wikipedia.org/wiki/Big_O_notation
    – present
    Apr 16, 2020 at 1:55
  • you can put the two adjacent observations into a box and call it a single, even more unlikely, event being observed. so if the alternative to dealing from the bottom of the deck is something as improbable as 1 out of 649740, that's believable, but if the alternative is 1 out of 649740^2, that's suspect? still seems like a quantitative issue to me. how lucky do you have to be to suspect that it wasn't just a randomized deck? Apr 16, 2020 at 7:08
  • Right, that's true for 2 or really any finite number of events -- you can always combine them into a single event. My comment is more about the rate of growth in the ratios. Big O notation illustrates that some types of growth really are, I would say, qualitatively different (e.g., polynomial vs. exponential growth). Here, what that indicates is that any reasonable prior will relatively quickly be overwhelmed by repeated observations that clearly point one way (even if they're individually not as extreme as a royal flush).
    – present
    Apr 16, 2020 at 11:38

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