# Are cards drawn from a uniformly shuffled deck considered independent events?

When playing poker I sometimes run into a situation where the board has 3 to a flush and I'm holding a 4th card of the same suit so I need the river to give the 5th. I understand that the probability of drawing it are `(13 - 4) / (52 - 6) = ~20%` [1]

But can we use the fact that a uniformly shuffled deck is unlikely to have long streaks of one suit to estimate a lower probability?

Here's a contrived example: Imagine you have a uniformly shuffled deck of cards. Miraculously you pull out 6 diamond cards in a row from the top of the deck. It seems intuitive that pulling another diamond would be very unlikely.

This is in contrast to a game like roulette where landing on red 6 times in a row in the past would have no effect on the outcome of the next spin because they are truly independent events rather than shuffled cards.

[1] Because there are 13 total of the given suit, 4 visible to us, 52 total cards and 6 total visible cards

Assuming a fair shuffle we cannot assume a deck is uniformly shuffled. Thus is the nature of a fair shuffle. Uniformly means and implies either or both that things will be the same in all cases at all time, or occur in equal amounts, evenly.

A fairly shuffled deck of cards has 52! number of combinations. Millions of those combinations will be 13 cards of the same suit in a row. What has come before in terms of cards you know to have pulled influence the probability of what is likely to occur, but has no baring on what the next card actually it. We can say with a greater degree of certainty that it is less likely to be a card of the same suit, but there is still chance of it bucking the trend. You can only make calculations on what information you have, you can make assumptions but they are just that, assumptions. Your assumptions have no baring on what will come next, but can and should be used to make correct decisions based on maths.

I think comparing a shuffled deck game that has multiple stages of play to roulette may not be the best example. Roulette is easy to identify that the independent event was the spin and putting the ball in. However in a card game you could think that the independent event is each street, and they are in a way, but in terms of the deck, the independent event is the shuffle. Assuming a fair game again, even if you don't make it to see all 5 community cards, the deck has been set from the beginning of the hand, even if we don't get to see those cards.

Anyway to answer the main question:

But can we use the fact that a uniformly shuffled deck is unlikely to have long streaks of one suit to estimate a lower probability?

Again we can't assume uniformity in a fairly shuffled deck, but yes what information we know influences the probability, which can be used to make decisions, however it does not change the reality of how the deck has been dealt.

Drawing two cards from a deck are definitely not independent events. Roulette can hit "0" twice in a row, but the deck cannot hand out the ace of spades twice in a row.

But can we use the fact that a uniformly shuffled deck is unlikely to have long streaks of one suit to estimate a lower probability?

No, because you're dealing with prior probabilities here. We can imagine all the possible shuffles. But many of them have been excluded by your knowledge. In the case of the diamond draws, you don't want to know "how likely is it that this deck has diamonds as the first 7 cards?". What you want to know is "given that that this deck diamonds as the first 6 cards, how likely is it that there is another one following?"

It turns out that after you get rid of all the decks that don't start with 6 diamonds, the chance of the next card being a diamond is exactly the same as the probability that you computed.

Yes a long streak is unlikely. But that fact isn't helpful when you've already got information that most of the streak is in place.

But can we use the fact that a uniformly shuffled deck is unlikely to have long streaks of one suit to estimate a lower probability?

Yes, but you already did that. It was represented as a `- 4` in your calculation.

(13 - 4) / (52 - 6) = ~20%

There is no need to try to factor this in twice.

A deck that is classified as completely random does not lose its property of being completely random in instances where arbitrary combinations of cards are pulled from the deck. So if the deck started the round as completely random, the chances you'll find the 5th card to that flush will be just as likely on the top of the deck as any other spot

"Miraculously you pull out 6 diamond cards in a row from the top of the deck. It seems intuitive that pulling another diamond would be very unlikely."

This is incorrect. All 46 cards that remain on the deck are equally likely to be the next one. 7 of those are diamonds, which gives you a 15.2% chance.