# Playing with an infinite copies of a deck

Let's assume we are no longer having a deck with 52 cards, but infinite cards, where every card has a probability of `1/52` to be received. With the logic, if you have two cards, the probability for a combination is no longer `1/52 * 1/51`, but `(1/52)^2` (that's the condition of the infinite deck).

Question: Will the order of hands be swapped? What will be the probabilities (and probabilities difference) for each hand? Which will be more often, which less?

• I guess the question would be clearer if you said "infinite copies of the deck" rather than an "infinte deck", which can be understood in several different ways. Hand like quads would definitely be easier to get. Straights get harder on the other hand – David Oct 3 at 12:47

The new rule makes the probability of each card 1/52. But you will still have the same behavior in the "hand rareness" sense of the game.

making one pair is easy because you only need 2 cards of the same number (each number has 4 occurrences in the same 52 card-set). So it will be (52/52)x(3/52)x(52/52)x(52/52)x(52/52) = (3/52).

3/52 only because I assume the first card was drawn in the first slot and there are 3 occurrences left. If you say drawing a card doesnt actually eliminate that card from the probability, then the probability is 4/52.

However, a royal flush is way harder because you will need 1 specific card in each slot of the 5 cards hand. So it will be (1/52)x(1/52)x(1/52)x(1/52)x(1/52) = (1/52)^5

• Actually it would be 4/52’s because the same suit can reappear again in an infinite deck, but thanks gor pointing that out. Wouldnt a royal flush be 4/52 * 1/52... due to the suits? – Anatoly Wein Oct 3 at 9:47
• Yes you are correct about the royal flush. Its (1/52)^5 for each suit. So it you dont care about the type of royal flush (and in a normal poker game, you dont), the actual odds will be (4/52)^5 – jackhammer Oct 3 at 9:58

Was hoping someone would do the math but what would happen is the convergence of rareness for 5-card hands. If the value of a hand is based on its improbability (which is not always the case), then the values of 5-card hands also converges.

The way i look at it is outs. In particular, hands which block their own outs increase in probability, while hands that don't, decrease in probability. For eg, a pocket pair is now 33% more likely than before, because after drawing your first card, you have 4 outs to make a pair, instead of 3. You are twice as likely to flop a set (4 outs instead of 2), and twice as likely to hit a boat/quads by the river. Hence, the value of boats/quads is significantly lower.

For a flush, its somewhat similar, but the difference in outs is less significant. For eg, a flush draw is now drawing to 13 outs instead of 9. Not double the probability like the previous scenario, but still about 45% more likely.

For a straight, the reverse is true. Straights dont block their own outs, so a OESD would still have 8 outs. Since you are now drawing from a 52 card deck instead of 47/48 card deck for turn/river, your odds actually decrease. Thus, the value of a straight should increase.

You also have to have extended rules for new hands, i.e. 5 of a kind, exact pairs, 5 of a kind flush?. Also, a royal flush is almost equally unlikely (kinda disagree with jackhammer that its way harder).