I thought this was pretty neat. I’ve been playing Texas Hold ‘em poker for a few weeks now to try to learn the game. I found the Yahoo! Games version to be pretty pointless. There are 10 or 11 players to a poker table, which makes bluffing impossible, and betting is also limited to raising the pot like 5% at a time. With these factors combined, there is hardly any opportunity for strategy. A quick Google brought me to AOL’s version, which I have to admit is pretty good. I’ve been playing a few hands every now and then. The particularly neat thing is that I was just dealt a royal flush of spades, arguably the best hand possible. The odds of this are 1 in 123,760. I’m not sure which, but I suppose the poker gods were either trying to coerce me to play more with an early gift of encouragement, or say, “alright here’s a royal flush of spades, now give it a break and go do something else already.” Either way, here’s a screenshot:
Not too bad after playing around what I guess to be about 700 hands. This is actually an interesting case of probability. At first, my naive self thought the probability of being dealt a royal flush of any suit out of the 7 cards dealt in Texas Hold ‘em would be 52/52 (any card) x 51/51 (any card of the remaining 51) x 20/50 (10, J, Q, K, A of any suit) x 4/49 (Another royal flush card of the same suit) x 3/48 x 2/47 x 1/46, or 0.0001888%
As it turns out, this is not the case since this doesn’t correctly account for a 10 – Ace being drawn before the third card (we have 2 “any” cards out of 7). If you’re given two chances, for example, of drawing a 3 from a numbered set of ten, you get two 1-in-10 chances which are additive. A chance of 1 in 9 assumes you botched the first try. Likewise, a 10-A of another suit could be dealt before the royal flush becomes restricted to a specific suit by the third card. Since these conditional probabilities can become too complicated, this almost has to be looked at through calculating possible outcomes.
We have to find out how many combinations there are for a royal flush, and then divide that number by the total number of possible 7 card combinations. For a royal flush in each suit, 5 cards out of the 7 have to be 10-A. There are now 47 cards remaining in the deck. 47 x 46 / (2×1) = 1081 possible combinations of those remaining 2 cards. We divide by 2×1 since we’re looking at unordered combinations, so, for example, 6,7 is no different from 7,6. 1081 x 4 suits equals 4324 royal flush combinations. Divide this by the total number of 7 card combinations, or C(52,7) = 133,784,560, and you get 4324 / 133784560 = 0.003232%. 0.003232%, or 1 in 30,940, of poker hands dealt will contain within them a royal flush. One fourth of those will be a royal flush of spades, or 1 in 123,760.
Of course, if you’re already dealt 2 cards out of a royal flush, your chance of complete your royal flush increases drastically. The total number of 5 card combinations dealt on the table, or C(52,5) is 2598960. Again, we’ve got 47 cards left to fill in the 2 non-royal flush cards, so there are 47×46/2 = 1081 royal flush combinations which use your hand, and 3 additional royal flush combinations which don’t. 1084 / 2598960 = .04171%, or roughly 1 in every 2,400 hands will deliver your royal flush.
I suppose the most awesome hand possible would be being dealt a 10, J of spades, then seeing Q,K,A of spades on the flop, all in order. Then two additional aces to boot. Cross your fingers an opponent has the fourth ace, and pretend like you’re trying to hide a bluff. You’ll be rich real quick. The probability of this hand would be a mere 1/52 x 1/51 x 1/50 x 1/49 x 1/48 x 3/47 x 2/46: 1 in 1123790304000, so don’t count on it! If this were heads up, your opponent would have 44 possible two card combinations with the last ace, out of 45×44/2, or 990 = 1 in 22.5. So, are there better options than A-A for the turn and river? Game theory-calculating Beowulf cluster asplode.
I guess, then, my royal flush is not as spectacular as I initially thought. And in retrospect, I don’t think the poker gods really cared. The math gods just wanted to give me a math problem. Darn you, Muhammad ‘Al-Jabr’ ibn Musa Khwarizmi.