Odds Of Picking Five Specific Cards From A Pool Of 20

An online poker game I play has this side game that players can qualify for, every few dozen or so hands. You’re given five cards needed to make a Royal Flush (10-J-Q-K-A, in one suit) and 15 random cards. Get the Royal Flush, win tens of millions of in-game chips.

What are the odds?

In order to win you need to get a winning card each draw:

1st draw: 5/20
2nd draw 4/19
3rd draw 3/18
4th draw 2/17
5th draw 1/16

Multiplying those fractions gives 120/1860480, or 0.000645.

Good luck!

What zimaane said.

The fractional version simplifies to 1 chance in 15,504, which may be the easiest way of thinking about it.

Or if you get 10 chances a day to play this game, you should win once every 4 years and 3 months, on average.

I think you’re missing a zero—0.0000645.

You are correct. My bad - hope no one got their hopes too high.

Strictly speaking, the chances are slightly better, because it’s possible that the 15 random cards contain another royal flush in a different suit.

I interpreted is as 15 random cards between 2 and 9 that could never be part of a royal flush. I could be wrong though.

Given that it’s an online poker add-on game, I’d strongly bet, either:
a) the other 15 cards are chosen to exclude the possibility of a royal flush, or

more likely, b) the programmer didn’t even think of that possibility and so the program doesn’t check for that. You could show 10-A of a different suit and the program would say “Sorry! Better luck next time”

On the distribution of the non-essential cards:

I don’t know the algorithm, but I do know that the cards are such that every possible winning hand (pair, two pair, etc.) can come up. I’ve seen face cards of other suits come up as well, but I don’t think I’ve seen enough so that it’s possible to get a royal flush in either of two suits. But the algorithm does make sure that it’s possible to get a royal flush in one suit.

Obviously, there’s no telling how well this is all coded up and, if real money is on the line, I would expect that the servers are outside of the US and that you likely can’t sue them if the implementation doesn’t follow a reasonable reading of the stated rules.

But, I would doubt that an online casino would be stupid about probabilities.

If they’re giving tens of millions of credits for a 1 in 15.5k probability in the worst case then either credits aren’t worth much money or they’re more interested in making sure that there are enough jackpots going on to entice people into playing their other games, that they don’t mind losing out the tens of millions. Since you can only play the game after playing a dozen other games, the frequency is pretty strongly throttled. They can lose at a slow enough rate that their wins far exceed it, even with what are relatively good odds for the players in terms of a payout/probability ratio.

sorry for the hijack, would this be the same method to figure the odds in Keno of picking exactly 20 drawn numbers correctly of 80 possible numbers, as 5/20 = 20/80? Something so rare I don’t think it has ever happened, but has a payout of not even a million?

Same “method”, sure, but picking 20 out of 80 gives astronomical odds compared to 5 out of 20.

ETA since the Keno odds are so long, I would use a different method involving approximations instead of trying to compute the exact value. Or just feed it into the computer.

Same method, and the odds are 1 in 3,535,316,142,212,174,320.

thanks, yes the odds are over a quintillion, but wasn’t sure of the method.

If I can continue the hijack once more, why is there a point when Keno odds do this:

1 1 in 46.446
2 1 in 20.115
3 1 in 8.009
4 1 in 4.877
5 1 in 4.287
6 1 in 5.258
7 1 in 8.826
8 1 in 20.055
9 1 in 61.420

More likely to get 5 of 20 than 4 or 20, if I understand correctly?

Roughly speaking, each of your 20 picks has a 1/4 chance of hitting a number, so on average the most likely result is that you would expect to get 5 out of 20.

I was trying to do a slightly more comprehensive, but still simplified explanation to expand on what DPRK wrote, and I don’t think any approach really adds much until you get to actually learning how the hypergeometric probability distribution is calculated.

Which isn’t all that hard, but requires understanding combinations and permutations.