Hi guys,

First poster, long time reader. I need some help for a project I'm doing. I'm designing a game and I need to know the probability of something.

For a standard 52 card deck - two colours including the jokers (so 54 cards) - and dealt with 5 cards at a time:

what is the probability of getting 4 aces and a joker?


I'm not good with the math solo but if you guys can help or direct me to where I can that would be greatly appreciated.

Thanks lads



Edit: I know thats not really a 5 of a Kind but in this game that's what it's referred to

Welcome to the Probability Forum!

These types of questions are easier than you might believe.

First tally how many different 5-card hands "qualify" for the hand you are seeking (in your example, how many different ways are there to hold 4 of the 4 aces and 1 of the 2 jokers?).

Then tally how many different 5-card hands are possible (in your example, how many different ways are there to select 5 cards from a 54-card deck?).

Feel free to take a crack at these two.

(Clearly, the probability is the first number divided by the second number.)

There are 2 combinations of AAAA+joker, and 54C5 5 card combinations in a 54 card deck.
2/54C4=1/1581255
Odds against: 1581254 to 1

Also, I would just mention to be sure you are defining your terms precisely.
If you're looking for 5 of a kind using all 4 Aces and only one wild card, see brown's post above.
If you are looking for any 5-ace combo (which can use either 4 aces and one joker, or 3 aces and 2 jokers), then there are 6 combinations, or 527,084 to 1.

Thanks for your help guys. I'll level with you, this is for a game Im designing which requires the probabilities of poker hands (from Royal flush to one-pair or high card) and one of the 'ultimate' goals is to get what I call '5 of a kind' in which it is simply 4 aces and one Joker.

@whosnext I did your first calculation in my first ever attempt at working this out haha:

54! / 5! x 49! (so the number of possible ways in a 54 deck divided by the amount needed and the amount left over) - all together I got 3,162,510 possible dealings in this game. Does that math sound right?

There's another outcome I need help with. What is the probability of dealing a flush consists of 5 cards of the same colour? (regardless of suite)

Yes, I and many other people write this as C(54,5) where the C is referred to as the "Choose" function. So you would say this as "54 choose 5". And the formula is what you typed above.

Do you mean either all 5 red cards or all 5 black cards (with no jokers)?

Let's tally it for red, and then multiply our result by 2 to include black (obviously the prob of 5 black cards = prob of 5 red cards).

How many red cards are in the deck?

How many do we need to select (choose)?

How many ways are there to have a "qualifying" hand (5 red cards)? [Hint: this will be a Choose function.]

Multiply by 2.

Remember the total number of possible 5-card hands that you derived above.

Divide the first number by the second number.

Awesome guys! I'm really having a lot of fun using maths to bring this game to life, the universal language.

So I had a crack at the first question I asked about AAAAJ and got 0.0000632% probability.


I also had a crack at the 5 cards of same colour and got this (please correct if wrong - still an amateur)

26! / 5! x 21! = 65780 for one colour

x 2 = 131560 for both.

/ 3,162,510 = a 0.0415% of getting 5 cards of the same colour.



I've seen on the internet others calculating the probabilities of all poker hands but they are based around the 52 card deck. But because I am including the Jokers in this game, does the values change? For example, I've seen the probability of getting Royal Flush is ~ 0.000154%. Since their are two extra cards, does that mean I need to re-calculate every poker hand?

First, 131,560 / 3,162,510 = 0.0415999 which is 4.15999% (not 0.0415999%).

Second, yes, adding additional cards to a deck (going from 52 to 54 cards) will affect all of the probabilities.

And, as mentioned above, how the extra cards (jokers) are treated can have a significant effect on the probabilities. As you undoubtedly know, in some games jokers are treated as "wild cards".

It sounds like you have added the jokers solely to create a super-hand that only works with four aces and one joker. I am not positive on that though.

Yes of course! My bad, I forgot the percentage conversion haha

As I thought, well I'm in a bit of a dark area with re-calculating every probability with 54 cards. Is there anywhere to seek help with tutorials?

Well that's right, Jokers can usually be wild cards but this game I'm designing, it isn't necessarily based around Poker rules or Poker itself, but you are certainly right that it's a super hand. I just need to know the maths before finalizing my project.

I am not sure exactly what you seek. There are many places on the internet that describe, explain, and give examples of using the Choose function (combinations). That may be what you are looking for.

In any event, you are more than welcome to pose any questions here or post any calculations you are wondering about. I promise that people on this forum are very forgiving and very helpful (a great combination).

It may not be exactly what you are looking for, since you said that your new game is not really poker-based, but the Wikipedia page on Poker Probabilities ( https://en.wikipedia.org/wiki/Poker_probability ) has a lot of useful information and derivations presented that may be helpful to you.

Nice, thanks whosnext you've been a lifesaver.

Just one more thing -the odds.

In 52 deck cards, Royal Flush is obviously ~0.00001539% and a 1:649 739 chance.

How do I calculate the 1:649 739? It's probably extremely simple but I ain't the sharpest tool in the shed (for all my Shrek fans out there)

I've had a lot of fun calculating these things but I need to know the odds.

A probability p tells you something happens p times out of 1.
A percentage x (= 100p) tells you something happens x times out of 100.

Dividing 1/p or 100/x for a result y would tell you that it happens 1 out of y times.

In odds form that would be (y-1) to 1 against.

Echoing heehaww's great post above.

Sometimes an example makes things crystal clear.

Suppose you know that the probability of selecting a spade from a regular standard deck of 52 cards is 13/52 (of course). This is easily seen and easily understood to be 1/4, since, of course, spades is one suit out of the four suits.

1/4 = 0.25 = 25% (all of this is super obvious)

Now, many people, especially in a gaming or betting environment, speak of "odds" rather than probabilities.

So what are the correct odds of selecting a spade from a regular standard deck of 52 cards?

Clearly, the odds are 3:1 against (again, obviously, there are three other suits that are "against" your one suit of spades).

A probability of 1/4 or 0.25 or 25% is clearly equivalent to 3:1 odds (against).

Coming at it another way, suppose you are told that some event has 3:1 odds (against). What is the inherent underlying probability associated with this event?

I hope it is patently obvious if some event is 3:1 against, that means that it is likely to occur 1 time in every (3+1) total occurrences. Or 1 time in every 4 times. Or, of course, 1/4 = 0.25 = 25%.

Okay, so it is easy to go from odds to probabilities. Some event that has T:1 odds against is likely to occur with probability P equal to 1/(T+1).

So P = 1/(T+1).

Applying some simple algebraic manipulations, we can see that this relationship can be written "in the other direction" (that is, T in terms of P).

T = (1/P) - 1

Let's confirm this with our example from above by plugging in P = 0.25 into that equation for T.

T = (1/0.25) - 1
T = 4 - 1
T = 3

So we confirmed that an event having 3:1 true odds (against) is likely to occur 25% of the time. And the true odds for an event that is likely to occur 25% are 3:1 against.

In summary, the true odds (against) associated with an event that is likely to occur with probability P are given by [(1/P)-1]:1 (against).


I apologize for this undoubtedly overly pedantic and silly post (already covered by heehaww's post), but I thought there might be some value in spelling this all out in painstaking and obvious detail.