In a standard 52-card deck, what are the odds that out of five cards drawn at random, exactly three will be hearts? That exactly four will be hearts? all five?

1. 13/52 x 12/51 x 11/50 x 39/49 x 38/48 x 10

2. 13/52 x 12/51 x 11/50 x 10/49 x 39/48 x 5

3. 13/52 x 12/51 x 10/50 x 10/49 x 9/48

That's the probability which is slightly different than the odds. Don't pay attention to any other method.

Thanks!

These type of questions always baffle me. Do you want someone to just tell you the answers? Do you want to learn how to figure these out for yourself? Have you ever had a class in probability and statistics?

Anyway, every serious poker player (presumably anyone who is a member of 2+2) should be able to work this out for themselves.

The answer(s) will be the number of 5-card hands that meet your condition(s) as a percentage of all possible 5-card hands. How many cards are in a standard deck? How many cards are drawn at random? So, for the denominator of the fractions, how many ways are there to randomly draw five cards out of a standard deck of cards? (Look up "combinations" if you don't know.)

Okay, for the condition(s) that you seek to meet (the numerator in the fraction), how many hearts are there in a standard deck? Exactly how many of these do you want to be included in the five cards drawn at random? So how many ways (possible combinations) are there to draw a specified number of hearts?

For the first two conditions, there are remaining non-heart cards to be drawn. How many do you need to fill out the five-card hand? How many non-heart cards in the deck are these drawn from? Since the heart draws and the non-heart draws can be considered independent, we need to multiply the respective heart and non-heart combination tallies in order to arrive at the overall tally.

P.S. David has provided the answer while I was typing up this post!

DS again insists that his direct probability method is the only one you should use to answer questions like this. I understand that it is intuitive for beginners to say there are 13 hearts out of 52 so the probability the first card is a heart is 13/52. Now there are 12 hearts left out of 51 so 12/51 is the chance the second card is a heart, etc. But what is the 10 for at the end and how do you calculate it?

Well, we know it is the number of ways to distribute 5 cards, 3 of which are of one kind (hearts) and 2 of another (non-hearts). So the number of ways is 5! / (3!* 2!) = C(5,3). Huh? Where did that come from? I thought DS didn’t like combinations and permutations. If you asked randomly selected poker players to figure out the number of orders for this simple problem, I would say that the vast majority could not do it unless they had the formula or spent considerable time writing out all possibilities HHHxx, HHxHx, HxHHx, etc. For 7 cards, they would have to write out C(7,3) = 840 ways.

Now, if you use combinations, the formula for exactly 3 hearts out of 5 drawn cards is C(13,3)*C(39,2 / C(52,5). The number of ways to select 3 of 13 cards times the number of ways to select 2 of 39 non hearts divided by the number of ways to select 5 of 52 cards. For 4 hearts, the probability is C(13,4)*C(39,1) / C(52,5) and for 5 hearts, C(13,5)*C(39,0) / C(52,5). There is a nice regularity in these formulas in that the sum of the first terms in C(x,y) in the numerators is 52 (e.g. 13+39), which is the first term in the denominator and the sum of the second terms in the numerators is 3+2 or 4+1 or 5+0 = 5, equal to the second value in the denominator.

Is this method better? For me, yes, for I don’t have to worry about order and it only involves one math form, combos. The other method often leads to errors in the order calculation, with order often being forgotten. Do I ever use the direct probability method? Sure, but to state that all other methods should be ignored is overboard IMO.

To that I'll add, combos are actually useful in poker at the table. When you're making a tough decision and thinking about Villain's range, it helps to be able to quickly figure out how many combos there are of each type of hand in the range.

Oh but Somehow you did 7 P 4.

Statman knows C(7,3) but to the readers who don't: C(7,3) = 7*6*5 / 3! = 7*5 = 35

If it's C(n, r), you multiply r numbers together (starting with n) and then divide by r factorial. As shown above with C(7,3).

When you do this stuff often enough, you start to know some "combo tables" by heart. By memory I know all the combos up to N=10, and then for higher N's I know some of the smaller R's. There are also ways to quickly compute one you don't know using one you do know.

There aren't as many as you might think, because C(n, r) = C(n, n-r)
For instance C(10,8) = C(10,2) so if you know a small r then you automatically know the equivalent bigger r.

That identity is also useful for unknown combos. Suppose we want C(21,19). That equals C(21,2) which I don't know either, but the latter is much easier to calculate: 21*20 / 2 = 21*10 = 210

I lied. Just trying to goad Bruce to post.

Good intention. I made the dumb calculation error for the same reason.*



*Not true but I have no better explanation

Ah. That makes good sense.

I wrote a response, but then I read your post and decided to not post my response.

Now that you've explained that you lied, I'll post my response.

First find the number of ways three hearts plus two non-hearts could be chosen from 52 cards.

(In a standard 52-card deck there are 13 hearts plus 39 non-hearts).

Step one: Find the number of ways three hearts can be chosen from 13 hearts when we don't care about the order in which they are selected:
There are 286 ways three hearts can be chosen from 13 hearts.
To find this for yourself, google "13 choose 3." (without the quotes).

Step two: Find the number of ways two non-hearts can be chosen from 39 non-hearts when we don't care about the order in which they are selected:
There are 741 ways two non-hearts can be chosen from 39 non-hearts.
To find this for yourself, google "39 choose 2." (without the quotes).

Step three: Find the number of ways 3 hearts + 2 non-hearts can be chosen (when we don't care about the order in which they are selected):
To do this, multiply the number of ways three hearts can be chosen by the number of ways two non-hearts can be chosen:
286*741=211,962

Step four: Find the number of ways any five cards can be chosen from 52 cards when we don't care about the order in which they are selected.
C(52,5)=2,598,960
To find this for yourself, google "52 choose 5." (without the quotes).

Step five: Divide the number of ways 3 hearts + 2 non-hearts can be chosen by the number of ways any five cards can be chosen:
211,962/2,598,960=0.0815.
0.0815 is the probability of being dealt exactly three hearts.

For completeness, when we don't care about the order in which they are dealt, there are
575757 ways to be dealt 0 H, 5 N
1069263 ways to be dealt 1H, 4N
712842 ways to be dealt 2H, 3N
211926 ways to be dealt 3H, 2N
27885 ways to be dealt 4H, 1N
1287 ways to be dealt 5H, 0N
2598960 ways to be dealt any five cards.

Check: 575757+1069263+712842+211926+27885+1287=2598960

For completeness:
P exactly none will be hearts: 575757/2598960=0.2215
P exactly one will be a heart: 1069263/2598960=0.4114
P exactly two will be hearts: 712842/2598960=0.2743
P exactly three will be hearts: 211926/2598960=0.0815
P exactly four will be hearts: 27885/2598960=0.0107
P exactly five will be hearts: 1287/2598960=0.0005

I'm not a mathematician.

Buzz

What kind of a number is 575757? Looks like a made-up number to me!

C(39,5)=575757.

or 39*38*37*36*35/5/4/3/2/1=575757.

Does that explain it?

Buzz