odds question

hi, this is my first post. i was calculating odds and one of my numbers seems to be a bit high. my question is, if 10 people are at the table and i pull one A, and 1 flops, what % chance that atleast 1 person has one of the other aces?
i came up with 72% and that seemed a bit high but semi believeable. thank you for your time

ok, there are 47 unseen cards 18 of them are in your opponents hands.

18/47 =~ 39% chance an opponent has one of them, and a 18/46 (if they don't have the other one) or a 17/46 (if they do) in either case that too is =~ 39%

i get 39% + (39% or 39%)= 78% chance.
(i don't know how to do the "or" math off the top of my head but both cases work out to very nearly the same value so to heck with it.

it's important to keep in mind some percenatage of players will fold that preflop. so, it may have been dealt to some player, but not made it to the flop.

for example if only 3 players see the flop 6/47 + 6/46 =~ 28% without respect to people calling based on card quality. Seems to me that people call more often with aces than with 3's so this number is probably low, but less than 78% because people will fold stuff like A2 utg.

The correct way to count is to begin with calculating the odds that one of your opponents DO NOT have any ace and then substract that number from 1. Then you don´t have to care about if someone has two aces and the calculation gets a lot easier.

You take one of your opponents cards at a time. There are 18 cards in their hands. Is the first one an ace? There are 2 unseen aces and 47 unseen cards, so the odds is 45/47 that it is not an ace. Next card, there are 46 unseen cards, still 2 aces: 44/46. The odds that the first player has no ace is

45/ 47 * 44/46 ~ 0,92 (92 %)

as there ar no +'s and -'es, we don´t have to care about ( or ) or which order in which we calculate things. 45*44/47/46 is the same. Just a little easier to calculate.

We go on with the rest of the cards:

45*44*43*42*41*40*39*38*37*36*35*34*33*32*31*30*29 *28/47/46/45/44/43/42/41/40/39/38/37/36/35/34/33/32/31/30

= 0.37557816836262719703977800177598 :)

1 - 0.3756 = 0.6244 = 62.44 %

So, a little more than 62 % chance (risk) that someone was DEALT another ace. As whileone said, you must consider the possibility of being folded before the flop.

W.

If you hold Ax and the flop is Ayz, there are 47 cards left in the deck. Only two of them are aces, which leaves 45 non-ace cards. The number of possible two card hands not containing an ace are therefore 45C2, which equals 990.

The total possible number of two card hands left in the deck is 47C2, which is 1081.

Therefore the probability of at least one person holding an ace is 1 - 990/1081 = 0.084..., or about 8%.

So is it 8% or around 70%?

Wouldn't 8% be great! I'd go all in every time with a single A paired to the board, kicker be dammed!

However, it is not 8% it would be 62.44%

Waggho’s method is the correct way to calculate the chance of someone else having an Ace. As he says, there are 18 cards in the hands of other players. 47 cards remain unseen and of these 2 are Aces. On the first of the 18 cards that makes the chance of no ace 45/47. Then we multiple the chance of an A not showing another 17 times. So we have:
45/47 x 44/46 x 43/45 ….. on to the 18th one which would be 28/30. So this works out to:
(45*44*43*42*41…..*29*28) / (47*46*45…*31*30). We would calculate this by using factorials. Just in case you don’t know what a factorial is (if you care?), it is simply the value of a series of numbers multiplied, i.e. 5 factorial (which is written 5!) is 1x2x3x4x5=120. To answer the above question the equation would be:
(45!/27!) / (47!/29!). This gives us the percentage chance that no Ace was drawn in the 18 cards as 37.56%. So the chance that at least one was drawn is 62.44%

Thanks,
BreakfastMan

See link off main page re: Proabilities & Aces

No offense to anyone, but scanning this thread quickly, it looks to me like there are many errors in the calculations.

e.g. How can you say: ""ok, there are 47 unseen cards 18 of them are in your opponents hands. 18/47 =~ 39% chance an opponent has one of them"

Those 18 cards might contain 0, 1, 2, 3 or 4 aces.

Nomdeplume did his calculation INCLUDING AN A ON THE FLOP!

That being said, I ain't going to check anyone's math.

Great stuff to know, but are there any shortcuts/tricks to calculating this on the fly in your head. I can work this out fine w/ a calculator, or even paper, but is there a simple way to figure out roughly what % 41/43 comes to. How does everyone else do it? Or do you just have to do the 45/47 * 44/46 * 43/45 ect...

Apologies, my calculation was assuming only ONE other opponent, my mistake.

What I should have done was:

Probability that none of NINE opponents has an ace =
[45C2/47C2] * [43C2/45C2] * ... * [31C2/33C2] * [29C2/31C2]
= 38%

which is equivalent to what has been posted. 1 minus this is indeed 62%.

Linker, I don't know about calculating this in your head, but if you use the nCr function on a scientific calculator it makes it quicker to calculate. Also less chance of making a mistake. :D