odds help for dealing 3 named cards...

this could be wrong as its been a long long time since ive attempted to work anything like this out but i think the equation would be

12*combin(40,2)/combin(52/3) which is 9360/22100

which works out as a 42.35% chance of getting 1 card right out of 3 drawn and decimal odds of 3.36

as i say, it might be wrong and there may be an easier way to work it out

Approx. 42.8 per cent, or 3 from 7 times, or 4/3, or 2.33

I did it 52/12 + 51/12 + 50/12. This way allows for more than one of the selected cards to be picked. Is this wrong?

or if you do it as the chance of the first card being a 2,7 or jack and the second and third card are not any of these then you would work it out as

a) the chance of the first card being one of 12 cards (2,7, jack) out of 52 possible cards = 12*52

b) the second card being anything other than those 12 cards (2,7, jack) would be 40 remaining cards out of a possible 51 as you have taken the first card out already so 52 cards - the 2s, 7s, and jacks out of a possible 51 remaining cards = (52-12)/51 or (40/51

c) the third card being anything other than those 12 cards (2,7, jack) plus the second card you have taken for a total of 13 cards out would be 39 remaining cards out of a possible 50 remaining cards = (39/50)

d) so this would be (12*52)*(40/51)*(39/50) which would be a 14.118% chance but as any of the 3 cards selected could in fact be a 2,7 or jack then multiply the 14.118% by 3 which gives you a chance of 42.353% and again decimal odds of 3.36

@tim, well you got close to what i got so the method you used seems to be fine and also a lot simpler than the ones i suggested.  unless someone else comes along and gets a different one to us

sorry, in the second method above it should read (12/52)*(40/51)*(39/50) and not 12*52, oops

We have evaluated a slightly different thing, so our answers should be slightly different. Which they are, so hopefully we are both right!

I think it is 11/9 fair odds.
1-P(0 correct)
1-(40*39*38/52*51*50)=app 0.55
Then fair odds A = P(A)/[1-P(A)]
But I am not an expert.

Of course I got it upside down
Fair odds against 9/11.

First selection has 12 outcomes in your favour out of 52
Second selection has 8 outcomes in your favour out of 51
Third selection has 4 outcomes in your favour out of 50

Therefor for all selection to be in your favour is the product of these three events

(52*51*50)/)12*8*4)
= 345.3

But then I'm no expert either

O sugar, I have misread the question.

Ok you have 12 cards in your favour throughout. The first selection is from 52 cards, the second is from 51 cards and the third is from 50.
You need to be correct at least once. Therefor it is the sum of the probabilities of the three selections
12/52 +12/51+12/50 which come to  a probability of .706 giving of about  odds of 3/7

Ther's a 55.3 % chance of it happening so the decimal odds would be roughly 1.81

I see now that dunlaying has the correct calculation in his first post and the correct fractional odds in his 2nd post.

I agree that there's a 55.3% chance of it happening.   My somewhat laborious calculations follow.  There are 12  2's, 7's, Jacks in the pack.

Chance of 2, 7 or Jack coming out as first card with none of the other 11 second or third is 12/52 x 40/51 x 39/50 = 0.1412 .
Chance of 2, 7 or Jack coming out as second card with none of the other 11 first or third is 40/52 x 12/51 x 39/50 = 0.1412 .
Chance of 2, 7 or Jack coming out as third card with none of the other 11 first or second is 40/52 x 39/51 x 12/50 = 0.1412 .

Chance of 2, 7 or Jack coming out first and second with none of the other 10 third is 12/52 x 11/51 x 40/50 = 0.0398 .
Chance of 2, 7 or Jack coming out first and third with none of the other 10 second
is 12/52 x 40/51 x 11/50 = 0.0398 .
Chance of 2, 7 or Jack coming out second and third with none of the other 10 first
is 40/52 x 12/51 x 11/50 = 0.0398 .

Chance of 2, 7 or Jack coming out first, second and third
is 12/52 x 11/51 x 10/50 = 0.0100 .

Adding all 7 answers = 0.5530 which is 55.3% .

Yopu have it correct there but have complicated the calculation much more than needed.   When calculating the probability of something happening it's very often easier to calculate the probability of it not happening.
  In this case that would just be  40/52 x 39/51 x 38/50.   The answer here is 0.447 and the chance of it happening is therefore 0.553 or 55.3 %

so have we come to an agreement that its 55% chance to happen, its a prop bet i have been using when playing poker and people always want to back against you when u say u only want 3 cards and they have the rest for the flop, but thought maybe i had been tucking myself up lol. thanks guys.

Age 21 can you explain what you mean?
I do not play cards at all and I have a feeling that there may be more to the question than appeared at first sighting.

would be different answer for poker though, because you would already know two cards in your hand , and wouldnt be drawing from a deck of 52

Oh dear.

Thanks Eddie.   Will remember your comment in future.

Regards, The Lanterne Rouge.

digger u make the bet before any cards are dealt.