probability question

[skolsuper]

Ok so calculate the above pls

Suited broadways are 3% of all hands.

The 6O question it is relevant whether you're asking for hands with at least a pair, or only hands with just 1 pair.

[George2Loose]

At least one pair

[skolsuper]

card 1: 1
card 2: 48/51 (3 cards pair us)
card 3: 44/50 (6 cards pair us)
card 4: 40/49 etc
card 5: 36/48
card 6: 32/47

multiply together to get the probability of being dealt an unpaired hand = 34.5%

therefore the probability of at least 1 pair in 6 cards (or 2 pair or trips or a full house or quads, but ignoring straights and flushes) is 65.5%

[Oxford_HRV]

if you took out any ace, then deal out 10 cards from a shuffled deck afterwards what is the probability one is an ace?
it's approx 20% of the deck with 3 aces in, it must be worked out! ty

[AdamM]

The way I'm seeing the original question, if your first card is an ace, of the 51 remaining cards 3 are aces.
9 cards are dealt before it comes back to you and the question is what are the chances of one of them being an ace.
each card has a 5.88% chance of being an ace (3/51*100)
multiplied by 9 shots, that's 52.92% chance that someone elses first card is an ace.

look right?

[AdamM]

Ok so calculate the above pls

Suited broadways are 3% of all hands.

The 6O question it is relevant whether you're asking for hands with at least a pair, or only hands with just 1 pair.

This

chances of card 1 being a suited broadway card 20/52
chances of second card being the same suit and aslo 10+ 4/51

(20/52) x (4/51) = 80/2652 = 3.02%

[AdamM]

if you took out any ace, then deal out 10 cards from a shuffled deck afterwards what is the probability one is an ace?
it's approx 20% of the deck with 3 aces in, it must be worked out! ty

that's a different question and the answer isn't 20%
not sure if I'm awake enough to work out what it is though.

the way you've phrased it, from a 51 card deck, containing 3 aces, what are the chances of 1of the next 10 being an ace (not 2 or 3)

[Oxford_HRV]

I got to assume it surely can't be odds in favour another ace is dealt out.

think possibly you will have to add all fractions up 3/51----- 3/43 and multiply by chances you not receiving an ace, in a tree diagram possibly?

I was saying 10 cards is approx 20% of 51 cards, not really written well :s

[Oxford_HRV]

if you took out any ace, then deal out 10 cards from a shuffled deck afterwards what is the probability one is an ace?
it's approx 20% of the deck with 3 aces in, it must be worked out! ty

that's a different question and the answer isn't 20%
not sure if I'm awake enough to work out what it is though.

the way you've phrased it, from a 51 card deck, containing 3 aces, what are the chances of 1of the next 10 being an ace (not 2 or 3)

that's the question I would like answered, I said 10 cards instead of 9 because I'm including the second card that would be dealt to me

[doubleup]

if you took out any ace, then deal out 10 cards from a shuffled deck afterwards what is the probability one is an ace?
it's approx 20% of the deck with 3 aces in, it must be worked out! ty

There are 12777711870 combunations of 10 cards in 51 and 6540715896 combinations of 10 cards in 48 (the 3 aces removed).  So the probability of no aces in the 10 cards dealt is 51% (6540715896/12777711870) and therefore of an ace being dealt 49%.

[AdamM]

I got to assume it surely can't be odds in favour another ace is dealt out.

think possibly you will have to add all fractions up 3/51----- 3/43 and multiply by chances you not receiving an ace, in a tree diagram possibly?

I was saying 10 cards is approx 20% of 51 cards, not really written well :s

your questions are all over the place.
In your OP you're asking what are the chances of one of your opponents getting an ace as their first card, knowing that one of them is in your hand.
Each of them has a 3/51 chance of getting an ace, and there's 9 shots.
3/51*100=5.88% * 9 shots = 52.94%

try it by taking an ace out of the dec and dealing 9 cards in a row
doesn't seem unreasonable that there'd be an ace around half the time

[AdamM]

if you took out any ace, then deal out 10 cards from a shuffled deck afterwards what is the probability one is an ace?
it's approx 20% of the deck with 3 aces in, it must be worked out! ty

that's a different question and the answer isn't 20%
not sure if I'm awake enough to work out what it is though.

the way you've phrased it, from a 51 card deck, containing 3 aces, what are the chances of 1of the next 10 being an ace (not 2 or 3)

that's the question I would like answered, I said 10 cards instead of 9 because I'm including the second card that would be dealt to me

your original question specifically says what are the chances of another ace being dealy BEFORE it comes back to you.
that's 9 cards

[AdamM]

Do you want to know if you have an ace as either of your cards, what are the chances of 1 or mor other aces being held by one of the other 9 players?

[AdamM]

doubting my method now

[kinboshi]

Do you want to know if you have an ace as either of your cards, what are the chances of 1 or mor other aces being held by one of the other 9 players?

I'm not sure he knows, as that question has been answered by doubleup (well he included all ten players as requested, but the method is there).