| | | # Probability of drawing a double pair from a deck of cards
Asked Feb 6, 2009, 04:39 AM
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** 3 Answers** | **Question:**
A deck of 52 cards, 2 cards are dealt. We assume that the card we have
in our hand are not a pair.
For example we get: '3 of hearts' and '2 of clubs'.
What is the probability that if you draw 3 cards at random from the remaining 50, you'll get at least another 2 and at least another 3? In other words: having two 'not pair' cards, what is the probability of making at least two pairs (or full house) if you draw 3 cards from a deck of the remaining 50 cards? **What I did so far:**
2 Cards, unpaired, are now in our hands. 50 cards remaining.
Currently 6 cards can make one pair. So when the first card hits the
table, the chance of 'not making' a pair is:
(50-6)/50
When the second card hits the table (that becomes the 4th visible card
in total), 49 cards are remaining, but only 3 of these can make a
pair. So when the second card is placed on the table, the chance of
not making a double pair is:
(49-3)/49
Now what confuses me is: I could multiply (50-6)/50 *
(49-3)/49 = 2024/2450
Then 1 - (2024/2450) would give me the chance on having (at least a
double pair): 426/2450... this would mean that once out of 5.75
times would result in a double pair... this looks pretty high and this
is only after 2 cards have been dealt... one is remaining, so I'm
wondering: What did I overlook?
What do I do with the third card? How do I include this in this
calculation?
Keep in mind: we need *at least* (in this example) a 3 and a 2 on the
table. If on the table would appear: 3-3-2, this should be included in
the probability. |
3 Answers
| **Expert** | |
Feb 6, 2009, 12:24 PM
| |
First of all, to clarify the game - if you happen to draw 3 of a kind but not a full house (say starting with a 2 and a 3 you then draw 2, 2 and 5) then that's considered losing, right? Also, if you match either one of the first two cards, and then draw a pair of a different value from the deck (for example start with a 2 and 3, then draw 2, 4, and 4) that's also considered losing? Assuming this is what you mean...
There are two ways to win - you either get two pairs and an unmatched card or you get a full house. The probability of drawing 2 pairs and an unmatched card is:
6/50 * 3/ 49 * 44/48 * 3
This is because to make the first match you have 6 chances out of 50 cards in the deck. If you make that then the probability of matching the other pair on the next card is 3 out of the remaining 49 cards. If you make that then the probability of having the next card be unmatched is 44/48. You then multiply by 3 because the unmatched card could have come with either the first, second, or third draw.
The probability of drawing the full house is:
6/50 * 3/49 * 4/48 + 6/50 * 2/49 * 3/48
The first group represents the probability of matching one of the cards on the first draw, then matching the other card on the second draw, then matching either of the cards on the third draw. The second group is the probability of matching one of the cards on the first draw, matchingtaht same card again on the 2nd draw, and then the other card on the third draw.
Add these togther to find the total probability:
(6*3*44*3 + 6*3*4 + 6*2*3)/(50*49*48) = **0.02112**
Another way to do this is to determine the probability of not getting two pairs, and then subtract that from 1. There are four ways to not get two pairs:
a. None of the three draw cards match the first two. The probability of this is: 44/50*43/49*42*/48
b. The first draw card matches one of the hole cards, and neither of the next two draws match the other hole card: 6/50 * 46/49* *45/48
c. The first draw doesn't match a hole card, the second does, and the third doesn't match the other hole card: 44/50 * 6/49 * 45/48
d. The first and second draw cards don't match either hole card and the third does: 44/50*43/49*6/48
Add these up and you get .97888. This is the probability of not getting two pairs. So the probability of getting two pairs or a full house is1-.97888 = 0.02112, which agrees with the first technique. | | | New Member | |
Feb 8, 2009, 04:27 PM
| | ebaines, Thank you very much for this answer. I've been looking for days now and many people told me different stories. Your results seem to match my empirical tests (program that generates random hands).
Could you verify this?
Having a 2 and a 3 in my hand, 50 cards remaining, what is the chance of hitting *exactly* one pair if I draw 3 cards?
Using what you wrote, I'd expect it would be:
6/50 * 44/49 (49 remaining - 5 cards that can still make a pair) * 43/48 (48 remaining - 5 cards that can still make a pair) * 3 = 0.2895, so about 29%
Is this correct?
What I am wondering then:
Having a 2 and a 3 in my hand, 50 cards remaining, what is the chance of hitting *at least* one pair if I draw 3 cards?
Using what you wrote, I'd expect it would be:
6/50 * 46/49 (49 remaining - 3 cards in our hands) * 45/48 (48 remaining - 3 cards in our hands) * 3 = 0.3168, so about 32%
Is this correct?
The difference between:
-> 29% of exactly one pair
and
-> 32% of at least one pair
this 3% difference, is the accumulation of double pair, tri's, full houses and quads. Is this correct?
Double pairs: 6/50 * 3/ 49 * 44/48 * 3 = about 0.02 or 2%
Tri's:
Full houses: 6/50 * 3/49 * 4/48 + 6/50 * 2/49 * 3/48 = 0.00095 or about 0.095%
Quads: 6/50 * 3/49 * 2/48 = 0.00031 about 0.031%
My question is: How do I calculate Tri's? | | | **Expert** | |
Feb 9, 2009, 09:37 AM
| | Originally Posted by **durexlw** Having a 2 and a 3 in my hand, 50 cards remaining, what is the chance of hitting *exactly* one pair if I draw 3 cards?
Using what you wrote, I'd expect it would be:
6/50 * 44/49 (49 remaining - 5 cards that can still make a pair) * 43/48 (48 remaining - 5 cards that can still make a pair) * 3 = 0.2895, so about 29%
Is this correct? Not quite. This is the probability of drawing exactly one pair and not three or four of a kind. For example, if you start with a 2 and 3, this would include the chance that you then draw three cards that are 2, 5, and 6. But if does not include the chance of drawing a 2, 2, and 6 (making three 2's), or 2, 2, and 2 (making 4 2's). So my question to you is - do you consider drawing three 2's as exactly the same as drawing 1 pair, or not? You need to be specific as to what you are trying to do here. | | Thread Tools | Search this Thread | | | ### Add your answer here. ##
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