# Thread: Chance of being dealt a pair if another player has a pair.

1. Basically each player is dealt a two cards from a standard pack.
One player drops his cards and you see that he has a pair (a pair of jacks for example).

Now, you have not looked at your cards yet, but given this information, is it
a) more likely your cards are a pair. (any pair)
b) less likelly your cards are a pair.(any pair)
c) makes no difference.(any pair)

2.

3. More likely:

Probs are:

no information about other hand - PA=3/51
information that other hand has pair - PB=P1+P2
P1= prob if your first card is same as pair =2/(50x49)
P2 = prob if your first card is different = (48x3)/(50x49)

calculation

PA = .058823
PB = .059592

4. Originally Posted by mathman
More likely:

Probs are:

no information about other hand - PA=3/51
information that other hand has pair - PB=P1+P2
P1= prob if your first card is same as pair =2/(50x49)
P2 = prob if your first card is different = (48x3)/(50x49)

calculation

PA = .058823
PB = .059592

If your first card is the same as the other players, pair. Your odds are higher then getting four of a kind in one hand. The fact that all four cards have to be in order, might take the odds way out there. Unless you get a lazy or lucky shuffle of a new deck.

Sincerely,

William McCormick

5. Ain't google great!

6. if your first card is the same as the other players, the odds of you getting a 4 of a kind is the same as his, and it depends on the community draws. this deals with Hold'um, not stud.

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