you draw 3 to a pair, what are your chances of improving and how much of that is to 3 of a kind? can you calculate the reverse implied odds of this play since your opponent will much of the time be starting with the bigger pair in this scenario

I’m not a five card draw expert. Some other posters to this forum are. Since nobody else has answered you, I’ll take a shot. This will be my good deed for the day.

The first part is relatively easy. Look in the Mike Caro tables referenced in the FAQ. Or go directly to
http://www.mikecaro.com/mcu/tables/Table16.asp
http://www.mikecaro.com/mcu/tables/Table17.asp

If you want to understand, following is a partial explanation.

You see 5 cards initially. Thus, assuming there is no joker in the deck, there are 52-5=47 cards you don't see.

2 of these cards you don't see are the same rank as the pair you keep. 45 of the cards you don't see are of other ranks.

This is probably easier to follow if we choose a specific five card hand for you to have been dealt. What we find for this particular five card hand will also be true of other five card hands with other pairs. Let's choose to make your starting hand
A, A, 9, 8, 7, and let's choose to keep the pair of aces.

You're going to be dealt three new (to you this hand) cards. Think of the 47 cards you cannot see as being arranged in three-card groupings. There are between fifteen and sixteen of these three-card groupings (actually 15.67) and unless both of the missing aces are contained in one of the three-card groupings, two of them have an ace. So roughly it's about 14 to 2 or ~7 to 1 that you won't draw an ace. I don't like to memorize numbers, preferring to approximate the odds while playing. Maybe you can do that math in your head like I do.

A better way is to calculate the probability none of the three cards you draw will be an ace. It's P=C(45,3)/C(47,3)=
P=45*44*43/47/46/45=0.875 which is the equivalent of the fraction 7/8. If the probability of not getting dealt an ace (or two) is 7/8, then the probability of getting dealt an ace (or two) must be 1/8. So it's 7/8 to 1/8 or 7 to 1 you won't get dealt an ace (or two).

From 47 cards, we can make 47*46*45/6=16,215 different three-card groups.

1*45=45 of these can have two aces.
2*45*44/2=1,980 of these can have exactly one ace.
16,215-45-1,980=14,190 of these must have no aces.
14,190/16,215=0.875
Notice this checks with our result from above.
This must be another way to find the probability of an ace or two.

Anyhow, there are different ways to think about it.

You also improve if you make two pair (by drawing a pair) or if you make a full house (by drawing trips or by drawing an ace plus a pair).

To just answer your question,
there are a total of 4,656 ways you can improve to two pairs or better (out of 16,215 possible three-card draws.
2064 of these are to trip aces.

16,215-4,656=11,559 are the number of ways not to improve.
Thus the odds against improving to two pairs or better are 11,559/4656=2.48 to 1 (against).

16,215-2,064=14,151 are the number of ways not to improve to exactly trip aces.
Thus the odds against improving to exactly trip aces are 14,151/2064=6.86 to 1 (against).

No. Not unless the cards both Hero and Villain are holding are known.

The second part is more complex because there are various possibilities, depending on the pair with which you start. And in addition your opponent may draw
• one card to two pairs, trips, a flush and/or straight, or actually anything at all.
• two cards to one pair, trips, a flush and/or straight, or anything at all.
• three cards to a pair or anything.

Furthermore, different opponents play and draw differently.

From my perspective, it’s a matter of knowing your opponent and estimating the range of hands he’s playing, partly dependent on how he’s playing the current hand and partly on the basis of past experience with this particular opponent.

Buzz

i love this forum

FYP