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In a standard poker game (no wild cards), suppose you are dealt five cards and your hand contains exactly one pair. You trade in the three worthless cards for new ones. What is the probability that your hand improves? meaning that there is a substantive transformation from one "kind" of hand into another

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You have a pair, so these are the (mutually exclusive) ways of improving your hand:

  • Being dealt two worthless cards and a third of the number in the pair, giving you three of a kind
  • Being dealt one extra card and the remaining two of the number in the pair, giving you four of a kind
  • Being dealt one extra card and another pair, giving you two pair
  • Being dealt three of a kind, giving you a full house

Note that you cannot get a flush or a straight. Also observe that the three cards you discarded all had distinct numbers, otherwise you would have had two pair.

Let us assume that there are no other players, so that there are $47$ cards remaining. (If there are other players, you need to do essentially the following for each possible hand they're dealt, then sum over all of their possible hands times the probability of those hands. Probably intractable by hand.) Fix the hand you were dealt. We count the number of ways of being dealt each of the preceding possibilities.

  • Think about picking two worthless cards. You can't pick the number you've been dealt, so you have $45$ options for the first. For the second, you can no longer pick the same number as the first, or you'd be dealt two pair, so you have $45 - 3 = 42$ options for the second. Now divide by $2$ (you double-counted each pair of cards). Now there are $2$ ways of picking a third card of the number of the pair you've been dealt, so there are $$\frac{45\cdot 42}{2}2=45\cdot 42$$ hands ways of getting three of a kind.
  • There is exactly one way of being dealt the remaining two cards of the number in your pair. There are $45$ options for the worthless extra card. So there are $$45$$ ways of being dealt a four pair.
  • There are $12$ numbers remaining for another pair. However, $9$ of them have four suits for each number, while $3$ of them are the same number as the three you discarded, so there are $\binom{9}{1}\binom{4}{2} + \binom{3}{1}\binom{3}{2}$ ways of choosing another pair. Then the extra card can't be from the same number as that pair, nor the same number as your original pair, so there are $43$ cards to select it from if you picked your pair from the $9$ full sets of numbers and $44$ cards to select it from if you picked your pair from the $3$ sets of numbers from which you discarded. Therefore, there are $$43\cdot \bigg( 9\binom{4}{2} + 3\binom{3}{2} \bigg)$$ ways of being dealt two pair.
  • To be dealt a full house, you can either be dealt a pair with extra card same number as your original pair, or you can be dealt three of a kind. For the former, replace $43$ by $2$ in the previous bullet point. For the latter, there are $9$ numbers to choose from with $4$ suits and $3$ numbers to choose from with $3$ suits, so $\binom{9}{1}\binom{4}{3} + \binom{3}{1}\binom{3}{3}$ ways of being dealt a three-of-a-kind. Adding, $$2\cdot \bigg( 9\binom{4}{2} + 3\binom{3}{2} \bigg) + \bigg(\binom{9}{1}\binom{4}{3} + \binom{3}{1}\binom{3}{3}\bigg)$$ ways of being dealt a full house.

Since there are $\binom{45}{3}$ ways of being dealt three replacement cards, we have the probability of improvement is: $$\frac{45\cdot 42 + 45 + 43\cdot 9\binom{4}{2} + 45\cdot 3\binom{3}{2} + 2\cdot 9\binom{4}{2} + 3\binom{3}{2} + \bigg(\binom{9}{1}\binom{4}{3} + \binom{3}{1}\binom{3}{3}\bigg)}{\binom{45}{3}}$$

Google's calculator says that is about $34\%.$

I think I covered everything here without error, but these counting arguments can be tricky.

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Can I also say that I want to find the possibility of the hand not improving, and use 1-P(not improved)=P(improved)? Are there any flaws in that?If not, what is the probability that we still get 3 worthless cards? – user25329 Feb 20 '12 at 19:14
Yes, you can certainly do that as well. You'll have to rule out various cases, which I thought would lead to about as much work as the direct computation. – Neal Feb 20 '12 at 20:19


I know nothing of poker; but I'm assuming the three cards you throw away are not added back to the deck before you choose three new cards from it. If this is not the case, the following isn't relevant (but you can suitably modify it, and in fact it becomes an easier problem):

You may assume that the "new deck" from which you choose the three new cards:

$\ \ $1) has a pair missing

$\ \ $2) has three other cards missing that have distinct face values and distinct from the face value of $\ \ \ \ \ \ $the original pair.

So, for example, the new deck is missing the two of hearts, the two of spades, the four of diamonds, the five of diamonds, and the ace of hearts. You may assume that this is the case.

Now you want to find the probability that selecting three cards from this new deck of 47 gives you a better hand. This can occur in, and only in, the following mutually exclusive ways:

$\ \ $1) you obtain a three of a kind whose face value is the same as your original pair (and nothing $\ \ \ \ \ \ $better).

$\ \ $2) you obtain a four of a kind.

$\ \ $3) you obtain an additional pair, distinct from the original pair, and nothing better.

$\ \ $4) you obtain a full house (note this can occur in two fundamentally different ways).

As stated above, the above list is complete. You better your hand if and only if one of the above occurs.

So, find the probability of each of the above cases ocurring and take their sum.

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As noted above, the easiest way is to calculate $P(\text{Not improving})$, and then $P(\text{Improving})=1-P(\text{Not improving})$.

If you already have a pair, you can't get a straight or a flush. You can only improve by making trips or two pairs. Once you have improved to trips or two pairs you can improve further, but we don't need to concern ourselves with that.

The rough way to calculate this is: $$ P(\text{First card won't give you trips}) = \frac{45}{47} $$ as there are two card that give you trips.

$$ P(\text{Second card won't give trips or two pairs}) = \frac{41}{46} $$ as there are two cards for trips + three cards for two pairs.

$$ P(\text{last card won't give trips or two pairs}) = \frac{37}{45} $$ i.e. $2 + 3 + 3$ cards that improve. $$ P(\text{Not improved}) = \frac{45}{47} \cdot \frac{41}{46} \cdot \frac{37}{45} \approx 0.702 $$ and hence $$ P(\text{Improved}) \approx 0.298. $$

This isn't exact, as you may have picked a card of the same rank that was earlier removed. In that case, there are only two and not three matches, which improves your odds slightly. Likewise, for your last card, you may have picked one or two cards of ranks that were earlier discarded. In all, this adds $\approx 1\%$ to your chances of not improving.

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