Simple question

[SA125]

What are the odds say sharing a Q with someone else when a Q flops. We have one , the flop has one and what's the odds someone else does too? In layman's terms please, if possible.

[the alex]

I'll explain it logically.

1.) 52 cards in the deck
2.) 4 of those cards are Queens.
3.) You have seen 5 cards. The 2 in your hand and the 3 on the board. 2 of those 5 cards are Queens.
4.) 52-5= 47 unseen cards.
5.) 2 of those 47 are Queens.
6.) Everyone at the table has 2 hole cards.

(2/47)*2= 0.085~

Your result depends on how many people are seated. Multiply 0.085 by the # of OTHER people at the table.

1(0.085)= 0.085= 8.5%
2(0.085)= 0.17= 17.5%
3(0.085)= 0.255= 25.5%
4(0.085)= 0.34= 34%
5(0.085)= 0.425= 42.5%
6(0.085)= 0.51= 51%
7(0.085)= 0.595= 59.5%
8(0.085)= 0.68= 68%
9(0.085)= 0.765= 76.5%

I don't think you'd be playing with more than 9 other people as tables are normally capped at 10 players. These are numbers that will tell you the percentage of the time that someone else at the table has or had a Queen. It doesn't tell you the chances that your opponent in particular has one.

[SA125]

Simple answer. Thanks.

[BruceZ]

[ QUOTE ]
Simple answer. Thanks.

[/ QUOTE ]

Simple yes, correct no. With enough opponents it isn't even close. You cannot simply multiply 0.085 times the number of opponents. If you had 12 opponents that would come out to 0.085*12 = 102%. There certainly isn't over 100% chance of there being a queen out.

This is a decent approximation for a small number of opponents, but for 9 opponents the correct answer is 62.4% not 76.5%.

To do this properly, it helps to know something about combinations. For 9 opponents, we have to deal 18 cards, and the number of ways to select 18 cards out of 47 is called C(47,18). You can compute this in Excel using the function called COMBIN. It is equal to 47!/18!/(47-18)! where ! means factorial. For example, 47! = 47*46*45*44...*3*2*1. There are 45 non-queens, so the number of ways to deal 18 non-queens is C(45,18). The ratio of C(45,18)/C(47,18) = 37.6% is the probability that a queen is dealt. 1 minus this is the probability that no queen is dealt, or 62.4%.

Another method makes use of the 0.085 * number of opponents approximation. The reason this isn't correct is that there are 2 queens remaining. This would be correct if only 1 queen were remaining, but with 2 queens it is possible that both queens get dealt, and it turns out that this method would count these cases twice. So if we could compute the probability of 2 queens being dealt, we could subtract this off to get the right answer. The probability that 2 particular cards are both queens is 1/C(47,2) since there are C(47,2) ways to deal 2 cards, and there is only 1 way they can both be queens. There are C(18,2) pairs of cards, so the probability that both queens are out is C(18,2)/C(47,2) = 14.2%. If we subtract this from 0.085*9 we get the correct answer of 62.4%.

Here are the correct answers.

<font class="small">Code:</font><hr /><pre>
#opponents probability of queen
1 8.4%
2 16.5%
3 24.1%
4 31.5%
5 38.4%
6 45.0%
7 51.2%
8 57.0%
9 62.4%
10 67.5%
</pre><hr />

[the alex]

He wanted layman's terms which I interpreted as, "I don't know what 'the ratio of C(45/18)/C(47/18)'" means.

He wants practical numbers, not exact numbers, and how to estimate those numbers so he can apply them to situations on the spot.

SA125, stop playing QT and QJ. Does that accurately answer the REAL question?

[SA125]

[ QUOTE ]
#opponents probability of queen
1 8.4%
2 16.5%
3 24.1%
4 31.5%
5 38.4%
6 45.0%
7 51.2%
8 57.0%
9 62.4%
10 67.5%

[/ QUOTE ]

[ QUOTE ]
and how to estimate those numbers so he can apply them to situations on the spot.

SA125, stop playing QT and QJ. Does that accurately answer the REAL question?

[/ QUOTE ]

Yeah, I was wondering how often you're likely to be dominated. This chart makes it as easy to understand as draw odds.

What I find interesting is the percentages (45-57%) from 6-8 opponents. The percentages that someone else has a Q is similar to an all-in race between a PP vs AK.

Funny you mentioned QJ and QT. With it 45-57% that a Q is even out, what are the percentages a Q is out with a higher kicker? AQ-KQ-QJ has to be much less likely than 2Q-9Q.

Excluding overcards and re-draws, just whether or not your kicker is good, doesn't that make holding QJ or QT to a Q high flop a better situation than the all-in PP vs AK race, which is accepted as a no brainer?

[BruceZ]

Approximations are great, but they need to be identified as such, and some estimate of the error should be given, especially when it is as large as it is here. It is a very common misconception that the method you presented is the exact way to do this type of problem, and when you say that it is a "logical" explanation, it only reinforces that misconception.

[ QUOTE ]
He wanted layman's terms which I interpreted as, "I don't know what 'the ratio of C(45/18)/C(47/18)'" means.

[/ QUOTE ]

That's how I interpreted it, which is why I gave a layman's explanation of what that means. However, this prompted me to consider how this problem could be done simply and exactly without combinations, and I came up with this:

There are 47 cards remaining, so the probability of each queen being among the 18 dealt cards is 18/47. If a queen is not dealt, then it must remain part of the 29 undealt cards in the deck which has probability 29/47. If one queen remains in the deck, the probability that the other queen remains in the deck is 28/46, so the probabilty that both queens remain in the deck is (29/47) * (28/46) = 37.6%. The probability that at least one queen is dealt is 1 minus this or 62.4%. Note that (29/47)*(28/46) is exactly what you get if you evaluate C(45,18)/C(47,18) after lots of cancellation.

Another way to do it is 18/47 + (1 - 18/47)*18/46 = 62.4%. This says a queen can be dealt one of two ways. Either the first queen can be dealt with probability 18/47, or the first queen is not dealt with probability (1 - 18/47) and then the other queen is dealt with probability 18/46.

A third way is 18/47 + 18/47 - (18/47)*(17/46). This adds the probabilities of the two queens being dealt, but since this would double count the times that both queens are dealt, we then subtract the probabilty that they are both dealt.

All this should look familiar since these are the same types of calculations we use to compute the probability of hitting an out on the turn or the river, but we have shown that they can also be used for a different type of problem.

[Stork]

BruceZ, you are my hero.

[BradleyT]

[ QUOTE ]

Funny you mentioned QJ and QT. With it 45-57% that a Q is even out, what are the percentages a Q is out with a higher kicker? AQ-KQ-QJ has to be much less likely than 2Q-9Q.

[/ QUOTE ]

In the math world this is correct, in the poker world its not because Q9-Q2 aren't going to see the flop very often at all (except in super loose B&amp;M games and micro limit online games). Remember we're concerned with # of players on the flop, whom most of the time have average or above average hands so you would see AQ-KQ-QJ more often than you would see Q9-Q2.

Eh, basically QT is trash anywhere except the SB/BB for no raise and the button.