Comments on: Why are there 169 starting hands in Hold 'em?
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em/
Comments on Ask MetaFilter post Why are there 169 starting hands in Hold 'em?Sun, 22 Aug 2010 17:39:06 -0800Sun, 22 Aug 2010 17:39:06 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: Why are there 169 starting hands in Hold 'em?
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em
Can someone explain to me why there are 169 starting hands in poker? <br /><br /> I understand that there are (52x51)/2 possible starting hands. Out of those 1326, there are (3x4)/2 x 13 pairs. Where I am confused is on the issue of probability relating to the 169 possible starting hands. I understand that the suits don't have relative value, but I still can't figure out how to get to 169. I know that of those 169 hands, 13 are pairs, but I'm confused on what 13/169 means if the probability of drawing a pair is actually 78/1326. Where am I going wrong?post:ask.metafilter.com,2010:site.162974Sun, 22 Aug 2010 17:35:58 -0800tomtheblackbearpokermathcardsgamesresolvedBy: chrisamiller
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340787
First of all, I assume you're talking about Texas Hold-em. (there are many different kinds of poker)<br>
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If we disregard the suits, there are 13 possibilities for the first card and 13 possibilities for the second card, so the total number of two-card possibilities is 13x13 = 169.<br>
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This number is not the probability of getting a specific starting hand.comment:ask.metafilter.com,2010:site.162974-2340787Sun, 22 Aug 2010 17:39:06 -0800chrisamillerBy: madcaptenor
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340795
chrisamiller: I don't actually know how the game works. Does it matter which order you get dealt the two cards in? In other words, if I get dealt an eight and then a five, is that different from getting a five and then an eight?comment:ask.metafilter.com,2010:site.162974-2340795Sun, 22 Aug 2010 17:50:53 -0800madcaptenorBy: logicpunk
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340798
chrisamiller's explanation doesn't work because it treats a starting hand of, say, K-Q as different than a starting hand of Q-K.<br>
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If you count all the possible combinations of starting hands but treat K-Q and Q-K as the same hand, that gives you 91 possible combinations. But you can also have suited hands, so K(spades)-Q(spades) is a different hand than K(spades)-Q(any other suit). If you count all the possible suited hands as well as their non-suited equivalent, that gives you 91-13 (since pairs can't be suited). So:<br>
<br>
91 possible non-suited combinations.<br>
+91-13 possible cominations<br>
=169.<br>
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But yeah, it doesn't reflect a probability.comment:ask.metafilter.com,2010:site.162974-2340798Sun, 22 Aug 2010 17:51:51 -0800logicpunkBy: tomtheblackbear
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340810
So just to clarify, knowing that there are 169 possible starting hands doesn't actually signify anything in the way of probability. If that's the case, why do poker books focus on there being 169 hands rather than 1326?comment:ask.metafilter.com,2010:site.162974-2340810Sun, 22 Aug 2010 18:04:52 -0800tomtheblackbearBy: Chocolate Pickle
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340819
1326 is for five-card hands. 169 is for the two dealt cards of Texas Hold-Em.comment:ask.metafilter.com,2010:site.162974-2340819Sun, 22 Aug 2010 18:09:14 -0800Chocolate PickleBy: tomtheblackbear
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340822
I realize I wasn't totally clear. I know that 169 by itself doesn't mean anything, but if there are 13 pairs, then that would mean that there ~7% chance of drawing a pair (13/169). I'm confused because the actual probability of drawing a pair is ~5% (78/1326). Am I wrong?comment:ask.metafilter.com,2010:site.162974-2340822Sun, 22 Aug 2010 18:11:09 -0800tomtheblackbearBy: Chocolate Pickle
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340825
You're comparing apples and oranges.<br>
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78/1326 is the probability of a pair in 5 cards. 13/169 is the probabiliy of a pair in 2 cards.comment:ask.metafilter.com,2010:site.162974-2340825Sun, 22 Aug 2010 18:13:31 -0800Chocolate PickleBy: Chocolate Pickle
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340829
Also, 13/169 isn't a real number because it's some of those combinations are more likely than others. The probability of a pair in 2 cards is 12/51.comment:ask.metafilter.com,2010:site.162974-2340829Sun, 22 Aug 2010 18:14:59 -0800Chocolate PickleBy: Chocolate Pickle
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340831
Ah, rats. I got that number wrong. It's 3/51.comment:ask.metafilter.com,2010:site.162974-2340831Sun, 22 Aug 2010 18:16:07 -0800Chocolate PickleBy: beagle
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340833
<em>why do poker books focus on there being 169 hands rather than 1326?</em><br>
Because, if you want to consider strategies for dealing with different hands, you need to come up with only 169 different strategies. If you tried to come up with 1326 strategies, you'd have a lot of duplicate strategies.<br>
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And, for a rather exhaustive rundown of probabilities <a href="http://en.wikipedia.org/wiki/Poker_probability_(Texas_hold_'em)">see the Wikipedia article</a>.comment:ask.metafilter.com,2010:site.162974-2340833Sun, 22 Aug 2010 18:17:26 -0800beagleBy: beagle
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340838
Also, <a href="http://www.top15poker.com/Strategy/Starting_Hands.htm">here</a> are the relative strengths of all 169 hands.comment:ask.metafilter.com,2010:site.162974-2340838Sun, 22 Aug 2010 18:20:04 -0800beagleBy: PercussivePaul
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340839
So I found <a href="http://www.top15poker.com/Strategy/Starting_Hands.htm">this</a>. Essentially they're grouping together hands that have an equivalent chance of winning at the beginning, before the flop makes some suits better than others. All unsuited combinations of the same set of number, e.g. 5H-8D, 5S-8D, 5S-8C, etc, have the same chance of winning. All suited combinations, e.g. 5H-8H, 5D-8D, etc, have a slightly better chance of winning because they could contribute to a flush. There are thus three classes of hands: pairs, suited non-pairs, and unsuited non-pairs. <br>
<br>
Pairs: <br>
[13 numbers * 1 number] * [4 suits * 3 suits / 2 orders]<br>
= 13 unique hands * 6 suit combinations = 78 hands<br>
<br>
Non-pairs, suited:<br>
[13 numbers * 12 numbers / 2 orders] * [4 suits]<br>
= 78 unique hands * 4 suit combinations = 312 hands<br>
<br>
Non-pairs, unsuited:<br>
[13 numbers * 12 numbers / 2 orders] * [4 suits * 3 suits]<br>
= 78 unique hands * 12 suit combinations = 936 hands<br>
<br>
Total unique hands: 13 + 78 + 78 = 169 unique hands<br>
Total hands: 936+ 312 + 78 = 1326 handscomment:ask.metafilter.com,2010:site.162974-2340839Sun, 22 Aug 2010 18:24:38 -0800PercussivePaulBy: Rhomboid
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340845
<i>but if there are 13 pairs, then that would mean that there ~7% chance of drawing a pair (13/169). I'm confused because the actual probability of drawing a pair is ~5% (78/1326).</i><br>
<br>
A single pair can represent multiple combinations. For example "pair of twos" represents six actual possibilities: 2C + 2H, 2C + 2S, 2C + 2D, 2H + 2S, 2H + 2D, 2S + 2D, so its probability is 13*6 out of the total number of two-card possibilities (1326). This is all summarized in <a href="http://en.wikipedia.org/wiki/Poker_probability_(Texas_hold_'em)#Starting_hands">this table</a>.<br>
<br>
<i>78/1326 is the probability of a pair in 5 cards.<i><br>
<br>
No. Co(52, 2) = 1326. This is the number of all possible combinations of two cards. No one is talking about 5 cards.</i></i>comment:ask.metafilter.com,2010:site.162974-2340845Sun, 22 Aug 2010 18:31:48 -0800RhomboidBy: madcaptenor
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340936
Chocolate Pickle: 78/1326 (which is the same as 3/51) is the probability of a pair in 2 cards. The probability of a pair in 5 cards is clearly <i>larger</i> than the probability of a pair in 2 cards. 13/169 isn't the probability of anything relevant; the 169 hands don't have equal probabilities.<br>
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Percussivepaul: you're right. But there's an easier way to find the total number of hands! There are 52 cards, so there are 52 ways to pick the first card, and 51 ways to pick the second card. But order doesn't matter so we have to divide by 2, giving (52*51)/2 = 1326.comment:ask.metafilter.com,2010:site.162974-2340936Sun, 22 Aug 2010 19:29:38 -0800madcaptenorBy: chrisamiller
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2340942
<em>chrisamiller's explanation doesn't work because it treats a starting hand of, say, K-Q as different than a starting hand of Q-K.</em><br>
<br>
By Jove, you're right. <br>
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<small><small>This is the part where I blame lack of sleep...</small></small>comment:ask.metafilter.com,2010:site.162974-2340942Sun, 22 Aug 2010 19:32:38 -0800chrisamillerBy: alfredwetch
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2341081
Prob to get pair is not 13/169. When you say 169 hands yes, but they do not have the same probability to occur. For instance pairs are less likely than non pairs.comment:ask.metafilter.com,2010:site.162974-2341081Sun, 22 Aug 2010 21:21:50 -0800alfredwetchBy: PercussivePaul
http://ask.metafilter.com/162974/Why-are-there-169-starting-hands-in-Hold-em#2341106
Yeah I realize that madcap (so does the OP presumably, it's in the question); my intent was to show the relationship between 169 and 1326, which depends on suit combinations.comment:ask.metafilter.com,2010:site.162974-2341106Sun, 22 Aug 2010 21:59:51 -0800PercussivePaul