Hold Em hand types

[donkeyradish]

How many hands should you "expect" to have to play to get dealt at least one of each of the 169 hand types?

[LetYouDown]

To get the probability of having an individual hand over 50% likely, I believe you'll need 117 hands (=BINOMDIST(0,116,1/169,FALSE or just 168/169^X < .5). The math beyond that should be straightforward.

[ QUOTE ]
To get the probability of having an individual hand over 50% likely, I believe you'll need 117 hands (=BINOMDIST(0,116,1/169,FALSE or just 168/169^X < .5). The math beyond that should be straightforward.

[/ QUOTE ]

Accounting for suit, there are 1326 possible pre-flop hands, which should occur with equal distribution. There are four of each suited hand, six of each pair, and 15 of each other unsuited hand.

The least likely of these hands occurs 2 in 663 times, so we're looking at (661/663)^n < whatever, which means that you need significantly more hands (about 528 for a 50% chance to get a particular suited hand).

[pzhon]

[ QUOTE ]
How many hands should you "expect" to have to play to get dealt at least one of each of the 169 hand types?

[/ QUOTE ]
22152693452048615566208387892588196577772269154856 639529
91746282324714374417714916646997830075946168757731 111588
89906014814606346530700790298782262899191127296942 921041
64832362234220134447012586708496340530013885521560 754171
65264726706700809312278679175464507344725164420708 678582
14735467
/
13482066915588051270383503078630756529337998275144 554865
99412051271878601088775621390053023199281122687988 303391
30574377418185995532449229115795194067742717399273 504617
51529465791075881682044449811701610019752811859627 383973
03449262651717258786959849452316408495928004045435 132846
65600
~
1643.12294

I did this by a 3-dimensional recursion indexed by the number of pairs, suited, and offsuit hands yet to be seen.

Mathematica code:

<ul type="square">
Clear[f];
f[a_, b_, c_] :=
f[a, b, c] =
If[a &lt; 0 || b &lt; 0 || c &lt; 0 || {a, b, c} == {0, 0, 0},
0,
1326/(6a + 4b + 12c) +
(6a/(6a + 4b + 12c) f[a - 1, b, c] +
4b/(6a + 4b + 12c) f[a, b - 1, c] +
12c/(6a + 4b + 12c) f[a, b, c - 1])
][/list]
By the way, the probability that the last hand to collect is a pair is 2.36%; suited, 97.56%; and offsuit, 0.08%.

[donkeyradish]

Wow, thats some calculation! Glad I didn't attempt it manually [img]/images/graemlins/smile.gif[/img]

[pzhon]

[ QUOTE ]
Wow, thats some calculation! Glad I didn't attempt it manually

[/ QUOTE ]
Actually, there is a lot of value in trying to estimate these manually. I couldn't think of a good estimate in the few minutes it took to type and debug the recursion, but I think there is a method.

One possibility is to estimate the amount of time it takes before all of the suited hands have been seen. As the second calculation indicates, that coincides with the time the last hand is seen over 97% of the time; if one could predict that ahead of time, you might guess that the two averages will be close, and you can estimate the average amoutn of time before every suited hand is seen relatively easily, as it is a 1-dimensional recursion and similar to the standard coupon-collecting problem.

The expected amount of time before you see all suited hands is

1326/(4*78) + 1326/(4*77) + ... + 1326/(4*1)

=
1326/4 (1/1 + 1/2 + 1/3 + ... + 1/78)
=
1637.7164...
~ 1326/4 (Euler-Mascheroni constant .5772157... + ln(78))
=
1635.5960...

These are not too far off of the exact answer I computed.

1643.1229...