Approximating the probablity of a higher pair

R Gibert · #1 09-03-2007, 11:50 AM

To approximate the probability of an opponent being dealt a higher pocket pair than yours, use:

P(p, q) = p*q*(50 - 2? - min(p,q))/100

where

p : number or opponents
q : number of higher pairs
x? : if p >= 5 and q >= 11, return x, else return 0
min(x, y) : the smaller of x and y

This accurate to within an absolute error of about 0.7%.

If the "- 2?" term is omitted, the maximum absolute error grows to only about 2.3%, so most practical purposes, it can be ignored altogether.

Here is a table generated by the above formula:
Code:<hr /><pre>
1 2 3 4 5 6 7 8 9
KK 0.5 1.0 1.5 2.0 2.5 2.9 3.4 3.9 4.4
QQ 1.0 1.9 2.9 3.8 4.8 5.8 6.7 7.7 8.6
JJ 1.5 2.9 4.2 5.6 7.0 8.5 9.9 11.3 12.7
TT 2.0 3.8 5.6 7.4 9.2 11.0 12.9 14.7 16.6
99 2.5 4.8 7.0 9.2 11.3 13.5 15.8 18.0 20.3
88 2.9 5.8 8.5 11.0 13.5 15.8 18.5 21.1 23.8
77 3.4 6.7 9.9 12.9 15.8 18.5 21.1 24.1 27.1
66 3.9 7.7 11.3 14.7 18.0 21.1 24.1 26.9 30.2
55 4.4 8.6 12.7 16.6 20.3 23.8 27.1 30.2 33.2
44 4.9 9.6 14.1 18.4 22.5 26.4 30.1 33.6 36.9
33 5.4 10.6 15.5 20.2 23.6 27.7 31.6 35.2 38.6
22 5.9 11.5 16.9 22.1 25.8 30.2 34.4 38.4 42.1
</pre><hr />

Here is the table of the absolute error differences:
Code:<hr /><pre>
1 2 3 4 5 6 7 8 9
KK 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
QQ 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 0.0 0.0 0.1
JJ 0.0 -0.0 -0.1 -0.1 -0.1 -0.1 -0.1 -0.0 0.1
TT 0.0 -0.0 -0.1 -0.3 -0.3 -0.2 -0.1 -0.0 0.1
99 0.0 -0.1 -0.1 -0.3 -0.5 -0.4 -0.3 -0.1 0.1
88 0.0 -0.0 -0.1 -0.2 -0.4 -0.6 -0.4 -0.2 0.1
77 0.0 -0.0 -0.1 -0.2 -0.3 -0.5 -0.7 -0.4 0.0
66 0.0 0.0 -0.0 -0.1 -0.2 -0.2 -0.4 -0.6 -0.1
55 0.0 0.0 0.0 0.0 0.1 0.0 0.0 -0.1 -0.2
44 0.0 0.0 0.1 0.2 0.3 0.4 0.4 0.5 0.5
33 -0.0 0.1 0.2 0.4 -0.5 -0.6 -0.5 -0.6 -0.6
22 0.0 0.1 0.3 0.6 -0.3 -0.2 -0.1 0.1 0.2
</pre><hr />
Here is the output of my Monte Carlo program, which I used to estimate the absolute error:
Code:<hr /><pre>
Number of iterations = 12345678

1 2 3 4 5 6 7 8 9
KK 0.0049 0.0098 0.0147 0.0196 0.0244 0.0293 0.0342 0.0390 0.0439
QQ 0.0098 0.0195 0.0292 0.0388 0.0484 0.0577 0.0672 0.0767 0.0859
JJ 0.0147 0.0292 0.0437 0.0578 0.0717 0.0857 0.0993 0.1129 0.1262
TT 0.0195 0.0389 0.0578 0.0764 0.0947 0.1128 0.1303 0.1476 0.1649
99 0.0245 0.0485 0.0719 0.0948 0.1170 0.1391 0.1603 0.1815 0.2016
88 0.0294 0.0580 0.0857 0.1128 0.1390 0.1646 0.1893 0.2134 0.2369
77 0.0343 0.0675 0.0995 0.1305 0.1606 0.1895 0.2175 0.2446 0.2709
66 0.0392 0.0768 0.1131 0.1479 0.1817 0.2136 0.2447 0.2748 0.3033
55 0.0440 0.0863 0.1267 0.1653 0.2019 0.2374 0.2708 0.3033 0.3342
44 0.0490 0.0955 0.1399 0.1821 0.2221 0.2601 0.2966 0.3312 0.3638
33 0.0539 0.1049 0.1530 0.1985 0.2418 0.2828 0.3212 0.3577 0.3923
22 0.0587 0.1140 0.1661 0.2149 0.2612 0.3040 0.3450 0.3833 0.4194
</pre><hr />
Note that each data point is the result of 12,345,678 iterations.

Here is my Monte Carlo program written in Python:
Code:<hr /><pre>
# Straightforward Monte Carlo program to determine probability of
# a higher pair being dealt to one of your opponents.

from random import *

deck = [2, 2, 2, 2,
3, 3, 3, 3,
4, 4, 4, 4,
5, 5, 5, 5,
6, 6, 6, 6,
7, 7, 7, 7,
8, 8, 8, 8,
9, 9, 9, 9,
10, 10, 10, 10,
11, 11, 11, 11,
12, 12, 12, 12,
13, 13, 13, 13,
14, 14, 14, 14]

card = "??23456789TJQKA"

ni = input("%26s" % ("Number of iterations = "))
plrs = input("%26s" % ("Range of opponents = "))
np = len(plrs)
ranks = input("%26s" % ("Range of ranks = "))

seed()
print ""
for p in plrs: print "%6u" % (p),
for r in ranks:
deck.remove(r)
deck.remove(r)
print "\n%s" % (card[r]*2),
for p in plrs:
c = 0
for i in range(ni):
shuffle(deck)
for j in range(p):
if deck[j] == deck[j + p] and deck[j] > r:
c += 1
break
print "%6.4f" % (float(c)/ni),
deck.append(r)
deck.append(r)
</pre><hr />

Elandriel · #2 09-03-2007, 02:49 PM

I love it