Cribbage Tournament Averages

This page is an expanded version of my article appearing in May 2004 issue of Cribbage World which stated that cut for deal indeed benefits the average player in tournament play.
This conclusion was reached by using CRIBBAGE TOURNAMENT AVERAGES, a program that computes probabilities of
cut wins game wins skunk wins game points cumulative game wins up to number of games in tournament cumulative game points up to number of games in tournament
for a cribbage tournament using the following parameters
N = number of games in tournament: 4 - 22 C = number of cut wins: 0 - N W = number of game wins: 0 - N D = first dealer win %: 54.0 - 58.0% A = skill adjustment %: 0.0 - 9.9% S = dealer skunk win %: 0.0 - 9.9%

Probabilities when N=22; W=13+; D=56.0; A=9.3 for expert =0 for average player; S=8.0
C=16 C=6 (16+6) avg C=11 Benefit
A. Player 35.13 18.35 26.74 - 26.01 = +0.73
Expert 69.79 49.44 59.62 - 59.89 = -0.27

The probability of an average player winning 13 or more games in an alternating deal tournament is 26.01%. When averaging 16 and 6 cut wins this probability increases to 26.74%, whereas the expert's probability decreases from 59.89% to 59.62% - a net benefit of 1% to the average player.
The average of all possible pair-wise cut wins appear in summary form below (in effect, 21+1 and 22+0 contribute 0%).
Note that 11 cut wins (alternating deal) is the reference base and confers no advantage.

Cuts
12+10
13+9
14+8
15+7

16+6
17+5
18+4
19+3
20+2

% Prob
15.42
11.86
7.62
4.07

1.78
0.63
0.17
0.04
0.01

Player
0.03
0.12
0.26
0.47
0.73
1.03
1.40
1.81
2.26

Expert
-0.01
-0.04
-0.10
-0.18
-0.27
-0.35
-0.49
-0.65
-0.86

Net
0.04
0.16
0.36
0.65
1.00
1.38
1.89
2.46
3.12

0.11% is the expected average net benefit to the average player over an expert by cutting for deal in a 22-game tournament. This was obtained by summing all the products of the net benefit and its corresponding probability of occurrence.

Methodology
D = 56% as per De Lynn Colvert's 100,000 game sample in "play winning Cribbage"
A = 9.3% results in an expert winning 13 or more games 60% of the time as per Larry Hassett in BOD report Sept 2002
S = 8.0% as per Dwight Christiansen Apr 2002 Cribbage World article; pone skunk rate = 55% of S
Binomial Distribution Probability equation for P = _NC_xp^x(1-p)^N-x
Screen shot uses the following which can be obtained from appendix tables in an introductory Mathematical Statistics textbook or repeated use of binomial function BINOMDIST in Microsoft EXCEL
N = 20 (values > 20 not available in textbook) C = 10 W = 10 D = 55% ie 0.55 (0.56 tables not available in textbook) A = 0% for A. Player Dealer p = D Pone p = (1 - D) for Expert Dealer p = D + A Pone p = (1 - D) + A Probability of 10 wins for an average player: P = 17.71% is sum of 11 different ways 10 games can be won in a 20 game alternating deal tournament Equation: P = dealer wins * pone wins use p=0.45 in probability text and reverse lookup entries for n = 10
Coincidentally, 17.71% is also the probability of an average player winning 11 of 20 games if player wins all 20 cuts in a cut for deal tournament. In this highly improbable case there is only one way of winning a specific number of games whether it be 10, 11, or any other number.

Alternative derivation using Microsoft EXCEL function BINOMDIST(number_s, trials, probability_s, cumulative) where:
number_s number of successes in trial use 11 trials number of independent trials use 20 probability_s probability of success on each trial use 0.55 cumulative logical value: either "TRUE" or "FALSE" use FALSE TRUE returns the cumulative distribution function FALSE returns the probability function

E-mail me if you would like a copy of Cribbage Tournament Averages or have any comments or questions.

Cuts	12+10	13+9	14+8	15+7	16+6	17+5	18+4	19+3	20+2
% Prob	15.42	11.86	7.62	4.07	1.78	0.63	0.17	0.04	0.01
Player	0.03	0.12	0.26	0.47	0.73	1.03	1.40	1.81	2.26
Expert	-0.01	-0.04	-0.10	-0.18	-0.27	-0.35	-0.49	-0.65	-0.86
Net	0.04	0.16	0.36	0.65	1.00	1.38	1.89	2.46	3.12