Mathematical Game Theory Probability

Probablitiy of Everyone Else Having Atleast 2 Cards of a Suit Given The Probability of you Having One Card of the same Suit

$(1-\frac{\mathrm{27\; choose\; 13\; +\; 26\; choose\; 13}}{\mathrm{39\; choose\; 13}})\times \frac{1}{3}=1-\frac{\left(\frac{\mathrm{27!}}{\mathrm{13!}\mathrm{14!}}\right)+\left(\frac{\mathrm{26!}}{\mathrm{13!}\mathrm{13!}}\right)}{\frac{\mathrm{39!}}{\mathrm{13!}\mathrm{26!}}}\times \frac{1}{3}=.999543\mathrm{=}99.95\mathrm{\%}$

for example: if your only club at the start of the game is a 3 you may count that as a trick because the second time clubs are played you may break spades with your lowest spade and win the hand 99.95% of the time

Probability of Someone Not On Your Team Missing Any Suit Givin That You Have 6 Cards of the Same Suit

$1-(1-{\mathrm{P(1\; player\; void\; suit\; |\; 6\; suited\; cards)}}^{\mathrm{amount\; of\; opponents}})=1-(1-\frac{\mathrm{39\; choose\; 13}}{\mathrm{46\; choose\; 13}}){\mathrm{}}^2=1-(1-\frac{\frac{39!}{13!26!}}{\frac{46!}{13!33!}}){\mathrm{}}^2=.153258=15.33\%\mathrm{your\; teammate\; missing\; that\; suit\; =\; 7.98\%}$

This allows you to anticipate your opponents bet before they speak, substantially increasing your odds of a "Nil Bet" success on your team. If an opposing team has a the void suit you are likely to lose at least two additional tricks. This concept leads to the conclusion that you may use negative numbers when summating the total tricks you can win, adding another layer of strategic depth to early game betting.

Probability of Having 6 Cards of the same Suit

$\frac{\mathrm{6\; favorable\; cards\; *\; other\; combinations}}{\mathrm{number\; of\; hands\; with\; 52\; cards}}\times 4=\frac{\mathrm{13\; choose\; 6\; *\; 39\; choose\; 7}}{\mathrm{52\; choose\; 13}}\times 4=\frac{\frac{13!}{6!7!}\frac{39!}{7!32!}}{\frac{52!}{13!39!}}\times 4=0.041564\times 4=.166256=16.63\%$

Probability of Having 5 Cards of the same Suit

$\frac{\mathrm{5\; favorable\; cards\; *\; other\; combinations}}{\mathrm{number\; of\; hands\; with\; 52\; cards}}\times 4=\frac{\mathrm{13\; choose\; 5\; *\; 39\; choose\; 8}}{\mathrm{52\; choose\; 13}}\times 4=\frac{\frac{13!}{5!8!}\frac{39!}{8!31!}}{\frac{52!}{13!39!}}\times 4=0.124692\times 4=.498768=49.88\%$

Probability of Drawing at Least Specific One Rank

$1-\mathrm{probability\; of\; not\; having\; that\; specific\; Rank}=1-\frac{\mathrm{number\; of\; hands\; with\; 48\; cards}}{\mathrm{number\; of\; hands\; with\; 52\; cards}}=1-\frac{\frac{48!}{13!(48-13)!}}{\frac{52!}{13!(52-13)!}}=0.696182=69.6182\%$

Number of Stacks to Separate Cards by Rank

$13\times 0.696182=9.05037Stacks$

Probability of Missing a Card of a Particular Rank

$1-0.303818=30.3818\%$

Probability of Missing an Entire Suit

$\frac{\frac{(52-13)!}{13!\left((52-13)-13\right)!}}{\frac{52!}{13!(52-13)!}}=\frac{\frac{39!}{13!26!}}{\frac{52!}{13!39!}}=0.0127909=1.27909\%$

When simulated this percentage approched 5.16%

Probability of Drawing an Ace and a King of the Same Suit

$\frac{4\left(\frac{\mathrm{50!}}{\mathrm{11!}\mathrm{39!}}\right)}{\frac{\mathrm{52!}}{\mathrm{13!}\mathrm{39!}}}=\frac{4\times \mathrm{15,817,620,178,664}}{\mathrm{67,526,966,271,912}}=0.2353=23.53\%$