birthday                package:stats                R Documentation

_P_r_o_b_a_b_i_l_i_t_y _o_f _c_o_i_n_c_i_d_e_n_c_e_s

_D_e_s_c_r_i_p_t_i_o_n:

     Computes approximate answers to a generalised _birthday paradox_
     problem. 'pbirthday' computes the probability of a coincidence and
     'qbirthday' computes the number of observations needed to have a
     specified probability of coincidence.

_U_s_a_g_e:

     qbirthday(prob = 0.5, classes = 365, coincident = 2)
     pbirthday(n, classes = 365, coincident = 2)

_A_r_g_u_m_e_n_t_s:

 classes: How many distinct categories the people could fall into

    prob: The desired probability of coincidence

       n: The number of people

coincident: The number of people to fall in the same category

_D_e_t_a_i_l_s:

     The birthday paradox is that a very small number of people, 23,
     suffices to have  a 50-50 chance that two of them have the same
     birthday.  This function generalises the calculation to
     probabilities other than 0.5, numbers of coincident events other
     than 2, and numbers of classes other than 365.

     This formula is approximate, as the example below shows.  For
     'coincident=2' the exact computation is straightforward and may be
     preferable.

_V_a_l_u_e:

qbirthday: Number of people needed for a probability 'prob' that 'k' of
          them have the same one out of 'classes' equiprobable labels. 

pbirthday: Probability of the specified coincidence

_R_e_f_e_r_e_n_c_e_s:

     Diaconis, P. and Mosteller F. (1989) Methods for studying
     coincidences. J. American Statistical Association, *84*, 853-861.

_E_x_a_m_p_l_e_s:

     require(graphics)

      ## the standard version
     qbirthday()
      ## same 4-digit PIN number
     qbirthday(classes=10^4)
      ## 0.9 probability of three coincident birthdays
     qbirthday(coincident=3, prob=0.9)
     ## Chance of 4 coincident birthdays in 150 people
     pbirthday(150,coincident=4)
     ## 100 coincident birthdays in 1000 people: *very* rare:
     pbirthday(1000, coincident=100)

     ## Accuracy compared to exact calculation
     x1<-  sapply(10:100, pbirthday)
     x2<- 1-sapply(10:100, function(n)prod((365:(365-n+1))/rep(365,n)))
     par(mfrow=c(2,2))
     plot(x1, x2, xlab="approximate", ylab="exact")
     abline(0,1)
     plot(x1, x1-x2, xlab="approximate", ylab="error")
     abline(h=0)
     plot(x1, x2, log="xy", xlab="approximate", ylab="exact")
     abline(0,1)
     plot(1-x1, 1-x2, log="xy", xlab="approximate", ylab="exact")
     abline(0,1)

