Then the distribution for the point count C for the hand can be found from the program NFoldConvolution by using the distribution for a single card and choosing n = 13. $$\begin{aligned}{} & {} {\widehat{F}}_Z(z) - F_{Z_m}(z)\\= & {} \left\{ \frac{1}{2}\sum _{i=0}^{m-1}\left( {\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) \right) \left( {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) +{\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) \right) \right\} \\{} & {} -\left\{ \frac{1}{2}\sum _{i=0}^{m-1}\left( F_X\left( \frac{(i+1) z}{m}\right) -F_X\left( \frac{i z}{m}\right) \right) \left( F_Y\left( \frac{z (m-i-1)}{m}\right) +F_Y\left( \frac{z (m-i)}{m}\right) \right) \right\} \\= & {} \frac{1}{2}\sum _{i=0}^{m-1}\left\{ \left( {\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) \right) \left( {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) +{\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) \right) \right\} \\{} & {} -\frac{1}{2}\sum _{i=0}^{m-1}\left\{ \left( F_X\left( \frac{(i+1) z}{m}\right) -F_X\left( \frac{i z}{m}\right) \right) \left( F_Y\left( \frac{z (m-i-1)}{m}\right) +F_Y\left( \frac{z (m-i)}{m}\right) \right) \right\} \end{aligned}$$, $$\begin{aligned}{} & {} {\widehat{F}}_Z(z) - F_{Z_m}(z)\nonumber \\= & {} \frac{1}{2}\sum _{i=0}^{m-1}\Big \{{\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) \nonumber \\\ \quad \quad \quad{} & {} +{\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) \nonumber \\ \quad \quad \quad{} & {} - F_X\left( \frac{(i+1) z}{m}\right) F_Y\left( \frac{z (m-i-1)}{m}\right) + F_X\left( \frac{i z}{m}\right) F_Y\left( \frac{z (m-i-1)}{m}\right) \nonumber \\ \quad \quad \quad{} & {} - F_X\left( \frac{(i+1) z}{m}\right) F_Y\left( \frac{z (m-i)}{m}\right) + F_X\left( \frac{i z}{m}\right) F_Y\left( \frac{z (m-i)}{m}\right) \Big \}\nonumber \\= & {} \frac{1}{2}\sum _{i=0}^{m-1}\Big \{\Big [{\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) - F_X\left( \frac{(i+1) z}{m}\right) F_Y\left( \frac{z (m-i-1)}{m}\right) \Big ]\nonumber \\ \quad \quad \quad{} & {} +\Big [ F_X\left( \frac{i z}{m}\right) F_Y\left( \frac{z (m-i-1)}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) \Big ]\nonumber \\ \quad \quad \quad{} & {} +\Big [{\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) - F_X\left( \frac{(i+1) z}{m}\right) F_Y\left( \frac{z (m-i)}{m}\right) \Big ]\nonumber \\ \quad \quad \quad{} & {} +\Big [ F_X\left( \frac{i z}{m}\right) F_Y\left( \frac{z (m-i)}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) \Big ]\Big \}\nonumber \\= & {} \frac{1}{2}\sum _{i=0}^{m-1}\left\{ A_i(z)+B_i(z)+C_i(z)+D_i(z)\right\} \quad (say).
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