Question 798500
You are missing one factor, a coefficient in the expansion of {{{(0.4929+0.5071)^10}}}. I'll explain so you understand why a crazy formula is applied. Once you understand, it will be easy to remember where the crazy formula comes from, and you will know how to calculate without memorizing the crazy formula. (Or so I hope).
 
In one game the probability of winning is {{{0.4929}}},
and the probability of loosing is {{{1-0.4929=0.5071}}}.
 
For one game,the probability of either winning or loosing (anythingthat happens is cool, as long as a game is played) is
{{{1=0.4929+0.5071}}}.
 
For 2 games, the probability of an outcome of any sort happening is
{{{1=1^2=(0.4929+0.5071)^2=0.4929^2+2*(0.4929)(0.5071)+0.5071^2}}}
where {{{0.4929^2}}} is the probability of winning both times,
{{{0.5071^2}}} is the probability of loosing both times,
and {{{2*(0.4929)(0.5071)}}} is the probability of winning once and losing once.
The coefficient 2 comes from the 2 equal terms that you get when you multiply
{{{(0.4929+0.5071)^2=red((0.4929+0.5071))*green((0.4929+0.5071))=red(0.4929)green(0.4929)+highlight(red(0.4929)green(0.5071)+red(0.5071)green(0.4929))+red(0.5071)green(0.5071)}}}
I made them different colors (red for first game, green for second) so that you could tell them apart, but the products highlighted are equal, so we add them up as {{{2*(0.4929)(0.5071)}}}, and that is where the coefficient 2 comes from.
 
For 3 games, you would have
{{{1=1^3=(0.4929+0.5071)^3=(0.4929+0.5071)(0.4929+0.5071)(0.4929+0.5071)=0.4929^3+3*(0.4929)^2(0.5071)+3*(0.4929)(0.5071)^2+0.5071^3}}}
 
For 4 games, you would have
{{{1=1^4=(0.4929+0.5071)^4=0.4929^4+4*(0.4929)^3(0.5071)+6*(0.4929)^2(0.5071)^2+4*(0.4929)(0.5071)^3+0.5071^4}}}
 
For 10 games, the probability of an outcome happening is
{{{1=1^10=(0.4929+0.5071)^10=(0.4929+0.5071)(0.4929+0.5071)}}}{{{..."}}}={{{0.4929^10+10*0.4929^9*0.5071+45*0.4929^8*0.5071^2)}}} {{{"+...+"}}}{{{0.5071^10}}}.
The terms in the middle (all except the first and last one) have coefficients that reflect the number of combinations (products) that result from picking the first term in some of the ten {{{(0.4929+0.5071)}}} factors and the second term in the rest of the factors.
The term with {{{0.4929^6*0.5071^4=(0.4929)^6 *(1-0.4929)^4}}} has the coefficient {{{10*9*8*7/(1*2*3*4)=210}}},
and {{{210*0.4929^6 *0.5071^4}}} is the probability of winning 6 and losing 4 of the 10 games.