Picking 20 different numbers, in the same draw $\;$ vs. $\;$ picking 10 different numbers, in 2 different draws.
Daily Keno lets you
Does Pr (probability) of winning jackpot differ between
1. picking 20 different #s, for the same draw
vs.
2. picking 10 different #s, for 2 different draws ?
Intuitively, purchase 1 results in a low Pr(winning jackpot). Why? In 1 ― you picked $\color{yellowgreen}{20}$ different integers for one draw. To win, your picked $\color{red}{20}$ numbers must match OLG’s drawn 20 numbers. Thus, this one draw has $\color{red}{20} – \color{yellowgreen}{20} = \color{DarkTurquoise}{0}$ degrees of freedom.
In 2 ― you picked $\color{yellowgreen}{10}$ different integers for 2 draws. Thus, each of your 2 draws has $\color{red}{20} – \color{yellowgreen}{10} = \color{DarkTurquoise}{10}$ degrees of freedom.
Doubtless, $\color{DarkTurquoise}{0 < 10}$. Thus, 2 has higher Pr(winning jackpot). Please correct my intuition. I DON’T want proof or formality! Please explain at my 18 year old level.
1 answer
Does Pr (probability) of winning jackpot differ between 1. picking 20 different #s, for the same draw vs. 2. picking 10 different #s, for 2 different draws ?
Yes. Consider the limiting case where there are only 20 possible choices. Playing all 20 numbers guarantees a win. Playing 10 out of 20 choices results in probability of winning of ½. Therefore playing 10 choices each in two different draws results in a probability of winning of ¾.
Note here that "winning" is used only to mean picking the winning number. It does not imply being better off at the end or a positive expected value. In any remotely competent lottery, it would cost more to play every possible number than you get by winning.
0 comment threads