4 ms·
huh? it's a random draw and each drawing is an independent event. The "statistically right bet" is zero.
by dschobel 17y ago
huh? it's a random draw and each drawing is an independent event.
The "statistically right bet" is zero.
- colonelxc 17y agoEven WOPR knows that. "The only winning move is not to play."
- electromagnetic 17y agoI was proposing a thought experiment: create a program that bets the lowest frequency numbers and see if it is even capable of breaking even. Right now the Ontario Pick 3 has a frequency deviation of 1/20th from norm, in that the number occurrences have been >5% off of a 1/10th share between the 10 numbers. Would betting the three most statistically likely to occur numbers change your lucky enough above random probability. To put it even simpler: after 100 coin flips and 57H/43T would betting T give you greater odds than betting H, or are you equally as likely to win as if you flipped a coin yourself to decide your bet?
- dschobel 17y ago"Would betting the three most statistically likely to occur numbers change your lucky enough above random probability." No. It would not. The drawings are statistically independent events. Same thing for your coin flip example. If after 100 flips of a coin you have 99H/1T, the probability of the next flip being H or T is still 0.5 (assuming your coin is fair).