I know a game with 97,5 % chance of winning
it is called roulette, just play 35 numbers of 37 and you have a chance of 97,5 % to win.
it is called roulette, just play 35 numbers of 37 and you have a chance of 97,5 % to win.
35/37 = 94.59% chance of winning.
I will keep an "ear" open for your answer.
D.
if you bet 18 numbers as a impair or pair, beside this, the advatage is soooo big.
If you bet 35 numbers, then either one of them hits or it doesn't.
35 out of 37 times (assuming single-0 but no double-0) you will win $35 on the bet you won, but you will lose your other 34 $1 bets. Thus you will come out $1 ahead.
Thus, you win $1 (in the long run) on 35 of every 37 bets.
If one of the other two numbers hits, you lose $35.
Thus, you lose $35 (in the long run) on 2 of every 37 bets.
Thus, out of 37 bets, you will:
win 35, at $1 each, for income of $35
lose 2, at $35 each, for a net loss of $70
Thus, you will lose $35 per 37 spins (on average), using this method, even though you are winning about 95% of your bets.
In particular, your total amount wagered in 37 spins will be 37x$35 = $1295. If you lose $35 for every $1295 wagered, you are losing at a rate of 2.7%, which is exactly as expected for single 0 roulette.
--Eliot
Roulette has 37 numbers but is payng as its has 36 numbers and your edge is - 0.027027027,but because you can cover a lost (minimum)bet in
1/0.027027027 bets you will get old .One more thing:the roulette does`n`t have a memory to keep a counting.
