Is this right on a single wheel for probability of a win?
Spin 1:
pick 2 numbers = 2/37
Spin 2:
pick 5 numbers = 5/37
Total chance to win 2/37 + 5/37 = 7/37?
I thought about multiplying them together, but wouldn't that just give me the chance that the event will occur? I don't need the event to occur to win because I can win during the building of the event.
Hello
You will lose 20 less coins in 74 moves than if you play 7 coins (every other time, to have an equivalent stake)
Hello Plolp,
That is cryptic. Can you put that into a formula?
Excuse me, I was wrong: in fact there is no difference.
2/37 + 5/37= 7/37
with the way you play you will lose 7 coins in 74 moves .
The chance of at least one win in the two spins is the chance of 1 â"€ both lose.
chance of first bet losing = 35/37
chance of second bet losing = 32/37
So, chance of at least one win = 1 â"€ 35/37 × 32/37 = 18.2%
â†' How'd you calculate the at least 2 or more .. & exactly 2
in an x interval of spins eg. 2Q?
& same, when combining two districts overall eg. Q & DS
Q=4 spins, DS=3 spins max
& another thing;
Qâ†'DS, shifting to DS for 3 spins max,only after Q hits:
Q spin1 hit up to spin4 for DS to hit, combined
Q spin2 ...
Q spin3 ...
Q spin4 ...
Show the principle & formulas so I can thdn apply to,any other combinations