Probability of Parachute success

[Let Me Win]

Could one of our resident math gurus please tell me the probability of success using Victor's Parachute.

In other words what chance that I will win at least one of the 36 spins using the same random number each spin.

[Firefox]

The chance of a number not winning is 36/37 or 37/38 US

In 36 spins the chance of it not coming up is (36/37)^36 = 0.373  or (37/38)^36 =0.383 US

So your chance of a profit or break even is 62.7% or 61.7% US wheel.

[Let Me Win]

I think this is wrong because it does not take into account that on my first spin I have an even chance of winning and on my second spin I have a 2/1 chance of winning and on spins 3,4,5 I have a 5/1 chance of winning etc etc

In other words my target number can never appear but I could still win each time.

[Firefox]

I think you may be right but your number does have to win after spin 18 when bet straight up.

In that case the chance you will not win is

(19/37) x (25/37)  x (31/37)^3 x (33/37)^3 x (34/37)^3 x (35/37)^6 x (36/37)^19

Does that seem more reasonable?

[Let Me Win]

Yes, I think that's correct. Thanks could  you tell me the result of that expressed as a percentage please.

[Firefox]

4.78% you lose

95.22% you will win or break even.

I only have mobile phone at moment which is not a very good calculator, so hopefully someone else can check and verify!

[Let Me Win]

Thanks.

The way I'm playing it is first spin bet one unit on corner 0,1,2,3

Then assuming I lose using the number spun as my target number for the remaining 35 spins.

Spin 2 I use the column so I play as per the chart just skipping the first even money bet.

I do this so I have a random method for selecting my target number each time.

[Bigbroben]

4.78% you lose

95.22% you will win or break even.

I only have mobile phone at moment which is not a very good calculator, so hopefully someone else can check and verify!

Yes, that is correct