Is there a difference between Probability and True Odds

Can 1300/1 odds be offered on a number doubling on a European Single Zero wheel due to the probability of the event being (1^37)*(1^37) giving 1369 as the probability of the event occurring?
I look at this as being (For true odds) 36/1 plus stake * 36 which equals 1332?
Can anyone help point out the error of my thinking please?

Sure, that works. It'd have about a 5% house edge. The odds and payout would be the same for any two-number bet (1 and then 2, 6 and then 9, 14 and then 14, etc.). You'd just have to book the bet(s) before the first spin, and have a marker or something to make sure that the hit on spin 1 was not paid off.

rdw4potus: What would the odds be if there was no payout for the 1st event as the bet only requires a bet to be placed on the 2nd event occurring? ie you nominate the 2nd number and only lose the stake for that, there is no odds or payout for the 1st event, so to speak?

If you mean that the first number has to happen, but results in no payment, then the odds of success are 1/37^2, or 1:1369. Example, the bet is on a repeated 23. The first 23 is recorded but not paid. The second 23 is paid at 1300:1. the house edge is 69/1369=5.04%.

If you mean that the first number is completely ignored, then the odds of success on the second spin are 1/37. I'm thinking that the first example is what you're going for.

thanks