Probability Question With Percentages, Please Help! Thanks!

Okay, what you wrote doesn't make a lot of sense to me. Here's what I think you're basically asking. If I'm wrong, please correct me. You have a gambling game set up where the three outcomes are Left, Right, and Fail. Left and Right each occur 47.5% of the time and Fail occurs the remaining 5% of the time. In this game, you can place a bet on either Left or Right. If what you bet on comes up, you win, and if not you lose. What you're asking, then, is: If you always bet on Left, what is the odds that you will lose 10, 20, and 30 times in a row?

If that's the gist of the problem, then you can approach it like this... If you lose your first game, what is the odds of losing the second game? Then what is the odds of losing both games? If you lose both the first two games, what is the odds of losing the third game? Then what is the odds of losing all three? As a hint, consider an easier example. If you have a standard deck of cards, the odds of the top card being a Jack is 1/13. And the odds of the top card being a Heart is 1/4. So what is the odds of the top card being the Jack of Hearts?

EDIT: Make a mistake in my odds of the card being a heart. Thanks to Subhotosh Khan for the correction.

Basically yes. If I bet LEFT there is a 47.5% chance that I will be right and win the bet. What are the chances that if I lose the first game, what are the chances of me losing the second, then third, forth, fifth, ect. What is the formula if there is one that I can use to solve that? Ill leave out the betting amounts as I think I can solve that myself. Thanks, and also first time on this forum

Well, intuiting the formula was what I was trying to lead to with the card example. You know that the odds of winning/losing any given game don't change no matter how many games have been played before, right? And the odds of losing any given game is the same as the odds of not winning. So, how would you formulate that as an expression? As a hint, the odds of any outcome happening is, of course, 100%. Now, let's look at probabilities a bit more, to see if you can figure out a method for solving it.

If we play some games, the odds of winning are 47.5%. That can be written as 0.475 or 475/1000, right? So, then what fraction represents the odds of losing? If each game has 1000 outcomes (where 475 of them are winners and ? of them are losers), then how many total outcomes do two games have? Well, no matter the outcome of the first game, we will play a second game. So there's 1000 outcomes of the second game for each of the 1000 outcomes of the first game. Now let's do something similar for the odds of winning both games. 475 of the 1000 outcomes for the first game will result in winning. For each of those outcomes, we have another 475 out of the 1000 outcomes where we win the second game. Do you see how to calculate the odds of winning both games?