Jason L. answered • 02/22/17
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I am assuming that you are betting on groups of 6 numbers out of 36 (I was a little unclear on that in your question but I think that's what you said).
There are 6 blocks of 6 numbers on the wheel (36/6 = 6). That means the odds a number is picked from that block is 1/6 (I'm also assuming things like the two "00" spots are not included in this question, although it might need to be added here to make the odds 6/38 instead of 6/36).
A.) On average, how many times will you need to play until you win this bet?
If you're betting on one different block, then you are covering 1/6 of the wheel. So you should pick a winner around every 6 turns.
B.) Find the probability that you win this bet at least once in 5 spins of the wheel
You can solve this with the binomial distribution formula.
(Combos of you winning once) * P(winning)^(times won) * P(losing)^(times lost)
5C1 * (1/6)^1 * (5/6)^4
Jason L.
Sorry, my formula is for exactly one. So to solve for at least once you could just say it's 1 - the probability of never winning in 5 spins (since that would count everything else, 1-5 wins). So take the same binomial formula, change it for 0 wins and then subtract that % from 1. You'll get a little over 59% if you do it right.
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02/22/17
Riley M.
Ok so it would be (1/6)^0 * (5/6)^5, which I then got 0.402 and then subtract that from 1 which I got to be 0.598. Does that look correct?
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02/22/17
Jason L.
Yep exactly. That's the same thing as doing the binomial formula for each number 1-5 individually and then adding them together.
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02/22/17
Riley M.
02/22/17