Calculate Probabilities in percent

Hi I'm new to the forum so please forgive the title, it may not be accurate!

Let me explain a little, this is probably more a maths problem but I want to do it in excel and I am struggling with getting the maths right, I'm hoping some one can help?

For years I have had a fascination with the lottery, purely hobby, and have had lots of fun over the years working different things out. The last 6 months though I have become fascinated with roulette & thought it would be a fun project to work out all sorts based around that, plus I don't have to wait for lotto results I can get instant numbers & results, however my latest attempts are hitting a brick wall!

I am trying to work out (in percentages) the increasing & decreasing % of 3, 11, 12, 22, possible outcomes

I have worked out the 2 possible outcomes initially for odd/even as follows

At the start they both have a 48.65% chance of hitting, then whatever is hit first the percentages are 76.33% and 23.67%. If you have 2 in a row of odd/even then the percentages are 88.49% & 11.51%, 3 in a row would give you 94.40% & 5.60% etc.

I have used the following formula for this (BM5 is where the totalhits for even are calc'd)

=IF(BM5=1,$C10*$C10,IF(BM5=2,$C10*($C10*$C10),IF(BM5=3,$C10*($C10*($C10*$C10)),IF(BM5=4,$C10*($C10*($C10*($C10*$C10))),IF(BM5=5,$C10*($C10*($C10*($C10*($C10*$C10)))),IF(BM5=6,$C10*($C10*($C10*($C10*($C10*($C10*$C10))))),IF(BM5=7,$C10*($C10*($C10*($C10*($C10*($C10*($C10*$C10)))))),IF(BM5=8,$C10*($C10*($C10*($C10*($C10*($C10*($C10*($C10*$C10))))))),IF(BM5=9,$C10*($C10*($C10*($C10*($C10*($C10*($C10*($C10*($C10*$C10)))))))))))))))))

and the same but in BM4 for tracking the even!

This works great, but I'm sure there must be a better way! BUT I have no clue how to advance this for 3 possible outcomes and more!

Can anyone help? I have been stuck on this for 2 months now and its really annoying!

First of all, thanks to NBVC thats a much simpler way with the 2 outcomes! I still can't work out how to make it work for more outcomes though!

Secondly thanks to SHG for BINOMDIST (BiNOMinal DIStribution?) I'm still trying to get it to work though I can see its use I am unable to make it work! LOL! I'm obviously thicker than I thought coz when I saw it I thought YAY that'll work easy! BUT I must be doing somethinh wrong coz it don't I'm still looking though!

OK Maybe this will make things a little clearer.

One of the reasons I have been attracted to working these things out for roulette is the number of possibilities.

In roulette you have many possible outcomes per spin for example it can be red odd & low or red even & high & thats just for the 1:1 bets.

I know the table/ball has no memory and this is purely a hobbyistic venture just for fun but as with tossing a coin eventually your results will level up I believe the same can be said of roulette! I'm already very successful at predicting the following outcomes within my parameters for roulette and its become a little party piece that I show off to friends occasionally, who keep wanting me to show them how to do it so they can gamble for real, but I want take it a little further for myself which is why I want a percentage for each possible outcome. So to elaborate further.

For each spin we have:

2 possible outcomes (Red/Black-Odd/Even-1-18/19-36)
3 possible outcomes (The Dozens(1-12,13-24,25-36), Columns/Rows)
11 possible outcomes (Line Bets(Sets of 6 Numbers))
12 possible outcomes (Street Bets (Sets of 3 numbers))
22 possible outcomes (Quarter Bets (Sets of 4 numbers))
54 possible outcomes (Splits(pairs of numbers))
37 Possible outcomes (Straight(0-36)

Each spin is recorded and whichever of the outcomes it falls into is reset to zero. i.e. if 5 is the first result then Red/Odd/1-18/1st Dozen/Column 2/Line 1-6/Street 4-6/Quarter 1,2,4,5/Quarter Bet 2,3,5,6/Quarter Bet 4,5,7,8/Quarter Bet 5,6,8,9/Split 2,5/Split 4,5/Split 5,6/Split 5,8 & Straight 5 is set to zero & All others are set to 1.

Each spin result that hits from then on will result in that possible outcome being set to zero, with all other outcomes having 1 added.

This is then taken and used in my previous formula (BM5) in my earlier posting and translated into a percentage representation. I have only managed to do the 2 possible outcomes thus far. I have no idea how to calculate 3 against each other any further than knowing that initially they all have a 32.43% chance.

For example All Columns Start at 32.43%. Col 1 hits so column 2 & 3's 32.43% chance increases and column 1's 32.43% chance decreases but how do I calculate this? Further next spin column 2 hits so column 3 increases further column 1 increases and column 2 decreases to below 32.43%. Etc

I hope you can fathom that out coz I found it bloody difficult to explain!

I'm still working on making use of the BINOMDIST though I'm unsure how to factor it in at the moment BUT I do see its uses now thanks to SHG!

What your missing (I think) is that over time eventually hits will even themselves up even though the table has no memory. I have seen somewhere, a long time ago when I first started doing this with the lottery, that mathematical studies showed that if you toss a coin an infinite no of times and record each toss then at some stage your results will level up. For example 500 tosses and you have 350 heads & 150 tails but by 900 you have 450 of each.

Now the no memory still applies to the coin toss but on numerous occassions these tests have turned out the same results. So as I applied the same theory to my lottery hobby, I want to apply the same theory to my new roulette fascination.

I never progressed this far, in fact it never applied, with the lottery. So though I can work out a percentage chance for the even bets (2 outcomes per spin) I am unable (don't posess the knowledge/experience) to do this for the 2:1 bets that have 3 possible outcomes.

I hope that explains it because I honestly do not know how I can make it clearer! The binomdist was great for working the odds out but again I can't work out how to apply it for 3 or more outcomes.

Originally Posted by mremixer

What your missing (I think) is that over time eventually hits will even themselves up even though the table has no memory. I have seen somewhere, a long time ago when I first started doing this with the lottery, that mathematical studies showed that if you toss a coin an infinite no of times and record each toss then at some stage your results will level up. For example 500 tosses and you have 350 heads & 150 tails but by 900 you have 450 of each.
.

Well the above is not actually true!

Over time the results as a proportion will tend to the result as actualy stated but in an absolute number the gap is likely to become larger (although it is a random sample)

Eg tossing a coin, the following reultsare made up to illustrate twhat could happen

100 trials 53 heads 47 tails p =0.53
500 trials 260 heads 24 tails p=0.52
100000 trials 50500 heads 49500 tails. p=0.505

but the absolute difference is 3 in the first instance and 10 in the second, and 500 in the third.

So whilst p is likely to get closer to 0.5 the absolute ifference from this expected result is likely to get larger. so if you gambled with this fair bet, it is likely that the longer you play the more likely one of the 2 parties will win more, it will not eventually break even if you play long enough!

Regards

Dav

I think I'll mark this as solved, though it isn't we seem to be slipping away from my original question!

If you don't know then say so, don't start with all the this says that stuff and wanting to know more about the why's & wherefores. I asked a question answer it or don't! YOU don't need to know why I asked or even what I'm using it for, I told you that from the off but you still want to know more, WHY?

Even when I give you more you argue the semantics of my statements instead of addressing the question, the semantics are only there to try and make my question clearer not to be examined and dissected and commented on!

And dav, though I get your point, you kinda actually proved it for me while you were trying to disprove it.

I stated that overtime things will even up, you say it doesn't matter how long there will always be a larger discrepancy! I get that honestly I do but everyone of your examples has a 0.5??? for a 2 outcome chance now in my world that shows that no matter how long you do these trials its always gonna be a 50/50. So to me if you have a bias of 90% on one side the other is, and I choose my wording here accordingly, a "safe bet" or maybe that should be a safer bet.

It's meant to be a fun hobby for me & you guys are taking the "FUN" out of my enjoyment of maths!

I shall take my question elsewhere, thanks for no help whatsoever!