can anyone program this?

[simon]

I want to generate 30,000 random number decisions (spins) but only the numbers 1, 2, 3, and 4.  then I want the computer to find the numbers 1, 2, and 3 anytime they follow each other in any order, in 3 decisions (spins).  then I want to know how many times the number 4 follows anytime 1 2 and 3 hit in 3 spins in any order (anotherwords, there is no repeat of 1, 2, or 3 on the 4th decision.)  I want to know what probability the computer comes up with for event 4 to hit on the 4th spin, after events 1, 2, and 3 follow each other successively in any order, in 3 spins.

This is the basic question I am trying to get an answer to:

We have 4 possible outcomes, which are randomly generated and each outcome always has a 1 in 4 chance to appear.  If in 4 trials, on the first trial outcome A appears, and on the second trial outcome B appears, and on the third trial outcome C appears, or outcomes A, B and C appear in any order in the first three trials, what is the probability that outcome D will appear in the fourth trial?

The simple answer would seem to be that outcome D has the same 1 in 4 chance to appear on the fourth trial that it did at the first trial.  However, I have been looking at something that indicates that in fact outcome D now has a less than 25% chance to appear on the 4th trial, due to the fact that the 3 previous outcomes A, B, and C also each have a 1 in 4 chance to re-appear, which ends up reducing outcome D's 25% chance to occur that it had at the first trial.

I would like to know what computer results show, and if it indicates that the fourth outcome showed less than 25% immediately following the other 3 outcomes, then that is very good news.

[gizmotron]

Let's try it without a program, just an algorithm.

On bet number 1 you have a 75% chance of hitting step number one. D is the only loss.

On bet two you have a 50% chance because whatever hit in step one (plus D) is opposite half the choices remaining.

On bet three, step three,  you have a 25% of hitting the remaining condition while avoiding the other two (plus D)

I think this last step is right. You have a 25% chance of hitting D on the forth step.

So you do this for four in a row:

75% X 50% X 25% X 25% = 2.34%

If I wrote a simulation that did millions of spins then it would be this outcome, if I did the algorithm right.

[simon]

hmmm, I did not really follow that but are you saying that the probability for D hitting on 4th spin following A-B-C in any order in 3 spins is only 2.34%?  No, you know that's not right, it is somewhere between 17% and 20 or 25%

what I am asking is, if we divide the wheel into four 9 number sections,  and 3 sections hit without repeat in 3 spins, what is the probability that the 4th section will hit on the 4th spin?

[gizmotron]

I'll try it this way:

Step one is 3/4

Step two is 2/4

Step three is 1/4

Step four is 1/4

The formula for the odds for four heads or tails in a row is:

(.5) times (.5) times (.5) times (.5)

Half of .5 = .25, (25%)

Half of 25% is 12.5%

Half of 12.5% = 6.25%

So there is a 6.25% chance for 4 heads in a row.

If there were a four headed coin and on each step one more was off limits until the fourth flip then that alone would even be more rare than a two sided coin. Yes, my algorithm is correct. Were you going to be rich?

[simon]

what is your final answer for the probabilty for the event I am asking about?

the payouts make it always a wash, with 4 possible outcomes, we can bet on the one vs the other three, or bet on all 3 vs the one (similar to betting 2 dozens against one dozen, or one dozen against two), it is always a wash and the house edge prevails.  BUT, if the events I am asking about are not a wash because the expected probability of the 3 events become greater and the expected probability of the one event becomes smaller, than yes you have a winning bet.

[gizmotron]

what is your final answer for the probabilty for the event I am asking about?

Do I get a "lifeline?"

This isn't Slum Dog Millionare --but, my final answer is 2.34%

[simon]

that was a great movie.  so you're saying that with 4 possible events A-B-C-D, the probability for event D to hit on the 4th spin following events A-B-C hitting in any order in the 3 previous spins has been reduced from the 25% probability it had at spin 1, to only 2.34% probability on the 4th spin after A, B, and C hit on the 3 previous spins?  Don't you think then, that if you divide the wheel up into 4 nine number sections, all you would have to do is wait for 3 sections to hit without repeat in 3 spins, and then bet on all 3 sections to hit on the following spin, since your probability to hit one of them would be so great, vs the very small probability of the final event to hit on the 4th spin?

[Tangram]

Hi Simon,

Gizmo's analysis is correct regarding that particular sequence (that of one of each of 1,2,3, in any order, followed by by a 4) but to answer your question - the probability is 25%, no more or less. It has to be, because you are talking about independent trials, so whatever sequence has come before the arrival of the 4 will make no difference.

[gizmotron]

If you were to try the bet thousands and thousands of times then you would win 97%+ times with a nine unit payoff for each. In 100 bet groups you would win 97 X 9 = 873. You would lose three times 27 for 81 lost. Yes, it looks very good. I'm going to have to write a new sim. It's an interesting problem looking for the rarity of an exact situation over four steps. The problem is that you will only see that precondition a little less than 10% of the time. I'm going to find a rarity that happens more commonly, and that has similar long shot odds. 27 numbers out of 37/38 is a good bet. It's funny that the single bet odds are 27% to lose that bet, but the conditional bet is 3% to lose, if you round the numbers a little.

[simon]

Hi Simon,

Gizmo's analysis is correct regarding that particular sequence (that of one of each of 1,2,3, in any order, followed by by a 4) but to answer your question - the probability is 25%, no more or less. It has to be, because you are talking about independent trials, so whatever sequence has come before the arrival of the 4 will make no difference.

I think it is not quite that simple, because you have 3 events each of which can repeat on the fourth decision, vs one event that has not shown yet.  In any case I would love to see what the computer would show.

[simon]

The problem is that you will only see that precondition a little less than 10% of the time. I'm going to find a rarity that happens more commonly, and that has similar long shot odds. 27 numbers out of 37/38 is a good bet. It's funny that the single bet odds are 27% to lose that bet, but the conditional bet is 3% to lose, if you round the numbers a little.

......... no it happens over and over, I have several ways I divide the wheel up into nine number sections for each of the even chance factors, and I can keep track of all of them as they hit and the sequence I am looking for comes in frequently (especially when playing a rapid roulette virtual machine), and often simultaneously under the 3 factors the way I code them.

[gizmotron]

Your exact condition of hitting three of four in any order for three spins without the forth position hitting and any of the other three not repeating is a little less than 10% You can assume that any breach of the condition must allow four steps to occur before the next sequence of four starts. I agree that on each spin a new four step sequence can be tested for. I wonder what that does to the odds. If you have a new sequence starting on each spin then I would think that it increases the odds of losing by 25%. You might lose 27% of the time. That is in line with the true single event odds. I think you must let all four spins happen before checking for a new precondition to occur.

[simon]

I check on a continuous basis not divided up, anotherwords if any 3 previous spins anywhere in the cycle of spins I am tracking shows 3 of the 4 events without repeat, that is the trigger to bet on the next spin for a repeat of any of the 3 events which just hit.  I guess you are saying that is a problem for what I am trying to do?  this is why I just want a computer to do what I originally asked, so I can see what it comes up with and if a greater or lesser than 25% comes out for the event I want to track.

[gizmotron]

The condition is a four spin sequence. The math is for exactly four completed spins. If you do it a thousand times then it reaches the statistical normality of 2.34% for it to occur if a thousand four spin sequences happen back to back. If you go by each spin then the odds are 25%.

So let's look at it from another point of view. Every four spins you will have a 2.3% of the D slot will hit on the fourth spin after each one of the other three have hit. After the first three have hit there is a 25% chance that the D section will hit. I think the sim will produce that 25%. I'm not sure what is missing here. I think it's an individual spin and so it must have an individual spin's odds. Tangram  is correct. There is no point in producing this sim.

[simon]

hmmm, that's not exactly what I want to hear but I suppose it's true.  the thing is I was looking at a very unique formula that shows that something a little different happens then the 25% expectation for the 4th event after the first 3 hit, although maybe you are right and you can't just look for any 3 spins that show 3 of the 4 events, the 4 spins have to be divided up back to back.  which is not a problem, I just want to know if the formula I was looking at has any validity.  if possible maybe I can scan the page of the book to show it to you.  basically the author says that if we watch 4 spins, and A hits then B hits then C hits, it is a better bet to bet on A or B or C, then D, and he has an interesting explanantion and formula to show why.  I just want to know what the computer shows.

[simon]

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