Probability that any 50/50 option will not have hit by the 15th spin
April 4th, 2020 at 4:01:24 PM
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I'm trying to calculate the probability that any 50/50 option (red,black,even,odd,high,low) will not have hit by the 15th spin. I saw an old post from the Wizard which brilliantly explains the probability of any single number not being hit by the 200th spin, and was wondering if that same formula applies here, and how I would need to adjust the formula to account for the 50/50 bets. Can someone please help me out?
This is the single number formula response I'm referring to:
The probability that any given number will not have hit is (37/38)200 = 0.48%.
With 38 numbers, we could incorrectly say that the probability that any one of them would not be hit is 38 × (37/38)200 = 18.34%.
The reason this is incorrect is it double counts two numbers not being hit. So we need to subtract those probabilities out. There are combin(38,2) = 703 sets of 2 numbers out of 38. The probability of not hitting any two given numbers is (36/38)200 = 0.000020127. We need to subtract the probability of avoiding both numbers. So we are at:
38×(37/38) 200 - combin(38,2)×(36/38) 200 = 16.9255%.
However, now we have canceled out the probability of three numbers not hitting. For any given group of three numbers, we triple counted the probability of any single number not being hit. We then triple subtracted for each way to choose two numbers out of the three, leaving with zero for the probability that all three numbers were not hit. There are combin(38,3)=8,436 such groups. Adding them back in we are now at:
38×(37/38) 200 - combin(38,2)×(36/38) 200 + combin(38,3)×(35/38)200 = 16.9862%.
Yet, now we have over-counted the probability of four numbers not hitting. For each of the combin(38,4)=73,815 groups of four numbers, each was originally quadruple counted. Then we subtracted each of the combin(4,2)=6 groups of 2 out of the 4. Then we added back in the 4 groups of 3 out of the 4. So, for each union of four numbers, it was counted 4 − 6 + 4 = 2 times. To adjust for the double counting we must subtract for each group. Subtracting them out we are now at:
38×(37/38) 200 - combin(38,2)×(36/38) 200 + combin(38,3)×(35/38)200 - combin(38,4)×(34/38)200 = 16.9845%.
Continuing in the process we would keep alternating adding and subtracting, all the way until missing 37 numbers. Thus the probability of at least one number never being hit is:
Sum i=1 to 37 [(-1)(i+1) × combin(38,i) × ((38-i)/38)38] = 16.9845715651245%
This is the single number formula response I'm referring to:
The probability that any given number will not have hit is (37/38)200 = 0.48%.
With 38 numbers, we could incorrectly say that the probability that any one of them would not be hit is 38 × (37/38)200 = 18.34%.
The reason this is incorrect is it double counts two numbers not being hit. So we need to subtract those probabilities out. There are combin(38,2) = 703 sets of 2 numbers out of 38. The probability of not hitting any two given numbers is (36/38)200 = 0.000020127. We need to subtract the probability of avoiding both numbers. So we are at:
38×(37/38) 200 - combin(38,2)×(36/38) 200 = 16.9255%.
However, now we have canceled out the probability of three numbers not hitting. For any given group of three numbers, we triple counted the probability of any single number not being hit. We then triple subtracted for each way to choose two numbers out of the three, leaving with zero for the probability that all three numbers were not hit. There are combin(38,3)=8,436 such groups. Adding them back in we are now at:
38×(37/38) 200 - combin(38,2)×(36/38) 200 + combin(38,3)×(35/38)200 = 16.9862%.
Yet, now we have over-counted the probability of four numbers not hitting. For each of the combin(38,4)=73,815 groups of four numbers, each was originally quadruple counted. Then we subtracted each of the combin(4,2)=6 groups of 2 out of the 4. Then we added back in the 4 groups of 3 out of the 4. So, for each union of four numbers, it was counted 4 − 6 + 4 = 2 times. To adjust for the double counting we must subtract for each group. Subtracting them out we are now at:
38×(37/38) 200 - combin(38,2)×(36/38) 200 + combin(38,3)×(35/38)200 - combin(38,4)×(34/38)200 = 16.9845%.
Continuing in the process we would keep alternating adding and subtracting, all the way until missing 37 numbers. Thus the probability of at least one number never being hit is:
Sum i=1 to 37 [(-1)(i+1) × combin(38,i) × ((38-i)/38)38] = 16.9845715651245%
April 4th, 2020 at 4:23:45 PM
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Quote: kahcwwI'm trying to calculate the probability that any 50/50 option (red,black,even,odd,high,low) will not have hit by the 15th spin.
It's far easier than that.
Since your so called 50:50 bets are not 50:50 at all, I'll do it for European single zero roulette.
Probability red = 18/37 = 0.48648648649
Probability black = 18/37 = 0.48648648649
Probability green = 1/37 = 0.02702702703
Probability red or black or green = 37/37 = 1 = 100%
Probability of NOT red = 19/37 = 0.51351351351
Those were probability for the next ONE observed spin.
For the next* TWO observed spins, the probability of NOT red on both occasions is (19/37) x (19/37)
or
For the next THREE observed spins, the probability of NOT red on all three occasions is (19/37) x (19/37) x (19/37) = (19/37)^3
For the next FIFTEEN observed spins, the probability of NOT red on all fifteen occasions is (19/37)^15 =0.00004552796
or, about 1 in 21964
In the absence of green 0, for a real 50:50 proposition, the chance of losing 15 consecutive wagers would be (0.5)^15 = 0.00003051758 or about 1 in 32768
When I say 'the next n spins' I mean any consecutive n spins starting from whenever you decide to start. That would be the same probability even if you said I'm not even going to think about betting until I've observed 10 consecutive non-red results.
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
April 5th, 2020 at 3:44:05 AM
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^ Good reply.
As a general rule when there's a question what are the chances of something occuring in the next somethings, it's generally easier to look at the thing not occuring in all those somethings and then use 1 - that.
A simple example is the chances of throwing one six (or more) in four rolls of a die. The chances of missing once is 5/6, so the chances of missing four times in a row is (5*5*5*5)/(6*6*6*6) = 625/1296. So the chances of at least one six is (1296-625)/1296 = 671/1296.
Thus you can plug in the figures (1/2, 18/37 or 18/38) as appropriate.
As a general rule when there's a question what are the chances of something occuring in the next somethings, it's generally easier to look at the thing not occuring in all those somethings and then use 1 - that.
A simple example is the chances of throwing one six (or more) in four rolls of a die. The chances of missing once is 5/6, so the chances of missing four times in a row is (5*5*5*5)/(6*6*6*6) = 625/1296. So the chances of at least one six is (1296-625)/1296 = 671/1296.
Thus you can plug in the figures (1/2, 18/37 or 18/38) as appropriate.
April 5th, 2020 at 4:28:01 AM
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Thank you for your response. I am aware of the (19/38)^15 formula, but based on the post I referenced, the Wizard says that only applies if you're watching a particular number, or in this case, a particular '50/50' bet like Red. But If I'm not watching any in particular, I'm calculating odds of any of the 1:1 to not hit for 15 turns, does the formula then change to the more complex option (Sum i=1 to 37 [(-1)(i+1) × combin(38,i) × ((38-i)/38)38] = 16.9845715651245%) which would need to be adjusted for the 1:1 19/38 instead of the single number that it's currently setup for. If so, can someone please show me how to adjust that formula, I'm not sure exactly what needs to change to apply to 1:1 or 1:2 bets.
April 5th, 2020 at 5:59:38 AM
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Quote: kahcwwThank you for your response. I am aware of the (19/38)^15 formula, ......I'm calculating odds of any of the 1:1 to not hit for 15 turns,
I'm sorry. I could help with this, but as an exercise in maths, I really cannot be bothered. Life is just too short.
I'll just remind you that the answer to this exercise will not yield a winning system.
Last edited by: OnceDear on Apr 5, 2020
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
April 5th, 2020 at 8:24:33 AM
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Quote: kahcwwThank you for your response. I am aware of the (19/38)^15 formula, but based on the post I referenced, the Wizard says that only applies if you're watching a particular number, or in this case, a particular '50/50' bet like Red. But If I'm not watching any in particular, I'm calculating odds of any of the 1:1 to not hit for 15 turns, does the formula then change to the more complex option (Sum i=1 to 37 [(-1)(i+1) × combin(38,i) × ((38-i)/38)38] = 16.9845715651245%) which would need to be adjusted for the 1:1 19/38 instead of the single number that it's currently setup for. If so, can someone please show me how to adjust that formula, I'm not sure exactly what needs to change to apply to 1:1 or 1:2 bets.
It's not as easy as that, for one reason; unlike with individual numbers, where each number wins on exactly one bet and that is the only number that wins on that bet, each number from 1 to 36 wins on three of the six even-money bets (and two of the six 2-1 bets), and there are 18 numbers that win on each even-money bet (and 12 on each 2-1 bet).
I don't think there's an easy formula for this - you may have to do something along the lines of a brute force search, where you check every possible combination of results from the 15 spins.
Example: start with the starting condition (0,0,0,0,0,0,0), which is zero spins, no reds yet, no blacks yet, no odds yet, no evens yet, no lows yet, no highs yet (the last six numbers are 0 if that bet has not won yet, and 1 if it has).
Assuming a double-zero wheel:
1, 3, 5, 7, 9 are Red, Odd, Low, so (1,1,0,1,0,1,0) has a probability of 5/38
19, 21, 23, 25, 27 are Red, Odd, High, so (1,1,0,1,0,0,1) has a probability of 5/38
12, 14, 16, 18 are Red, Even, Low, so (1,1,0,0,1,1,0) has a probability of 4/38
30, 32, 34, 36 are Red, Even, High, so (1,1,0,0,1,0,1) has a probability of 4/38
11, 13, 15, 17 are Black, Odd, Low, so (1,0,1,1,0,1,0) has a probability of 4/38
29, 31, 33, 35 are Black, Odd, High, so (1,0,1,1,0,0,1) has a probability of 4/38
2, 4, 6, 8, 10 are Black, Even, Low so (1,0,1,0,1,1,0) has a probability of 5/38
20, 22, 24, 26, 28 are Black, Even, High, so (1,0,1,0,1,0,1) has a probability of 5/38
0, 00 are Green, so (1,0,0,0,0,0,0) has a probability of 2/38
Now, for the second spin, check each of those nine results - for example, if the first spin was 1, you are at (1,1,0,1,0,1,0)
Another red, odd, low is (2,1,0,1,0,1,0); the probability = 5/38 (for the spins up to this point) x 5/38 (for this spin)
A red, odd, high is (2,1,0,1,0,1,1); the probability = 5/38 x 4/38
and so on for the other seven results
Note that, on the second spin, a green is also (2,1,0,1,0,1,0), so add 5/38 x 2/38 to the other (2,1,0,1,0,1,0) value you calculated earlier.
Repeat this for the remaining 13 spins, but note that there are now 28 possible sets of results for a particular number of spins.
When you are done, the value for (15,1,1,1,1,1,1) is the probability that all six even-money bets hit at least once in 15 spins; take 1 minus this value to get the probability that at least one does not. Alternatively, you can add up the 31 (15,x,x,x,x,x,x) values where at least one x = 0.
You will find that the probability on a double-zero wheel that at least one of the six even-money bets never wins in 15 spins is 1 / 2531.
Last edited by: ThatDonGuy on Apr 5, 2020
April 5th, 2020 at 10:52:14 AM
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Speaking from some first-hand experience with developing odds for games....your best bet (no pun intended) is to program a simulation and run it for as many trials as possible. Then create a program that reads the simulation that can search for the exact scenario you're looking for, then the easiest part is figuring out how many times that occurred over the whole data set.
April 5th, 2020 at 2:04:09 PM
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Don’t call them 50/50. Call them even money.
Um, that should be either (20/38)^x for double zero or (19/37)^x for single zero. Let’s not talk about triple zero.Quote: kahcwwThank you for your response. I am aware of the (19/38)^15 formula, ...
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April 5th, 2020 at 2:38:03 PM
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Thanks ThatDonGuy, I really appreciate you taking the time to write that out! I have just one follow-up question if you don't mind. Is it possible to do a quick equation to generate the final ratio (1 / 2531 for example)?
Let's say I choose the option to select 1 / 2500, following your formula I'd get 15 spins because that the closest total spin value to 1 / 2500. But what if I chose 1 / 1200 odds, is there a way to have your formula calculate how many spins it would need to get to 1 / 1200?
I'm not sure how I'd systematically setup an equation to check the probability at each spin other than 15, but I think you were leading to that answer with this paragraph:
When you are done, the value for (15,1,1,1,1,1,1) is the probability that all six even-money bets hit at least once in 15 spins; take 1 minus this value to get the probability that at least one does not. Alternatively, you can add up the 31 (15,x,x,x,x,x,x) values where at least one x = 0.
Thanks again!
Let's say I choose the option to select 1 / 2500, following your formula I'd get 15 spins because that the closest total spin value to 1 / 2500. But what if I chose 1 / 1200 odds, is there a way to have your formula calculate how many spins it would need to get to 1 / 1200?
I'm not sure how I'd systematically setup an equation to check the probability at each spin other than 15, but I think you were leading to that answer with this paragraph:
When you are done, the value for (15,1,1,1,1,1,1) is the probability that all six even-money bets hit at least once in 15 spins; take 1 minus this value to get the probability that at least one does not. Alternatively, you can add up the 31 (15,x,x,x,x,x,x) values where at least one x = 0.
Thanks again!
April 5th, 2020 at 3:00:12 PM
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I can do this: here are the probabilities of at least one of the six even-money bets losing all N times in N spins, for N from 6 to 40:
| 6 | 1 / 8 |
| 7 | 1 / 15 |
| 8 | 1 / 29 |
| 9 | 1 / 54 |
| 10 | 1 / 103 |
| 11 | 1 / 195 |
| 12 | 1 / 370 |
| 13 | 1 / 702 |
| 14 | 1 / 1332 |
| 15 | 1 / 2531 |
| 16 | 1 / 4808 |
| 17 | 1 / 9135 |
| 18 | 1 / 17,356 |
| 19 | 1 / 32,975 |
| 20 | 1 / 62,651 |
| 21 | 1 / 119,036 |
| 22 | 1 / 226,168 |
| 23 | 1 / 429,718 |
| 24 | 1 / 816,462 |
| 25 | 1 / 1,551,277 |
| 26 | 1 / 2,947,425 |
| 27 | 1 / 5,600,105 |
| 28 | 1 / 10,640,198 |
| 29 | 1 / 20,216,373 |
| 30 | 1 / 38,411,105 |
| 31 | 1 / 72,981,095 |
| 32 | 1 / 138,664,066 |
| 33 | 1 / 263,461,689 |
| 34 | 1 / 500,577,107 |
| 35 | 1 / 951,096,160 |
| 36 | 1 / 1,807,082,285 |
| 37 | 1 / 3,433,455,170 |
| 38 | 1 / 6,523,557,114 |
| 39 | 1 / 12,394,727,995 |
| 40 | 1 / 23,549,834,119 |
