rumba434
rumba434
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August 23rd, 2020 at 5:39:56 AM permalink
If 37 players were logged in to the same single zero roulette table and they all flat bet $1 every turn on their favourite number, each one different, how many turns average would it take before they all had less money than they started with?
ChumpChange
ChumpChange
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August 23rd, 2020 at 7:54:31 AM permalink
I used to play an electronic double zero wheel last century. It never really let me win more than 3 bets ahead on even money bets, same with the electronic dollar BJ machines.
ThatDonGuy
ThatDonGuy
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ChumpChange
August 23rd, 2020 at 8:43:26 AM permalink
I get 40,500 in simulation.

It becomes 18,750 on a double-zero wheel, and 11,100 on "the devil's wheel" (triple-zero)

OOPS - those were only paying 34-1. See below for corrected numbers.

Bonus math question: how many would it take if there were no zeroes on the wheel?

It would never get to the point where every number was behind, since, in order for that to happen, the sum of the profits would have to be negative, since every individual profit was negative, but on a no-zero wheel, every spin loses 35 on the losing numbers but wins 35 on the winning number, so the sum of the profits is always zero.

Last edited by: ThatDonGuy on Aug 23, 2020
sabre
sabre 
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RogerKint
August 23rd, 2020 at 2:19:26 PM permalink
Quote: ChumpChange

I used to play an electronic double zero wheel last century. It never really let me win more than 3 bets ahead on even money bets, same with the electronic dollar BJ machines.



Great post. Very on topic. Would read again 10/10
charliepatrick
charliepatrick
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August 23rd, 2020 at 5:42:29 PM permalink
I get totally different numbers, so possibly have a bug in my code.
Average length of run before a spin where everyone is broke at the same time = 168346 (average of different runs had 167016-170128, actual runs went from 48420-493344)
Average length of run, where broke people drop out and don't play on, before everyone has goes broke is quite a range 992 (average of diffrent runs 641-1506, actual runs from 37-109513).

In the second analysis (I added the logic before thinking about it much) the person who found the first number stays in, but everyone else went broke. So most often after 37 spins that person goes broke. It's interesting that starting with +35 that the person usually lasts nearly 1000 spins, but sometimes much longer. Thus on normal runs, after say three spins, three people might be up +33 or one doing really quite well. So they're going to last quite a while; with a chance that someone else has a run of luck. Though I'm surprised it's so long.
ThatDonGuy
ThatDonGuy
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charliepatrick
August 23rd, 2020 at 6:11:39 PM permalink
Quote: charliepatrick

I get totally different numbers, so possibly have a bug in my code.


I had the bug in my code - I was only paying 34-1.

At 35-1, I get:
170,000 spins on a single-zero wheel
43,500 spins on a double-zero wheel
21,000 spins on a triple-zero wheel
charliepatrick
charliepatrick
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August 24th, 2020 at 9:10:22 AM permalink
I did an overnight run and got an average of 169423 (26497-995421).
Ace2
Ace2
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August 25th, 2020 at 10:12:42 PM permalink
This problem got me thinking about a somewhat similar problem

If you start rolling a standard die and record the results, how many rolls will it take, on average, for all six numbers to have an equal number of hits?
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
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August 26th, 2020 at 6:38:23 AM permalink
Quote: Ace2

This problem got me thinking about a somewhat similar problem

If you start rolling a standard die and record the results, how many rolls will it take, on average, for all six numbers to have an equal number of hits?


I have a feeling the number approaches positive infinity.

Change the problem from a 6-sided die to the toss of a coin.
Let E(n) be the expected number of tosses needed to reach equality where one side has been tossed n times more than the other.
E(0) = 0
E(n) = 1 + 1/2 E(n-1) + 1/2 E(n+1) for n > 0
The starting point is 1 + E(1), since the first toss will result in one side being tossed 1 time more than the other.
Without getting into math details, E(n) = n + n / (n + 1) x E(n+1)
E(1) = 1 + 1/2 E(2)
= 1 + 1/2 (2 + 2/3 E(3)) = 2 + 1/3 E(3)
= 2 + 1/3 (3 + 3/4 E(4)) = 3 + 1/4 E(4)
= 3 + 1/4 (4 + 4/5 E(5)) = 4 + 1/5 E(5)
..., which diverges
Last edited by: ThatDonGuy on Aug 26, 2020
DJTeddyBear
DJTeddyBear
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August 26th, 2020 at 10:08:36 AM permalink
I take a different approach to this question.

Let's assume for a moment, that there is one number that is a complete loser. I.E. It never hits. In that case, similar to ThatDonGuy's argument that with a non zero wheel, at least one of the other numbers must be positive, or at least not all numbers be negative.

With that in mind, I choose to calculate the odds of a specific number not hitting in X number of spins. The formula for that is simple:

( 36 / 37 ) ^ X

The chart below shows the results.

Interestingly, it takes 26 spins for a number to pass the 50% mark.

Obviously, it never gets all the way to zero, but gets exceedingly close.

It takes 25,855 spins before it exceeds Excel's capabilities and incorrectly returns a value of zero.


Spins
Odds
1
0.972 972 972 973
2
0.946 676 406 136
3
0.921 090 557 321
4
0.896 196 217 934
5
0.871 974 698 531
10
0.760 339 874 878
25
0.504 103 157 284
26
0.490 478 747 628
50
0.254 119 993 184
75
0.128 102 690 893
100
0.064 576 970 936
150
0.016 410 299 414
200
0.004 170 185 175
300
0.000 269 297 927
400
0.000 017 390 444
500
0.000 001 123 022
750
1.190 097 E-009
1,000
1.261 179 E-012
2,000
1.590 572 E-024
3,000
2.005 996 E-036
4,000
2.529 920 E-048
5,000
3.190 682 E-060
7,500
5.699 349 E-090
10,000
1.018 045 E-119
12,500
1.818 481 E-149
15,000
3.248 258 E-179
17,500
5.802 194 E-209
20,000
1.036 416 E-238
22,500
1.851 296 E-268
25,000
3.306 873 E-298
25,200
1.379 027 E-300
25,400
5.750 799 E-303
25,600
2.398 190 E-305
25,800
1.000 089 E-307
25,825
5.041 483 E-308
25,850
2.541 427 E-308
25,851
2.472 740 E-308
25,852
2.405 909 E-308
25,853
2.340 885 E-308
25,854
2.277 618 E-308
25,855
0
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁

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