Dice odds

[Shadow00Fox]

Recent game calculation. The odds of failing every time in a row like happened were 1 in 18,000, before getting one number set that went double the expected average of 6, stopped at 11 dice rolls. Other person won on the next turn.

2 games later, failure odds I believe 1/14,400. With one turn where I stopped less than halfway to expected average.


I'm just wondering if the dice aren't as random as math says they should be, or are odds like this normal? Or am I calculating them wrong? I'm just going by the simple average dice rolls for the given number then multiplying the fractions.

Reviewing a game makes it very easy to see.

[Romain672]

You didn't explained much what you did, what are the games, what are you calculations, so not that's not very easy at all to see what you mean.

About maths, if I understood you (which I'm not sure at all) you can multiply those probabilities, but only the probability for you to miss at any point of your turn, not the probability to miss the last throw.

Please send us/me number of the game or a link to it if you are interrested by the math part.


For the rest others persons should explain it better, but human are bad at perceiving randomness, so this is normal, you are just biaised.

[Shadow00Fox]

You didn't explained much what you did, what are the games, what are you calculations, so not that's not very easy at all to see what you mean.

About maths, if I understood you (which I'm not sure at all) you can multiply those probabilities, but only the probability for you to miss at any point of your turn, not the probability to miss the last throw.

Please send us/me number of the game or a link to it if you are interrested by the math part.


For the rest others persons should explain it better, but human are bad at perceiving randomness, so this is normal, you are just biaised.

Thank you! I'm very interested in how much bias comes into it. I'll find the game link here.... I never took statistics in school.

See if this works: not one of the prime examples but I'm curious
https://boardgamearena.com/table?table=541445098

[Jellby]

Even if everything is correct, there are many more than 18000 games played, so it's not unexpected that it happened. And even if it were unexpected, it's not proof of non-random dice.

[euklid314]

See if this works: not one of the prime examples but I'm curious
https://boardgamearena.com/table?table=541445098

Hopefully Romain is not doing the game analysis in parallel to me...

In the above game you, Shadow00Fox, always rolled till you failed which simplifies the analysis. In your first 6 attempts you failed (much) earlier than expected. In your last, 7th attempt you failed (much) later than expected.

I will give a table of which 3 numbers you had, on which attempt (after the 3 numbers were determined) you failed, and how likely it was that you should fail by this time (or earlier):

1) 6-9-12, 1st attempt failed, probability of 17,1%
2) 6-7-9, 2nd attempt failed, probability of 16,5%
3) 8-9-12, 1st attempt failed, probability of 23,0%
4) 4-7-12, 2nd attempt failed, probability of 30,6%
5) 5-7-8, 5th attempt failed, probability of 36,2%
6) 2-7-8, 1st attempt failed, probability of 11,0%
7) 4-5-7, 12th attempt failed, probability of 89,9%

Summary: None of the events for itself is a very rare case. The most improbable case was No. 6, when you had 2-7-8 and fell down immediately. That happens only in 11% of all cases. Falling down on the 1st or 2nd attempt with 4-7-12 (No. 4), however, is quite likely - with 30,6%. Still below average but will happen every 3rd time. Surviving with 4-5-7 till the 13th attempt (No. 7) is rare. In 89,9% you fall down earlier.

The difficult part is how to evaluate the sum of all 7 events happening in the same game. I will argue in a separate thread (and probably on another day) why it is not correct to just multiply all probabilities 0.171 * 0.165 * 0.23 * ... * 0.899 = 0.000067 =1/14700 and treat it like a huge outlier, comparable to a lotto win. Quite on the contrary: Yes, this sequence is unlucky but such sequences are very likely to happen - not every 15000 games!

[euklid314]

Some inbetween analysis since the maths behind Cant Stop is fascinating and sometimes counterintuitive to us humans...

It is well known that 6-7-8 is the best combination, you will succeed in 92,0% of all tries if these 3 numbers are open. In other words you fail only in 8% of your tries.

Rolling till you fail the 6-7-8 is - mathematically speaking - modelled by a geometric probability function (i.e., forget everything what you know about normal distributions, binomial distributions, Gaussian bell curves,...).

The expected value for 6-7-8 is, that you fall with your 12,5th attempt.

*) The first misconception might be to feel "on the safe side" during your first 12 attempts and feel very daring and bold for playing on after the 13th attempt.

*) Surely the probability must be 50% for failing during the first 12 attempts and 50% for failing in the 13th attempt or later. True? False. Even if the expected failure is at the 12.5th attempt you will survive till the 12th attempt only in 36.8% of all cases. In 63.2% of all cases you wont reach the 13th attempt.

*) 8 consecutive successes for 6-7-8 have a probability of approx. 50%. Thus, if you have rolled 8 times successfully you should better stop because it is getting dangerous now! True? False. The dice have no memory.

*) If you have rolled 8 times successfully you still have a 50% chance to also roll the next 8 times sucessfully, resulting in a 16-series. Even if you have reached such a 16-series (unlikely? well 25% likely!) you could add another 16-series of tries and you have again a 1in4 chance to survive (if only CantStop ropes were longer...).
Thus a series of 32 consecutive sucesses has a probability of approx 7%. Failing at the first attempt has a probability of 8%! Thus, very short series (1 roll) and very long ones (33+ rolls) are quite common and equally likely!

People lament about undeserved bad luck (immediate falls) and undeserved good luck (ultralong success series) but both are to be expected with a geometric distribution!

Disclaimer: The relatively high probability of ultralong sequences is the reason why there is a 50% probability for succeeding 8 times for 6-7-8, but the expected failure roll is 12.5 times, i.e. much later.

[Shadow00Fox]

Some inbetween analysis since the maths behind Cant Stop is fascinating and sometimes counterintuitive to us humans...

It is well known that 6-7-8 is the best combination, you will succeed in 92,0% of all tries if these 3 numbers are open. In other words you fail only in 8% of your tries.

Rolling till you fail the 6-7-8 is - mathematically speaking - modelled by a geometric probability function (i.e., forget everything what you know about normal distributions, binomial distributions, Gaussian bell curves,...).

The expected value for 6-7-8 is, that you fall with your 12,5th attempt.

*) The first misconception might be to feel "on the safe side" during your first 12 attempts and feel very daring and bold for playing on after the 13th attempt.

*) Surely the probability must be 50% for failing during the first 12 attempts and 50% for failing in the 13th attempt or later. True? False. Even if the expected failure is at the 12.5th attempt you will survive till the 12th attempt only in 36.8% of all cases. In 63.2% of all cases you wont reach the 13th attempt.

*) 8 consecutive successes for 6-7-8 have a probability of approx. 50%. Thus, if you have rolled 8 times successfully you should better stop because it is getting dangerous now! True? False. The dice have no memory.

*) If you have rolled 8 times successfully you still have a 50% chance to also roll the next 8 times sucessfully, resulting in a 16-series. Even if you have reached such a 16-series (unlikely? well 25% likely!) you could add another 16-series of tries and you have again a 1in4 chance to survive (if only CantStop ropes were longer...).
Thus a series of 32 consecutive sucesses has a probability of approx 7%. Failing at the first attempt has a probability of 8%! Thus, very short series (1 roll) and very long ones (33+ rolls) are quite common and equally likely!

People lament about undeserved bad luck (immediate falls) and undeserved good luck (ultralong success series) but both are to be expected with a geometric distribution!

Disclaimer: The relatively high probability of ultralong sequences is the reason why there is a 50% probability for succeeding 8 times for 6-7-8, but the expected failure roll is 12.5 times, i.e. much later.

Wow, thanks!!

[Shadow00Fox]

See if this works: not one of the prime examples but I'm curious
https://boardgamearena.com/table?table=541445098

Hopefully Romain is not doing the game analysis in parallel to me...

In the above game you, Shadow00Fox, always rolled till you failed which simplifies the analysis. In your first 6 attempts you failed (much) earlier than expected. In your last, 7th attempt you failed (much) later than expected.

I will give a table of which 3 numbers you had, on which attempt (after the 3 numbers were determined) you failed, and how likely it was that you should fail by this time (or earlier):

1) 6-9-12, 1st attempt failed, probability of 17,1%
2) 6-7-9, 2nd attempt failed, probability of 16,5%
3) 8-9-12, 1st attempt failed, probability of 23,0%
4) 4-7-12, 2nd attempt failed, probability of 30,6%
5) 5-7-8, 5th attempt failed, probability of 36,2%
6) 2-7-8, 1st attempt failed, probability of 11,0%
7) 4-5-7, 12th attempt failed, probability of 89,9%

Summary: None of the events for itself is a very rare case. The most improbable case was No. 6, when you had 2-7-8 and fell down immediately. That happens only in 11% of all cases. Falling down on the 1st or 2nd attempt with 4-7-12 (No. 4), however, is quite likely - with 30,6%. Still below average but will happen every 3rd time. Surviving with 4-5-7 till the 13th attempt (No. 7) is rare. In 89,9% you fall down earlier.

The difficult part is how to evaluate the sum of all 7 events happening in the same game. I will argue in a separate thread (and probably on another day) why it is not correct to just multiply all probabilities 0.171 * 0.165 * 0.23 * ... * 0.899 = 0.000067 =1/14700 and treat it like a huge outlier, comparable to a lotto win. Quite on the contrary: Yes, this sequence is unlucky but such sequences are very likely to happen - not every 15000 games!

Thank you for this one as well!
Do note, I think it was nearly in two consecutive games.

[Shadow00Fox]

Some inbetween analysis since the maths behind Cant Stop is fascinating and sometimes counterintuitive to us humans...

It is well known that 6-7-8 is the best combination, you will succeed in 92,0% of all tries if these 3 numbers are open. In other words you fail only in 8% of your tries.

Rolling till you fail the 6-7-8 is - mathematically speaking - modelled by a geometric probability function (i.e., forget everything what you know about normal distributions, binomial distributions, Gaussian bell curves,...).

The expected value for 6-7-8 is, that you fall with your 12,5th attempt.

*) The first misconception might be to feel "on the safe side" during your first 12 attempts and feel very daring and bold for playing on after the 13th attempt.

*) Surely the probability must be 50% for failing during the first 12 attempts and 50% for failing in the 13th attempt or later. True? False. Even if the expected failure is at the 12.5th attempt you will survive till the 12th attempt only in 36.8% of all cases. In 63.2% of all cases you wont reach the 13th attempt.

*) 8 consecutive successes for 6-7-8 have a probability of approx. 50%. Thus, if you have rolled 8 times successfully you should better stop because it is getting dangerous now! True? False. The dice have no memory.

*) If you have rolled 8 times successfully you still have a 50% chance to also roll the next 8 times sucessfully, resulting in a 16-series. Even if you have reached such a 16-series (unlikely? well 25% likely!) you could add another 16-series of tries and you have again a 1in4 chance to survive (if only CantStop ropes were longer...).
Thus a series of 32 consecutive sucesses has a probability of approx 7%. Failing at the first attempt has a probability of 8%! Thus, very short series (1 roll) and very long ones (33+ rolls) are quite common and equally likely!

People lament about undeserved bad luck (immediate falls) and undeserved good luck (ultralong success series) but both are to be expected with a geometric distribution!

Disclaimer: The relatively high probability of ultralong sequences is the reason why there is a 50% probability for succeeding 8 times for 6-7-8, but the expected failure roll is 12.5 times, i.e. much later.

Say, can you explain how you'll only reach 12.5 rolls of 6-7-8 on only 36% of the cases?

[euklid314]

Say, can you explain how you'll only reach 12.5 rolls of 6-7-8 on only 36% of the cases?

Calculation: 0.92^12 = 0.368

Explanation: If you want to survive the first 12 rolls then you have to fulfill the 0.92 chance of rolling 6-7-8 successfully for 12 times in a row.

Conclusion: Thus you fail on your "13+"th attempt (i.e. 13-infinity attempts) in 36.8% of all cases and you fail on your "12-"th attempt (i.e. 1-12 attempts) in 63.2% of all cases.