Dice odds

[euklid314]

You can test your mathematical intuition on a theoretical game, Shadow00Fox:

Lets say i start with columns 6-7-8 and then I give you the option to win 100 Euro if you correctly guess when I will fall (I will continue to roll until I fall). As an example, if you guess that I will fall at my 3rd attempt and if indeed I exactly fail at my 3rd attempt I give you 100 Euro.

On which number should you bet in order to maximize your chance to win the 100 Euro?
A) 1st attempt
B) 5th attempt
C) 8th attempt
D) 12th attempt

The answer is easy if you calculate but not that trivial if you trust your instincts, I guess...

[Game]

its just A or not

[Shadow00Fox]

You can test your mathematical intuition on a theoretical game, Shadow00Fox:

Lets say i start with columns 6-7-8 and then I give you the option to win 100 Euro if you correctly guess when I will fall (I will continue to roll until I fall). As an example, if you guess that I will fall at my 3rd attempt and if indeed I exactly fail at my 3rd attempt I give you 100 Euro.

On which number should you bet in order to maximize your chance to win the 100 Euro?
A) 1st attempt
B) 5th attempt
C) 8th attempt
D) 12th attempt

The answer is easy if you calculate but not that trivial if you trust your instincts, I guess...

My guess is that C, the 8th attempt should be the most accurate in the long run based on the numbers you've mentioned previously? That will be close to 50% odds?

[euklid314]

My guess is that C, the 8th attempt should be the most accurate in the long run based on the numbers you've mentioned previously? That will be close to 50% odds?

You misunderstood my mathematical riddle.

Falling at the 8th attempt or earlier(!) is 50% (i.e., falling at the 1st, 2nd, 3rd, 4th, 5th, 6th, 7th OR 8th attempt).
But in my "game" you must guess the falling time exactly! The probability that you will fall exactly on the 8th attempt is obviously much lower than falling on the 8th attempt or earlier.

Still wanna bet on C? Is falling exactly at the 8th attempt the likeliest moment?

[Shadow00Fox]

A, first roll, because every roll has an equal chance to fail?

[euklid314]

A, first roll, because every roll has an equal chance to fail?

Answer A is correct.

The probability to fall at
1st roll: 8.0%
2nd roll: 7.4%
3rd roll: 6.8%
4th roll: 6.2%
5th roll: 5.7%
6th roll: 5.3%
7th roll: 4.9%
8th roll: 4.5%
9th roll: 4.1%
10th roll: 3.8%
...
20th roll: 1.6%
...
30th roll: 0.7%

This special kind of decreasing series of numbers is called a "geometric" series, thus the name of a "geometric random distribution".

If you add all the above numbers for 1-30 you get 91,8%. Thus failing later as the 30th attempt still has a probability of 8.2%.

[Shadow00Fox]

A, first roll, because every roll has an equal chance to fail?

Answer A is correct.

The probability to fall at
1st roll: 8.0%
2nd roll: 7.4%
3rd roll: 6.8%
4th roll: 6.2%
5th roll: 5.7%
6th roll: 5.3%
7th roll: 4.9%
8th roll: 4.5%
9th roll: 4.1%
10th roll: 3.8%
...
20th roll: 1.6%
...
30th roll: 0.7%

This special kind of decreasing series of numbers is called a "geometric" series, thus the name of a "geometric random distribution".

If you add all the above numbers for 1-30 you get 91,8%. Thus failing later as the 30th attempt still has a probability of 8.2%.

That's pretty cool, thanks!

[Ceaseless]

Lets say i start with columns 6-7-8 and then I give you the option to win 100 Euro if you correctly guess when I will fall (I will continue to roll until I fall). As an example, if you guess that I will fall at my 3rd attempt and if indeed I exactly fail at my 3rd attempt I give you 100 Euro.

On which number should you bet in order to maximize your chance to win the 100 Euro?
A) 1st attempt
B) 5th attempt
C) 8th attempt
D) 12th attempt

The answer is easy if you calculate but not that trivial if you trust your instincts, I guess...

The answer should be A, assuming that the chance of success and failure are each less than 100% and that each of the rolls have an identical chance of said success/failure. The first roll has a chance of failure, and that's it. Any future rolls have an identical chance of failure plus a prerequisite that all of the previous rolls succeeded, reducing their chances of being reached in the first place and so making them less likely to be the failed roll than the first roll.

[Gulchen]

getting back to why it is not correct to just multiply all probabilities:





Suppose the probabilities were

1) 10%
2) 90%

.


Getting ​ ​ ​ a failure in the earliest 10% of the time for turn 1 ​ and ​ a failure in the
earliest 90% of the time for turn 2 ​ ​ ​ is _a_ way of having the product be at most 9%.

However, getting ​ ​ ​ a failure in the earliest 90% of the time for turn 1 ​ and ​ a failure
in the earliest 10% of the time for turn 2 ​ ​ ​ also gives a product of at most 9%.

By the ​ ​ ​ ​ ​ ​ ​ P(A or B) ​ ​ ​ = ​ ​ ​ P(A) ​ + ​ P(B) ​ + ​ P(A and B) ​ ​ ​ ​ ​ ​ ​ formula, the probability of at least one of those
occurring is 17%, so there is at least a 17% chance that the product of the two probabilities will be at most 9%.

Furthermore, it's actually even higher, since there's also, for example, ​ failure in the earliest 30% of the time ​ for each of the two turns.




The effect is more extreme when there are more turns:



Let n be that number, and assume ​ ​ ​ ​ ​ ​ ​ n >= 1 ​ ​ ​ .


If failure was determined by a continuous process - so that every probability in the interval (0,1) was possible - then
there would be _at least_ a half chance of the product of the n probabilities being at most ​ ​ ​ ​ ​ ​ ​ 1 / 2^n ​ ​ ​ , ​ ​ ​ ​ ​ ​ ​ since

by symmetry, there is a half chance of the sum of the n probabilities being at most n/2
and
by ​ https://en.wikipedia.org/wiki/AM%E2%80%93GM_inequality , ​ for a fixed sum of
non-negative real numbers, the product is maximized when the numbers are all equal
so
there is at least a half chance of the product of the n probabilities being (1/2)^n

.


For discrete probabilities, one can use ​ ​ ​ n ​ * ​ something slightly greater than 1/2 ​ ​ ​ instead of n/2, to get the same lower
bound on the product of the probabilities being at most ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ 1 ​ ​ ​ / ​ ​ ​ ( something slightly greater than 2 ) ​ ^ ​ n ​ ​ ​ ​ ​ ​ ​ .

​

[davemenc]

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.

Thanks euklikd314!

I just wanted to emphasize that this is the key to understanding the probability of series of events: don't multiply the probability of success, multiply the probability of failure!

For example, if you want to know how likely 3 sixes in a row are on a dice, you would multiply 5/6^3 = 0.5797.

I just wanted to emphasize this (since euklikd314 kindly explained it) because I find it so useful in figuring these sorts of things out.

Dave