How odds and probabilities are calculated (Split from “Fire history”)

[mrelwood]

Just now, The Brahan Seer said:

I think this means you are getting frustrated so I will let you go. 

Oh come on! That’s no way to give up!

[Chriull]

If the chance for once cell to fail is x =1:10 million the chance for at least one out of 144 to fail can be calculated as follows:

Chance for one cell not to fail: 1-x

Chance for 144 cells not to fail: (1-x)^144

Chance for one or more out of 144 to fail:

1-(1-x)^144

[Zrex90]

Whoa, I didn't expect starting a math topic on statistics on my own. Maybe I should review them, but one thing is sure, if you don't get a ticket there is no chance to win the lottery, if you buy many of them there is more chance to have the winner. This can be applied to euc fires, but in that case you can mitigate the risk by checking that your battery is healthy. 

[Chriull]

Btw - for the interested

(1-x)^144 can be written (https://en.wikipedia.org/wiki/Binomial_theorem) as 1-144*x+10296*x*x...

As 1 / 10 000 000 gets much smaller once squared, cubed, ... these terms can be neglected as first approximation.

So (1-x)^144 can be approximated as 1-144*x, so the chance for one or more cell to fail out of a 144 cell compound can be approximated as 1-(1-144*x)=144*x.

So we are again with a chance of 144 : 10 000 000.

So

144 : 10 000 000 = 1 : 69444.4 is a very nice approximation.

The difference to reality ( 1 : 69444.9) of this approximation ( 1: 69444.4) is very small.

[The Brahan Seer]

35 minutes ago, Chriull said:

144 : 10 000 000 = 1 : 69444.4 is a very nice approximation.

So the approximation is about the odds of picking up a dodgy cell rather than the odds of it failing. If you have 144 cells that have a 1 in 10 million chance of being faulty you are saying thats equivalent to picking 1 cell in 69444 being faulty. 

So in the gun analogy the odds of losing is 1 in 6 but if you play again the odds of losing is now 1 in 3. Play again and its 1 in 2, play again 2 in 3, play again and its 5 in six and if you play it again 6/6= 1/1 means that you will definitely lose on the 6 round. But we all know that isn't how it works. Sorry I am obviously very stupid and missing something very fundamental but don't worry its not that important and I will have to just take your word for it.   

[mrelwood]

2 hours ago, The Brahan Seer said:

If you have 144 cells that have a 1 in 10 million chance of being faulty you are saying thats equivalent to picking 1 cell in 69444 being faulty.

Yes.

2 hours ago, The Brahan Seer said:

So in the gun analogy the odds of losing is 1 in 6 but if you play again the odds of losing is now 1 in 3.

Only if you don’t roll the barrel in between. That’s when you use two of the bullet slots out of 6.

If you do roll the barrel again, the chances to lose AT the second pull is again 1 in 6. (Chances if survival are 5/6 = 1 in 1.2 = 83%.) But the chances of you surviving TWO separate tries is 5/6 * 5/6 (surviving first AND second try). That’s 25/36 = 1 in 1.44 = 69%. Three tries is 5/6 * 5/6 * 5/6 = 125/216 = 1 in 1.73 = 58%.

2 hours ago, The Brahan Seer said:

Sorry I am obviously very stupid

You are not. You are just used to thinking in odds ratios, which in my opinion are harder to calculate.

You can find several probability calculation examples by just asking Google the probabilities you want to calculate.

[Mono]

On 5/29/2023 at 3:15 PM, The Brahan Seer said:

um a way of looking at it is... if you have a 6 chamber gun and you play russian roulette and you play once you have a 1 in six chance of losing. If you then played it again the odds don't change its still 1 in six chance of losing but could be written as 2 in 6  but this doesn't mean its now 1 in 3. Hope this helps.

I see, in this case the chance to get shot (at least once) is 1 - (5/6)^2 = 0.3055... rather than (1+1)/6 = 0.333..., which is about a 10% difference.

However, in the case of 144 in 10 million, computing for simplicity 144 / 10 million is just fine because the error compared to the true probability is tiny (because the denominator is large): 144 divided by 10 million equals 1:69444.44... where as the true probability for at least one cell catching fire (assuming these are independent events) is 1 - ((10,000,000-1) / 10,000,000)^144 = 1:69444.941, so the error is less than 0.001%.

Edited May 31, 2023 by Mono

[Mono]

On 5/30/2023 at 5:33 PM, mrelwood said:

You are not. You are just used to thinking in odds ratios, which in my opinion are harder to calculate.

Actually, depending on the situation, odds can be extremely useful and give much easier calculations.

[mrelwood]

1 minute ago, Mono said:

Actually, depending on the situation, odds can be extremely useful and give much easier calculations.

I’m sure you are right. But once one gets a mindset of either method, it might be difficult to switch. At least it is for me.

[Mono]

My 2 cents regarding probabilities and odds:

If p is a probability, then p / (1-p) is the according odds (not an odds ratio though which is a ratio between two odds and another extremely useful notion). The odds is (obviously) computed as a ratio or fraction (to my understanding of these two words as a nonnative speaker). There is a one-to-one back and forth mapping (bijection) between probability p and odds = p / (1-p) in that p = odds / (1 + odds), so we can use one or the other without losing any information, if we accept that infinity is a valid value (for computing or assigning the odds when p=1). While we often express the odds in two numbers, a nominator and a denominator, it is also perfectly legit to talk about the odds of 3.2 or 0.8.

If we say a chance of 1 in 10 million this suggests a probability (to me). However, it doesn't matter in this case, as the corresponding odds is (1 / 10 million) / (1 - 1 / 10 million) = 1 : (10 million - 1) which still equals to 1 : 10 million up to an error of about 0.00001% (1 / 10 million).

[woke rider]

How do we allow for the reality that the chance for euc battery failure decreases over time for the first few years then increases for failure after a few years?

[mrelwood]

10 hours ago, earthtwin said:

How do we allow for the reality that the chance for euc battery failure decreases over time for the first few years then increases for failure after a few years?

I believe how the batteries are handled is a much bigger factor in the risk department. Calculating the risk for either of those gets quite complicated.