# Question on calculating a probability/expectation **URL:** \ **Category:** Factual Questions\ **Created:** [March 15, 2016, 2:59pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070 "2016-03-15T14:59:32Z")\ **Posts on this page:** 14\ **Page:** 1
**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\ **Post date:** [March 15, 2016, 2:59pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/1 "2016-03-15T14:59:32Z")
I am pretty sure my mathematics correct and it is so simple but there is something bugging me like I’m missing something. Suppose the expectation of an event E(x) = n. Now we add some variables. The first variable increases the probability of the event p(x) by 25%. The second variable decreases the probability of the event by 7%. What is the new E(x)? I get E(x) = n x 1.25 x 0.93 Is that correct? ---
**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\ **Post date:** [March 15, 2016, 3:11pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/2 "2016-03-15T15:11:01Z")
You said both “expectation” & “probability”. Are you using those as synonyms or do you have a more detailed understanding of the terms that you’d care to share? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [March 15, 2016, 3:21pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/3 "2016-03-15T15:21:48Z")
You also want to be clear about what is meant by “increase the probability by 25%”. Does that mean adding 25 percentage points to the probability (e.g., 40% to 65%), or does it mean multiplying the probability by 1.25 (e.g., 40% to 50%)? I’ve seen it used both ways, in different contexts. If possible, ask your source to clarify. ---
**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\ **Post date:** [March 15, 2016, 3:41pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/4 "2016-03-15T15:41:34Z")
> [@Chronos](#): > > I’ve seen it used both ways, in different contexts. If possible, ask your source to clarify. Umm … the source is me. Increasing the probability by 25% means multiplying by 1.25 so that p(x)=0.5 increased by 25% means the new p(x) is 0.675 I don’t get LSLguy’s question. Let’s say I have a 20% chance to win $500. p(x) is 0.2 and E(x) is $40. Pretty standard definitions. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [March 15, 2016, 4:01pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/5 "2016-03-15T16:01:55Z")
Oh, OK, then, you obviously know what you mean. If it’s important for anyone else to know, though, you might consider more careful phrasing. ---
**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\ **Post date:** [March 15, 2016, 4:14pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/6 "2016-03-15T16:14:11Z")
If you can write expectation as as a function of the probability E(x) = f(p(x)) you only get what you want if f is linear map. If f(p(x)) = c p(x) you’re golden. If f(p(x)) = p(x) + 1 you’re not. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [March 15, 2016, 4:41pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/7 "2016-03-15T16:41:30Z")
To be fair, the expectation being proportional to the probability only requires that the payoff be unaffected by the changes to the probability, which one would expect to be a fairly common situation. ---
**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\ **Post date:** [March 15, 2016, 5:00pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/8 "2016-03-15T17:00:18Z")
It is indeed a fairly common situation and most likely the only situation the OP is considering. I only included my answer for completeness. ---
**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\ **Post date:** [March 15, 2016, 5:03pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/9 "2016-03-15T17:03:26Z")
And that’s the difficulty I am having Chronos. For example (and greatly simplified) p(insurance claim) = 0.1 Average value of repair/coverage = $2500 E(claim) = $250 But I have a fast expensive car so I have a 30% larger chance of filing a claim. And the value of my car is 50% more than the average. So is MY E(claim) = 250 x 1.3 x 1.5 = $487.50 ---
**Author:** ![Buck\_Godot](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/buck_godot/32/6573_2.png) [@Buck\_Godot](https://boards.straightdope.com/u/Buck_Godot)\ **Post date:** [March 15, 2016, 5:10pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/10 "2016-03-15T17:10:45Z")
> [@Saint\_Cad](#): > > And that’s the difficulty I am having Chronos. For example (and greatly simplified) > > p(insurance claim) = 0.1 > Average value of repair/coverage = $2500 > E(claim) = $250 > > But I have a fast expensive car so I have a 30% larger chance of filing a claim. And the value of my car is 50% more than the average. So is MY E(claim) = 250 x 1.3 x 1.5 = $487.50 Yes given your formulation, that would seem correct. To write it out, before your expectation was E = p_V =0.1_2500=250 now E2=(p_1.3)_(V_1.5)=p_V_1.3_1.5=E_1.3_1.5. ---
**Author:** ![CookingWithGas](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cookingwithgas/32/485_2.png) [@CookingWithGas](https://boards.straightdope.com/u/CookingWithGas)\ **Post date:** [March 15, 2016, 6:03pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/11 "2016-03-15T18:03:46Z")
> [@Saint\_Cad](#): > > And that’s the difficulty I am having Chronos. For example (and greatly simplified) > > p(insurance claim) = 0.1 > Average value of repair/coverage = $2500 > E(claim) = $250 > > But I have a fast expensive car so I have a 30% larger chance of filing a claim. And the value of my car is 50% more than the average. So is MY E(claim) = 250 x 1.3 x 1.5 = $487.50 This is different than your original question. This question has one factor that affects probability and another factor that affects value. So they would multiplied together as you show here. But your OP asks about two factors that both affect probability. If they are independent then you should consider them separately: E(x) = n x (0.25 - 0.07) For example, if the probability of getting a heart attack if I am 50 is 0.08 and the chance of being hit by lightning if I live in Virginia is 0.001, then if I am a 50-year-old living in Virginia the chance of either having a heart attack or being hit by lightning is 0.081, the sum of the probabilities. It is not clear from the formulation of your original question what calculation should be applied. ---
**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\ **Post date:** [March 15, 2016, 6:14pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/12 "2016-03-15T18:14:46Z")
> [@Saint\_Cad](#): > > And that’s the difficulty I am having Chronos. For example (and greatly simplified) > > p(insurance claim) = 0.1 > Average value of repair/coverage = $2500 > E(claim) = $250 > > But I have a fast expensive car so I have a 30% larger chance of filing a claim. And the value of my car is 50% more than the average. So is MY E(claim) = 250 x 1.3 x 1.5 = $487.50 That’s all good logic. Which is why P(higher premiums for such a car) = 1.0 🙂 FYI for various folks above: a 50/50 raffle is probably the common situation where E is not a linear function of P. Buying two tickets doubles your chances, but also changes the prize amount. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [March 15, 2016, 9:39pm UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/13 "2016-03-15T21:39:37Z")
Except buying another ticket doesn’t double your chances, either, since your chance also depends on the total number of tickets sold. In fact, your expectation in a 50-50 raffle is directly proportional to the number of tickets you buy. ---
**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\ **Post date:** [March 16, 2016, 1:40am UTC](https://boards.straightdope.com/t/question-on-calculating-a-probability-expectation/749070/14 "2016-03-16T01:40:07Z")
Hmmmmm. D’oh! :smack: