# Easy probability question. **URL:** \ **Category:** Factual Questions\ **Created:** [May 22, 2007, 11:17am UTC](https://boards.straightdope.com/t/easy-probability-question/405055 "2007-05-22T11:17:37Z")\ **Posts on this page:** 15\ **Page:** 1
**Author:** ![Revenant\_Threshold](https://avatars.discourse-cdn.com/v4/letter/r/f1d935/32.png) [@Revenant\_Threshold](https://boards.straightdope.com/u/Revenant_Threshold)\ **Post date:** [May 22, 2007, 11:17am UTC](https://boards.straightdope.com/t/easy-probability-question/405055/1 "2007-05-22T11:17:37Z")
My brain has decided to conk out on me, and i’m finding myself with a mental block on probabilities. I’m revising for my Parapsychology exam, specifically the use of Zener cards. If there are five types of cards, with five of each in a deck, randomly shuffled, and each card is called out (25 runs); what percentage of the runs will give correct answers if purely going on chance? ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [May 22, 2007, 11:43am UTC](https://boards.straightdope.com/t/easy-probability-question/405055/2 "2007-05-22T11:43:16Z")
20%, surely? Each guess is a 1 in 5 shot, assuming that the guesser isn’t told what any previous card was. (there are exams in parapsychology now?) ---
**Author:** ![Revenant\_Threshold](https://avatars.discourse-cdn.com/v4/letter/r/f1d935/32.png) [@Revenant\_Threshold](https://boards.straightdope.com/u/Revenant_Threshold)\ **Post date:** [May 22, 2007, 11:46am UTC](https://boards.straightdope.com/t/easy-probability-question/405055/3 "2007-05-22T11:46:31Z")
Essay-based exams. “Critically examine…”, “Compare and Constrast…”, that kind of thing. Unless you were questioning the subject rather than the examining method… in which case, it is just a voluntary module. 😉 And thanks. My brain’s response to it was pretty much “fleh”, even though I knew it was easy. ---
**Author:** ![Dunderman](https://avatars.discourse-cdn.com/v4/letter/d/ecb155/32.png) [@Dunderman](https://boards.straightdope.com/u/Dunderman)\ **Post date:** [May 22, 2007, 12:12pm UTC](https://boards.straightdope.com/t/easy-probability-question/405055/4 "2007-05-22T12:12:09Z")
**Usram** ’s answer is correct if and only if the revealed card is shuffled back into the deck each time. If it isn’t, the probability depends on the examinee, as he can maximize his chances by guessing the type of card that has the largest number left in the deck. Run 1: The examinee says “star” and has a 20% chance of getting it right. One card is revealed and removed from the deck. Run 2: The examinee says one of the symbols that wasn’t on the revealed card and has a 20.83% chance of getting it right (5/24). One card is revealed and removed from the deck. Run 3: The examinee says one of the symbols that was on neither of the two revealed cards and has a 21.73% chance of getting it right (5/23). One card is revealed and removed from the deck, And so on. ---
**Author:** ![Revenant\_Threshold](https://avatars.discourse-cdn.com/v4/letter/r/f1d935/32.png) [@Revenant\_Threshold](https://boards.straightdope.com/u/Revenant_Threshold)\ **Post date:** [May 22, 2007, 12:24pm UTC](https://boards.straightdope.com/t/easy-probability-question/405055/5 "2007-05-22T12:24:09Z")
Card-counting. Yeah, I should have mentioned that in the OP as well. I have considered it, but thanks for the reminder anyway - every little helps. 🙂 ---
**Author:** ![Dunderman](https://avatars.discourse-cdn.com/v4/letter/d/ecb155/32.png) [@Dunderman](https://boards.straightdope.com/u/Dunderman)\ **Post date:** [May 22, 2007, 12:29pm UTC](https://boards.straightdope.com/t/easy-probability-question/405055/6 "2007-05-22T12:29:23Z")
Even without conscious card-counting, the chances are affected by what the examinee says. Without card replacement, there is no one answer to the question. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [May 22, 2007, 12:35pm UTC](https://boards.straightdope.com/t/easy-probability-question/405055/7 "2007-05-22T12:35:17Z")
> [@](#): > > Usram’s answer is correct if and only if the revealed card is shuffled back into the deck each time Surely if the examinee isn’t told whether (s)he guessed right, (s)he has no reason to favour any symbol for the following cards, regardless of whether the cards are replaced or not? ---
**Author:** ![Dunderman](https://avatars.discourse-cdn.com/v4/letter/d/ecb155/32.png) [@Dunderman](https://boards.straightdope.com/u/Dunderman)\ **Post date:** [May 22, 2007, 12:41pm UTC](https://boards.straightdope.com/t/easy-probability-question/405055/8 "2007-05-22T12:41:09Z")
[QUOTE=Usram] Surely if the examinee isn’t told whether (s)he guessed right, (s)he has no reason to favour any symbol for the following cards, regardless of whether the cards are replaced or not? [/QUOTE] You’re right; I didn’t even consider that. If the examinee isn’t shown the card, then your answer is absolutely correct. ---
**Author:** ![Dunderman](https://avatars.discourse-cdn.com/v4/letter/d/ecb155/32.png) [@Dunderman](https://boards.straightdope.com/u/Dunderman)\ **Post date:** [May 22, 2007, 12:46pm UTC](https://boards.straightdope.com/t/easy-probability-question/405055/9 "2007-05-22T12:46:36Z")
[QUOTE=Priceguy] You’re right; I didn’t even consider that. If the examinee isn’t shown the card, then your answer is absolutely correct. [/QUOTE] Addition: Also provided the examinee isn’t an idiot and in fact says each symbol five times. ---
**Author:** ![Quercus](https://avatars.discourse-cdn.com/v4/letter/q/7ab992/32.png) [@Quercus](https://boards.straightdope.com/u/Quercus)\ **Post date:** [May 22, 2007, 12:53pm UTC](https://boards.straightdope.com/t/easy-probability-question/405055/10 "2007-05-22T12:53:58Z")
[QUOTE=Usram] Surely if the examinee isn’t told whether (s)he guessed right, (s)he has no reason to favour any symbol for the following cards, regardless of whether the cards are replaced or not? [/QUOTE] I’d phrase it as “if the examinee has no way of knowing whether (s)he guessed right”. Since we’re discussing parapsychology, it is a real distinction – there’s a long history of subjects picking up unconscious cues from testers. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [May 22, 2007, 5:48pm UTC](https://boards.straightdope.com/t/easy-probability-question/405055/11 "2007-05-22T17:48:54Z")
> [@](#): > > Addition: Also provided the examinee isn’t an idiot and in fact says each symbol five times. No, it’s still aa expectation of 20 percent correct. For example, if the examinee always answers the same symbol 25 times, he’ll get exactly 20 percent correct. (Well, assuming he’s answering with one of the symbols in the set. If he always says “fish” and that’s not one of the symbols, he’ll get zero.) ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [May 22, 2007, 6:12pm UTC](https://boards.straightdope.com/t/easy-probability-question/405055/12 "2007-05-22T18:12:17Z")
OTOH, if you wished to demonstrate ESP, and that were defined as scoring say 25% or more, you’d maximise your chances by guessing 5 of each symbol. ---
**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:** [May 22, 2007, 6:16pm UTC](https://boards.straightdope.com/t/easy-probability-question/405055/13 "2007-05-22T18:16:57Z")
There are other possible avenues for cheating. For instance, the pattern on the back of standard Zenner cards is not symmetric. If the deck were shuffled such that all the squiggles, circles, and squares were “right side up”, and the triangles and blanks were all “upside-down”, and if the subject knew this, the subject would score considerably better than 20%. ---
**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:** [May 22, 2007, 6:28pm UTC](https://boards.straightdope.com/t/easy-probability-question/405055/14 "2007-05-22T18:28:29Z")
There is a 20% chance of picking a \*single \* card correctly based on random chance alone. However, the issue when applied to such experiments with multiple trials is a little more complex. The answer is not a single number. You can randomly select 25 cards and have total result anywhere from 0 to 25 right. The question requires providing a confidence interval. For example (and I’m making these numbers up), there is a 95% confidence that a person selecting in the interval 22-24 out of 25 cards correctly is not making random choices but is demonstrating some other ability. When a person can pick only 5 out of 25 correctly, the confidence level drops dramatically (probably gets close to 0). The probability of guessing by pure random chance any number of cards from 0 to 25 can be calculated. For example, the chances of randomly guessing 25 out of 25 is 0.2^25, or 0.0000000000000000033554432, or something like 3 out of a quintillion. So if a person performed that well you could be very confident, but not 100% confident, that the performance is not due to random chance. IANAStatistician but that is the general idea. ---
**Author:** ![Dunderman](https://avatars.discourse-cdn.com/v4/letter/d/ecb155/32.png) [@Dunderman](https://boards.straightdope.com/u/Dunderman)\ **Post date:** [May 22, 2007, 6:28pm UTC](https://boards.straightdope.com/t/easy-probability-question/405055/15 "2007-05-22T18:28:53Z")
[QUOTE=ZenBeam] No, it’s still aa expectation of 20 percent correct. For example, if the examinee always answers the same symbol 25 times, he’ll get exactly 20 percent correct. [/QUOTE] Yeah, I realized that after posting. Call it a brainfart.