Probability wrong or randi talking about a different scenario/game/problem

chance of 1 in 5 to guess exactly one card out of the two is simply wrong. for the reasoning see my comments in the comments section of the original video...

say the symbols are 1,2,3,4,5: then if order does not matter, there are 10 combis: 12, 13, 14, 15, 23, 24, 25, 34, 35, 45, to get both right is 1/10 chance then, now each symbol figures in exactly 4 of the 10 combis, so you have 6/10 chance of getting the first wrong and then a 2/4 chance of getting the second symbol right (or wrong) (because two rights and two wrongs are still in the other four symbols left), so the chance of getting first wrong and the second right (or none right) is 6/10*2/4 = 3/10, the probability of first being right and then wrong is 4/10*3/4 (only one more correct symbol is among the four symbols left) = 3/10. We add and have 6/10=3/5.

if order is important, we have 12, 13, 14, 15, 21, 23, 24, 25, 31, 32, 34, 35, 41, 42, 43, 45, 51, 52, 53, 54; so as randi correctly says, to get both right, you have 1/20 chance, now you see that for each place you start with you have a 4/20 chance that a given symbol is in that place, so you have a chance of 16/20 to get the first guess wrong.
the chance of getting just one out of the 2 cards right, is 3/10. i can use symmetry reason here: consider first guessing right and then wrong: 4/20*3/4=3/20 (there is no controversy here as the first symbol was right, and so exactly 1 of other four must be in), now because of symm. getting the 1st wrong and 2nd right must have same chance, we add and have 3/10.
now, for fun, I give the other way around (order matters, first guessing wrong, then right) with conditional prob. now if you first guessed wrong (with p=16/20) you have 1/4 chance to get the 2nd place right, but only in case the symbol you were wrong with for the 1st place is not in 2nd place, so we must still multiply 16/20*1/4 by the prob(symbol 1st guessed not at 2nd place given it was not at the 1st place)=p(symbol nowhere)/p(symbol not at 1st place)=(6/10)/(16/20)=3/4.
16/20*1/4*3/4=3/20.

interestingly, as a tip if you have to play a game like this, and order is important, you are better off not excluding the possibility of repeating a symbol in your two guesses, if you choose them randomly while not ruling out your first guess, you increase your chance for having at least one right by 1%, and for having exactly one right by 2%!

even although you know that 11, 22, 33, 44, 55 cannot be the right answer.... :-) this has almost something of a pseudoparadox to it.... :-))

Category:

Education

Tags:

• probability
• theory
• order
• permutations
• guess
• card
• lottery
• james
• randi
• supernatural
• conjuring
• conman

[NFG49]

lets make this a little simpler, if I pick a number from 1 - 10 and ask you to guess it the odds are 1 in 10 if i'm not mistaken, but if I tell you to pick a number from 1 - 10 and there are 2 possible solutions, then I would believe the answer is 2 in 10, or 1 in 5, but what is the correct answer?

[Jackies1979]

the point simply is that the number of possibilities in the "pick a/one number out of 1-10" is 10; but the number of possibilities in the case where two are picked is just not ten, but 45 if order does not matter, and 90 if order matters...

but then still the odds are not 2/45 or 2/90 because the frequencies of the numbers in these combinations are a little different...

[Jackies1979]

@Jackies1979 and such... to calculate the probability of a simple problem with even writing down all the options is actually not what math seems to be all about to me.... but if i took a little bit of time, of which i have none right away, even i could probably arrive at a very general answer, but i gave every obvious detail to find the answer in the very special case i discussed...! Mercy upon me...!

[Jackies1979]

@Jackies1979 then you wanna look at the distribution of each symbol in these combinations, for just two symbols picked, this is relatively easy to see, for more symbols picked the task will probably be more "involuted" but can presumably be reduced to the simple case of just two symbols... but I seem to fail to see through the issue enough to give you a very general formula for more than two symbols picked... in this way I am not good at math, this is what math is all about, becoming general...

[Jackies1979]

@NFG49 i didn't use a general formula but actually wrote down all the possibilities in the description box, but in order to give you the answer for your other question I somehow had to generalize the issue in my head also deriving a more general formula for this problem.... if you have X symbols and wish to pick Y symbols out of them (without repeating) you have X!/(X-Y)! combinations for order matters and X!/((X-Y)!*2) combinations for order not important...

[Jackies1979]

@NFG49 Simpler? No, in that case (let's say it does not matter what number comes first; and numbers do not repeat) there are 10*9/2=45 combinations.

now each number figures in exactly 9 of these 45 combinations. now the probability for getting exactly one of the two "possible solutions" right is the probability of getting the first solution right and the second wrong + that of getting the first wrong and the second right =

9/45 * 8/9 + 36/45 * 2/9 = 16/45 = 0,355...!!

[NFG49]

as I said, i'm not too good at math, and I'll admit I do not know the formula in which is needed to calculate probability, but the way I rationalised it was 2 cards are right, and there are 5 choices, so I thought that the chance of him guessing the right card would be 2 in 5, I now realise that this thinking is wrong, and would like to know the formula that you used to formulate your answer, so that I could be more accurate next time, I appreciate your quick response to my previous post

[Jackies1979]

yes, i fail, i cannot make you see, nfg, why your answer is wrong, it most certainly is, so if you cannot tell me why mine should be incorrect, do a series of 100-500 runs of the game with a partner and you will see that the frequency of getting just one right is closer to 6/10 (for order does not matter) and 3/10 (for order matters) than to your 40 %; for order not important 100 runs will probably suffice for you to see the difference...

this is also how i convinced someone in the goat case

[Jackies1979]

@Jackies1979 Now how can I expose the incorrectness of nfg48's reasoning, this was a nice try but wrong...

[Jackies1979]

@Jackies1979 sorry, it is, if I just calculated correctly, it is even higher, almost 4 in a thousand... so my intuition pump was kind of totally wrong...