Given that \( w, x, y, z \) take on values \( 0 \) and \( 1 \) with equal probability, what is the probability that \( w+x+y+z \) is odd?

Which of the following arguments is correct?

Furthermore, can you generalize this result?

**Argument 1:**

If all 4 numbers are even, the sum is even.

If 3 numbers are even, the sum is odd.

If 2 numbers are even, the sum is even.

If 1 number is even, the sum is odd.

If 0 numbers are even, the sum is even.

In 2 of the 5 cases, the sum is odd. Hence, \( w + x +y + z \) is odd with probability \( \frac{ 2}{5} \).

**Argument 2:**

\( w + x + y + z \) is either odd or even.

In 1 of the 2 cases, the sum is odd. Hence, \( w + x +y + z \) is odd with probability \( \frac{ 1}{2} \).

**Argument 3:**

By listing out all the \( 2^4 \) cases, we find that there are \( { 4 \choose 0} + { 4 \choose 2} + { 4 \choose 4 } = 8 \) cases with an even sum, and \( { 4 \choose 1} + {4 \choose 3} = 8\) cases with an odd sum.

Hence, \( w + x +y + z \) is odd with probability \( \frac{ 1}{2} \).

Context: In a recent problem, Argument 1 was given, and many people agreed with it.

## Comments

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TopNewestThe problem with the first argument is that the probabilities of each of the cases is different. In reality, the first case has one way, the second has 4 ways, the third has 6 ways, the fourth has 4 ways, and the last one has only 1 way. Thus the actual probability is \(\dfrac{4+4}{1+4+6+4+1}=\dfrac{1}{2}\). – Daniel Liu · 3 years ago

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I disagree with Argument 1:

In case 1 there is only 1 outcome to make an even sum.

In case 2 there are 4 outcomes that make an odd sum

In case 3 there are 6 outcomes that make an even sum

In case 4 there are 4 outcomes that make an odd sum

In case 5 there is only 1 outcome to make an even sum.

Thus out of the 16 equally likely outcomes 8 will yield an odd sum and 8 will yield an even sum.

Therefore the probability that w + x + y + z is odd is 1/2. – Guiseppi Butel · 3 years ago

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Reminds me of Bertrand paradox, where three ways of choosing a seemingly the same uniformly random chord on a circle give different probability distributions (which means they are different). – Ivan Koswara · 3 years ago

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– Snehal Shekatkar · 2 years, 10 months ago

I am not very sure if these two are similar indeed. The problem stated above seems to have objective solution and other two are simply counting errors.Log in to reply

– Ivan Koswara · 2 years, 10 months ago

They are not similar, yes. It's just the fact that there are three solutions arriving at different answers immediately reminded me to that.Log in to reply

Well, this is how i solve this problem.

Assume that we choose w, x, y first.

w + x + y = A.

Doesn't matter whether A is odd or even. z will be the one that decide whether the sum of w, x, y, z (assume w +x+y+z= B) is odd or even

=> With that the possibility for B to be odd = 50%

=> Argument 1: Wrong

Argument 2: Correct (but lack at explaining)

Argument 3: Correct – Alight Be · 3 years ago

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See @SAMARTH M.O. Comment. – Calvin Lin Staff · 3 years ago

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Argument 3 is arguably the right one. Arguments 1 and 2 assume that their mentioned respective events are equally likely. – Samarth M.O. · 3 years ago

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Argument 1 is wrong because it assumes they all happen with equal probability, which they don't. – Frodo Baggins · 3 years ago

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Argument 1 is incorrect because each case is not equally likely. Argument 2 gives the correct answer but for the wrong reason. Consider the probability of drawing the Ace of spades from a deck of cards. The card drawn is either the ace or it is not. One out of two cases is favourable hence 1/2. (?) Argument 3 would be correct if it was asserted that each case was equally probable. Which of course it is. J – John Pickering · 2 years, 1 month ago

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– Calvin Lin Staff · 2 years, 1 month ago

Great :)Log in to reply

Total number of cases = 2

22*2 Even cases are If number of ONEs is 2 or 4, So Favourable number of cases : 4P2 + 4P4 = 6 + 1 = 7 So Probability for odd is (16 - 7)/16 = 9/16 – Manish Arya · 2 years, 2 months agoLog in to reply

When dealing with even cases, you forgot that 0 is an even number. – Calvin Lin Staff · 2 years, 2 months ago

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The first arguement is wrong as each case doesnot have equal probability .. The second solution (though gives right answer ) is an insufficient one. ARGUEMENT 3is descent – Devansh Shringi · 2 years, 2 months ago

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– Calvin Lin Staff · 2 years, 2 months ago

The second solution doesn't give a correct argument. The point isn't to "get to the correct answer". The point is to "substantiate your argument".Log in to reply

1 – Shashikanth Chavvan · 2 years, 11 months ago

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An even easier argument which is easily adapter to be generalisation for the number of: picking the first 3 numbers randomly, you'll find either an even or an uneven number. The last term will change it to either even or uneven, with an equal chance of either.

This will generalise to any number of variables easily, and also adapt to the odds of the result being k (mod n) if your variables can have values 1, 2, ... , n.

One problem it can't help with is: if the variables are binary, what are the odds of the result being a multiple of k (for a k greater than 2). – Bart Nikkelen · 2 years, 11 months ago

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We have 2^{4} case. Odd sum is 2.\frac{4!}{3!}=2^{3} case. There for sum is odd with probability \frac{1}{2} – Phúc Lê · 2 years, 11 months ago

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let a(n) = P(Bin(n,0.5) is even), then a(n+1) = 0.5

a(n) + (1-0.5)(1-a(n)) = 0.5 – Mattapalli Ram · 3 years agoLog in to reply

Obviously universal space= 2^4 Sample space= 4C0+ 4C2+ 4C4= 8 So probability= 8/16 =1/2 – Shubhashish Banerjee · 3 years ago

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2 – Prince Cirujano · 3 years ago

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1\2 – Khyati Srivastava · 3 years ago

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Argument 3 is perfect. – Venkateswara Reddy Kunduru · 3 years ago

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.5 – Manish Arya · 3 years ago

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argument 3 feels more logical – Avid Santiko Adji · 3 years ago

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The last argument is correct. In general for n numbers the probability is \(\frac {{2}^{n-1}}{{2}^n} = \frac {1}{2}\) – Adrian Neacșu · 3 years ago

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Was this problem perhaps my problem All That Glitters Is Gold? – Sharky Kesa · 3 years ago

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The problem that I was referencing had a similar title to my note. It has since been deleted. – Calvin Lin Staff · 3 years ago

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– Finn Hulse · 3 years ago

Dude Calvin, I just found this. You didn't even tag me in it! But thanks, it's still awesome! :DLog in to reply

1 – Yash Aggarwal · 3 years ago

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now since every single variable counts and is equally important we should be going with the the third one .. ARGUMENT 3 is correct .... nearly similar way to solve is .. there are two possibilities:-

1.) - three variables are 1

2.)- only one variable is 1 .. both ways we get odd

hence- 4C3[ any three numbers]

(1/2)^4[the probability of getting 1 is 1/2 and zero is also the same AND we need to get 1 in three cases AS WELL AS 0 in the other case] => 4C3 * (1/2)^4 +4C1[only one variable](1/2)^4 AND HENCE4C3 * (1/2)^4 + 4C1 * (1/2)^4 = 1/2 .. – Rohit Tamidapati · 3 years ago

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The issue with the first argument is that of frequency distribution – Hemant Yadav · 3 years ago

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10 – Abdul Gani · 3 years ago

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