3 Numbers

[BigToe]

Hi All,
If a number 0-9 is chosen randomly three times, is the probability of only 2 odd numbers or only 2 even exactly 75%?
If that is the percentage, then if I am paid 51 cents for every $2 bet, is this in my favour?
Cheers Big Toe

 

[tkhunny]

As worded, your number will be repeated three times, exactly. p(exactly 2 odd) = p(exactly 2 even) = 0.

 

[BigToe]

A random number is chosen 3 times...each time it can be anything between 0-9. e.g 123, 945, 666, etc...

 

[BigToe]

To be clearer sorry, a random digit is chosen 3 times to give a three digit number...each time the random digit can be anything between 0-9. e.g 123, 945, 666, etc... In effect it would become any number between 000 and 999.

 

[Dr.Peterson]

Trying to make sense of this, I'll suppose you mean this:

Three integers from 0 through 9 are chosen randomly and independently (that is, repetition is allowed). What is the probability that exactly two of the numbers are odd, or exactly two of them are even?​

There are 10^3 = 1000 possible outcomes. If the event of interest does not occur, then either all three numbers are odd, or all three numbers are even; there are 5^3 = 125 ways for each of these, so the probability you ask for is (1000 - 2*125)/1000 = 0.75 as you say.

 

[HallsofIvy]

I presume that "a number 0-9 is chosen randomly three times" means that each of the numbers is equally likely and that the choices are independent. Since there are five even and five odd numbers the probability of "even" (E) is 1/2 and the probability of "odd" (O) is 1/2.

"Two even" is EEO or EOE or OEE so the probability is 3(1/2)(1/2)(1/2)= 3/8. By symmetry the probability of "two odd" is also 3/8. The probability of "two even or two odd" is 3/8+ 3/8= 3/4= 0.75.

(I initially misread the problem and thought that it was asking for "probability of two even" and "probability of two odd" separately.)

 

[pka]

If a number 0-9 is chosen randomly three times, is the probability of only 2 odd numbers or only 2 even exactly 75%?

Note that the probability of \(\displaystyle EEE\text{ OR }OOO\) is \(\displaystyle 2\left(\frac{1}{2}\right)^3=\frac{1}{4}\)
The compliment is \(\displaystyle 1-0.25=0.75\)

 

[BigToe]

Thank you all for these replies. I assumed the same.
Now, I make a $2 bet that either "two even" or "two odd" digits will be the outcome.
If I win, I get my $2 back plus 51 cents. If I lose, (3 digits the same) I lose my $2.
Over time am I a winner or loser.

 

[Dr.Peterson]

Your expected value is 0.75(0.51) + 0.25(-2.00) = -$0.1175. So in the long term, you lose.

This is because you are gaining a little more than 1/4 as much as you lose per bet, but you will win only 3 times as often as you lose. The odds are 3:1 in your favor (0.75:0.25), so to break even you would need to gain 1/3 as much on a bet as you can lose.

 

[BigToe]

Many thanks Dr.Peterson, and everyone else who helped.

Saved me from losing my money.