# Chances of winning - multiple lanes

I assume this is stats 101 level of question.

If you have PC that is set to "randomly" pick a number between 1 and 100 each time a button is pushed, the chances of getting a 50 is 1 in 100, right?

So if you have two PCs doing the exact same thing, and you can press a button on either, but you can only pick one PC, what are the chances of getting a 50?

Also if you have one person at each PC, each pressing the respective button, what are each of their chances that one of them will get a 50?

• "Randomly pick a number" is not precise enough to allow this to be answered. If the computer picks a number from a discrete uniform distribution, then the odds of picking a 50 are indeed, 1 in 100. But if it picks a number from any continuous distribution, the odds of picking a 50 exactly are infinitesimal. If it picks from some other discrete distribution, the odds depend on the distribution. I suspect you meant the first thing. As for your other question: Do you mean exactly one person picking a 50? Or at least one person picking a 50? – Peter Flom - Reinstate Monica Aug 27 '13 at 19:16
• Thanks for the clarifying questions, and the time you've taken to expose my flawed premise. I do mean 1 in 100 integers. and I mean , 1 person, taking one attempt at pressing the button. – beauk Aug 27 '13 at 21:40
• The odds that any particular person will pick a 50 is 1 in 100. – Peter Flom - Reinstate Monica Aug 27 '13 at 21:59
• @AncientAnt All the questions seem identical. 1. 1/100. 2. Same because who cares if there is another PC? 3. Again, why would you care about anyone pressing button on some other PC? They are independent. – sashkello Aug 27 '13 at 23:28

If you have PC that is set to "randomly" pick a number between 1 and 100 each time a button is pushed, the chances of getting a 50 is 1 in 100, right?

Assuming by "randomly pick" you mean "with equal probability", then yes.

So if you have two PCs doing the exact same thing, and you can press a button on either, but you can only pick one PC, what are the chances of getting a 50?

Yes. Law of total probability ($\Pr(A)=\sum_i \Pr(A| B_i)\Pr(B_i)$). Let's call getting 50 a success ("S") and let $C_A$ and $C_B$ be the events that you picked computer A or B respectively.

$P(S)=P(S|C_A)\cdot P(C_A)+P(S|C_B)\cdot P(C_B)$

$\quad= \frac{1}{50}\cdot P(C_A)+\frac{1}{50}\cdot (1-P(C_A))$

$\quad= \frac{1}{50} (P(C_A)+ (1-P(C_A)))$

$\quad= \frac{1}{50}$

Also if you have one person at each PC, each pressing the respective button, what are each of their chances that one of them will get a 50?

Depends if you mean "exactly one" or "at least one". If the PC's work independently, then

a) $P(\text{At least one }S) = 1-P(\text{no } S) = 1-(\frac{49}{50})^2 = 0.0396$

(i.e. just under 2/50)

b) $P(\text{Exactly one }S) = P(\text{success on }A, \text{fail on }B) + P( \text{fail on }A, \text{success on }B)$
$\quad = \frac{1}{50}\cdot \frac{49}{50} + \frac{49}{50}\cdot \frac{1}{50}$
$\quad = 0.0392$

Alternate method:

$P(\text{Exactly one }S) = P(\text{At least one }S)-P(\text{Both }S)$
$\quad = 0.0396 - \frac{1}{50}\cdot \frac{1}{50}$
$\quad = 0.0392$