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 June 29th, 2017, 08:45 AM #1 Newbie   Joined: Jun 2017 From: Earth Posts: 2 Thanks: 0 Probability approaches 1? Say there are 10 cards face down, 1 is the ace, if I pick a card I have a 1 in 10 chance of finding the ace. Now assuming each time the 10 cards are shuffled so past knowledge has is no help in finding the ace of subsequent attempts, each attempt remains a 1 in 10 chance. So in my probably naive thought, if I play the game twice, I have a 2 in 10 chance of finding the ace at least once? But that can't be right, because if I play 10 times my chances of finding the ace at least once are not 1 in 1, I know, in theory at least I could play the game every second for the rest of my life and never find the ace. So how do probabilities stack? If I play the game 10 times, what is the probability I will find the ace at least once?
 June 29th, 2017, 08:55 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,940 Thanks: 2267 Math Focus: Mainly analysis and algebra The easiest approach is to determine the probability of not finding the Ace in $n$ attempts ($\frac1{10^n}$). Alternatively, in the case that $n=2$, the ways of finding an ace are AX, XA, AA (where X is a non-Ace). You have to sum those three probabilities to cover all cases. Thanks from CheeseAndBranston
June 29th, 2017, 09:29 AM   #3
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 Originally Posted by v8archie The easiest approach is to determine the probability of not finding the Ace in $n$ attempts ($\frac1{10^n}$).
OK, former computer programmer here, who struggles with mathematical notation

In my speak, I think you just said:

probability of not finding the ace = 1 / (10 * number of attempts)

Which yields 0.01 (for 10 attempts)

So my chances of finding the ace in 10 attempts is 0.99?

 June 29th, 2017, 09:42 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,940 Thanks: 2267 Math Focus: Mainly analysis and algebra No. Probability of not finding the ace is 1/(10 to the power $n$). Prob of not finding the A in one attempt is $\frac1{10}$ Prob of not finding the A in two attempts is $\frac1{10} \times \frac1{10}$ Prob of not finding the A in two attempts is $\frac1{10} \times \frac1{10} \times \frac1{10}$ etc. Thanks from CheeseAndBranston
 June 29th, 2017, 01:54 PM #5 Global Moderator   Joined: May 2007 Posts: 6,307 Thanks: 526 Probability of not finding the ace is 0.9. Therefore probability of finding the ace at least once in ten tries is $\displaystyle 1-.9^{10}=.65..$. Thanks from v8archie

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