# What is the probability distribution for this question? [duplicate]

I tried to find the probability distribution for this problem and then calculate the expectation of $n$. I suppose the total states of our problem would be $52^n$ and our desired states would be $52$ combination of n with permutation $52!$ for distinct cards and our probability function will be $p(n)=\frac{n!}{(n-52)!(52^n)}$ with $n = 52,53, ....$ But the summation over $p(n)$ is not $1$. I don't know what is wrong with this solution. Can anybody help? Thank you so much

## marked as duplicate by joriki probability StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 13 '16 at 13:12

• "I suppose the total states of our problem would be 52^n" ?? The probability space cannot depend on n the value of the random variable defined on it. – Did Dec 24 '14 at 11:16
• If you are only asked to calculate this expectation then I advice you not to use the distribution. – drhab Dec 24 '14 at 12:38
• Why does the space not depend on n? @ Did – user203021 Dec 24 '14 at 13:12
• How can we calculate expectation without distribution? @ drhab – user203021 Dec 24 '14 at 13:13
• As @drhab wrote, it's almost never a good idea to go for the distribution if you're only interested in the expectation; the expectation is usually far easier to determine than the distribution. I marked this as a duplicate of a question asking for the expectation; if you're still interest in the distribution, see this question as well as the links in the comments there. – joriki Jun 13 '16 at 13:14

Have a look at $p(53)$: $$p\left(53\right)=\frac{52}{52}\frac{\color{red}1}{\color{red}{52}}\frac{51}{52}\frac{50}{52}\cdots\frac{1}{52}+\frac{52}{52}\frac{51}{52}\frac{\color{red}2}{\color{red}{52}}\frac{50}{52}\cdots\frac{1}{52}+\cdots+\frac{52}{52}\frac{51}{52}\frac{50}{52}\cdots\frac{2}{52}\frac{\color{red}{51}}{\color{red}{52}}\frac{1}{52}$$
This is not the expression for $p(53)$ that you suggest.