# X=the number of cards that are drawn until the 1st ace is chosen. Find E[X] [duplicate]

Consider a deck of $52$ cards. Let $X$ be the number of cards that are drawn until the first ace is chosen (e.g., if the first two cards are not an ace and the third card is an ace then $X=2$). Find $E(X)$.
Hint: Let $I_j$ denote be the indicator variable that is $1$ if the $j$th non-ace is chosen before the first ace and $0$ otherwise for $j=1,…,48$.
## 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(); } ); }); }); May 8 '16 at 0:41
• To approach the problem, do you see a way to express $X$ in terms of the $I_j$? – snarfblaat May 8 '16 at 0:29
• One could find the probability that $X=k$ for the various possible $k$ fom $0$ to $48$, and then use that to find an expression for $E(X)$. Simplifying that expression to $\frac{48}{5}$ is quite painful, while indicator random variables give us the answer quickly. There are other approaches to calculating the expectation, but none as quick as the suggested method. – André Nicolas May 8 '16 at 0:34