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What is the probability of winning the max amoun
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<strong>Assumption</strong></p><p data-children-count="0">To rephrase my assumption in the answers heading, I'm assuming we can only play through the deck one, otherwise the trivial solution of a probability of 0 applies (you will pick up a red card when playing the XXXX forever).</p><p data-children-count="X"><strong>Solution</XXXXXX></p><p XXXX-children-XXXXX="0">The key XXXX is to note that XXXXXX XXX XXX independent. Hence, the XXXXXXX probability of XXXXXXXXXXXX XXXXXXX changes to:</p><p data-children-XXXXX="0"> X(A ∩ X) = X(B|X)*P(X)</p><p data-XXXXXXXX-XXXXX="X">instead XX</p><p data-children-count="0"> P(X∩ X) = X(B)*X(A).</p><p XXXX-children-count="0">For the XXXXX XXXXX (XXXX it X if you XXXX), XXXXXXX probability XX easily XXXXXXXXXX XX 26/52.</p><p XXXX-children-count="0">Winning the XXXXXX XXXXX provided XXX first XXX won (XXXX it B|X), we note that XX XXX XXXXXXXXXXX the XXXXXXXXXXX based XX XXXXXXX X less black XXXX in our deck. XXXX XXXXX, because of A, XXX XXXX XXX 25 XXXXX XXXXX XXX XXXXX 51 cards in total. XXXX gives us a XXXXXXXXXXX XX 25/51 for winning the XXXXXX XXXXX.</p><p data-children-count="X">By applying XXXX logic XXXXXXXXXXX, we XXX a pattern XXXXXXXX that on XXX i-th XXXXX, the XXXXXXXXXXX XX XXXXXXX XX (26-i)/(52-i). XX, the probability of winning XXX maximum XXXXXX is XXXXXX XXX probability of XXXXXXX XXX the XXXXXX which XX this XXXXXX chain XX 'product rules':</p><p XXXX-children-XXXXX="X">X(XXX won) = (26 x XX x 24 x ... x X x X x 1)/(XX x 51 x ... x XX x 28 x XX) </p><p XXXX-children-XXXXX="X"><strong>XXXX</strong></p><p XXXX-XXXXXXXX-count="0">If you are XXXXXXXXXX, this XXXXXXXXXXX simplifies XX X/(52 XXXXXXXXXXX 26). XXXX, XXXXX XX a much shorter argument XXXXX XX combinatorics XXXXXX in the details, XXXXX XX a nice XXXXXXXX for XXXXXXXX XXXXXXXXXX with combinatorics.</p><p data-XXXXXXXX-XXXXX="X"> </p>">
