Question 1146498
There are 16 cards that have a value of 10: the 10, J, Q, and K of all four suits.
<pr>
Chances both of the first two cards have a "10" value = 16/52 * 15/51 = 20/221
<pr>
Chances first card has a "10" value and second card does not have a "10" value = 16/52 * 36/51 = 48/221
<pr>
Chances first card does not have a "10" value and second card has a "10" value = 36/52 * 16/51 = 48/221
<pr>
Chances neither of the first two cards have a "10" value = 36/52 * 35/51 = 105/221
<pr>
Continuing...
<pr>
20/221 of the time, two "10" cards have been drawn, leaving 14 "10" cards remaining out of the 50 remaining cards.  Or, a 14/50 chance the third card will be a "10" card.
<pr>
96/221 of the time, one "10" card has been drawn, leaving 15 "10" cards remaining out of the 50 remaining cards.  Or, a 15/50 chance the third card will be a "10" card.  <i>(The 96/221 number comes from adding together the chance where a "10" card is drawn first and the chance where a "10" card is drawn second.)</i>
<pr>
105/221 of the time, zero "10" cards have been drawn, leaving all 16 "10" cards remaining out of the 50 remaining cards.  Or, a 16/50 chance the third card will be a "10" card.
<pr>
So...
<pr>
20/221 of the time, there is a 14/50 probability the third card will be a "10" card.
<pr>
96/221 of the time, there is a 15/50 probability the third card will be a "10" card.
<pr>
105/221 of the time, there is a 16/50 probability the third card will be a "10" card.
<pr>
The math is as follows:
<pr>
(20/221) * (14/50) = 28/1105
(96/221) * (15/50) = 144/1105
(105/221) * (16/50) = 168/1105
<pr>
Add together 28/1105 + 144/1105 + 168/1105 and you get 340/1105.  This fraction reduces to 4/13.
<pr>
So, there is a 4/13 probability the third card will be a "10" card.