by Karrsen B
Last Updated May 23, 2020 01:20 AM

Lets say you're trying to draw n cards, each higher than the last. How likely are you to succeed in consecutively drawing higher for a given starting number & number of iterations?

Somebody said to add my thoughts to the question! We've had a lot (my brother and i are trying to solve it!) so wait just a few minutes while I write it all out!

So first we tried by thinking about some easy cases. Lets say you only wanted 3 consecutive higher draws. If you drew a king or a queen you're done right off the bat, the odds of that are 8/52. If you drew a Jack, you have to get the sequence of JQK to succeed. That's 4/52* 4/51*4/50. From this we can't yet establish too much, but we found a general way to answer this problem for three draws. If you start with a Jack, there is one way, a 10 in three ways, a 9 in six ways, an 8 in 10 way... each drop in card number adds one more way than the last drop did (J-10 added two, 10-9 added three etc). So the equation is 1(4/52 * 4/51 * 4/50) + 3(4/52 * 4/51 * 4/50)... + 66(4/52 * 4/51 * 4/50). (<- if we aren't mistaken)

Any ideas of where to go from here to apply this to higher numbers of draws?

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