## Stacked Deck (part 4)

Posted by DP on September 17, 2015 – 12:47 pm

Andy, Bob, Charlie, and Dave are ready to play another round of cards. This time it is Charlie’s turn to deal.

The cards are sorted and Charlie decides to only use the Red cards (Hearts & Diamonds).

Three [3] cards are dealt one-at-a-time to each of the four [4] players. Highest pair wins; Aces are low.

If no one wins, the cards are shuffled and re-dealt until there is a winner.

What are the odds of Charlie dealing himself a pair of 10’s AND **winning**?

September 18th, 2015 at 8:03 am

As I developed this problem, I realized that the solution and its calculations would be pretty tough. To give some magnitude – Even with only 26 of the 52 cards being used, there are 26! (read “26 factorial”) possible ways of the cards being arranged in the shuffled deck. That is over 403 Septillion (or 4.03×10^26) !!!

If I’m not wrong (and I probably am), the probabilities I see as necessary to complete the problem are:

– At least [1] Pair

– Pair of 10’s

– Pair of 10’s AND Ks and/or Qs and/or Js

– Charlie having the winning hand (I think I know this one)

I tried to also think of this from the cards’ perspective, and quickly saw the difficulty of these calcs as well. Something like:

“Card positions 4&8 or 4&12 or 8&12 are 10s, & card positions 1&5 are NOT Ks/Qs/Js, nor 1&9, nor 5&9, nor 2&6,…, nor 7&11.”

At any rate, I’m sure there is someone much smarter than me that can find the solution.

September 21st, 2015 at 12:59 am

ToM lives! I only found by accident that a new puzzle had been posted. So welcome back, DP! I hope you inspire other contributors to come out of the woodwork and get some momentum going with more puzzles.

As for this one, I haven’t even started to think about it. I’ll let you know when I do. I just thought I’d let you know that another ToM regular has also emerged from hibernation.

September 21st, 2015 at 7:58 am

Hi Wiz! Yes, that was my point; to get others inspired to turn the ol’ brains back to the “on” position. I would love to see others post problems, riddles, or just whatever to get me thinking again.

September 21st, 2015 at 10:19 am

Let me be the first in with a partial (and certainly wrong) answer:

The probability of Charlie getting both 10s with two cards is 2/26 * 1/25 = 2/6500. With three cards this can occur three ways, so we get 6/6500.

Likewise the chances of any player getting pairs of Ks, Qs or Js is 6/6500 in each case. Any of the other three players can win any of these three possibilities, so the probability of this is 9 times more likely than Charlie getting a pair of 10s, i.e. 54/6500.

So, in any one game, Charlie’s chances of winning when none of the others get higher pairs are (1 – 54/6500) * 6/6500 = 6446*6/6500*6500 = 38676/42250000 = 0.0009154…

But this is just for one deal. Presumably the question requires us to take account the likelihood that before Charlie gets to win there could be an earlier deal where one of the others wins with ANY pair, which could be from Ks (high) to As(low).

This is where it gets beyond me. Nice to be back but this one’s too hard, I think.

September 21st, 2015 at 11:06 am

Thanks Wiz. I’ll leave it to someone smarter than me to congratulate and/or correct you.

Doesn’t the 6494/6500 other ways of dealing (where Charlie doesn’t get both 10s) include other pairs showing up in someone’s hand (including Charlie’s) ?