April 11th, 2016 at 8:15:57 AM
permalink

And no, I can't take credit for it...

Somebody is counting an entire 8-deck shoe that, for whatever reason, has all 416 of its cards dealt.

The counter notices that, after every dealt card except for the last one, the number of red cards dealt up to that point was greater than the number of black cards dealt up to that point.

What is the probability of this happening?

Enclose answers in spoilers - if you don't know how:

[spoiler=The Spoiler Button Text]Superman is really Clark Kent[/spoiler]

Somebody is counting an entire 8-deck shoe that, for whatever reason, has all 416 of its cards dealt.

The counter notices that, after every dealt card except for the last one, the number of red cards dealt up to that point was greater than the number of black cards dealt up to that point.

What is the probability of this happening?

Enclose answers in spoilers - if you don't know how:

[spoiler=The Spoiler Button Text]Superman is really Clark Kent[/spoiler]

Batman is really Bruce Wayne

April 14th, 2016 at 12:28:05 PM
permalink

Quote:ThatDonGuyAnd no, I can't take credit for it...

Somebody is counting an entire 8-deck shoe that, for whatever reason, has all 416 of its cards dealt.

The counter notices that, after every dealt card except for the last one, the number of red cards dealt up to that point was greater than the number of black cards dealt up to that point.

What is the probability of this happening?

Enclose answers in spoilers - if you don't know how:

[spoiler=The Spoiler Button Text]Superman is really Clark Kent[/spoiler]Batman is really Bruce Wayne

Hey, I learned something new today! (cancellation brackets). Thanks, TDG! I knew there was a way, but it hadn't "clicked" before.

If the House lost every hand, they wouldn't deal the game.

April 14th, 2016 at 2:03:37 PM
permalink

Never mind. I misread the problem. This would be a tough one. I'd have to solve it by random simulation, which is no fun.

Last edited by: Wizard on Apr 14, 2016

It's not whether you win or lose; it's whether or not you had a good bet.

April 14th, 2016 at 2:35:00 PM
permalink

Black begins 50% of the time. Red leads all the way must be less than 50% of the time.Quote:WizardOr maybe I misunderstand the question being asked.

So much bullshit; so little time!

April 14th, 2016 at 2:42:22 PM
permalink

The first card must be red and the last card must be black. Red leads all the way must be less than 25%.

April 14th, 2016 at 3:03:41 PM
permalink

414 cards to go.

So much bullshit; so little time!

April 14th, 2016 at 4:00:38 PM
permalink

I got 8 decks going, like super fast.

So far red is winning, and black is winning.

And I'm losing.... eff me, again ;-)

So far red is winning, and black is winning.

And I'm losing.... eff me, again ;-)

Youuuuuu MIGHT be a 'rascal' if.......(nevermind ;-)...2F

April 14th, 2016 at 4:44:36 PM
permalink

Quote:WizardNever mind. I misread the problem. This would be a tough one. I'd have to solve it by random simulation, which is no fun.

First, we need to solve a slightly different problem:

What if there were only 415 cards - 208 red, and 207 black?

Each of the 415! permutations can be divided into one of four groups:

(a) First card red, and reds > blacks after every card

(b) First card red, but at some point, reds = blacks

(c) First card black, but at some point, reds = blacks

(d) First card black, and blacks > reds after every card

Note that (d) is impossible, as reds > blacks after card 415

For each permutation in (b), let N be the first card where blacks = reds; notice that it has to be the opposite color of the first card, as otherwise there would be some number < N where blacks = reds

Switch cards 1 and N; the first N cards are the same, so blacks = reds after card N, but now the first card has changed color, so if it was a type (b) permutation, it is now type (c), and vice versa.

Thus, for every type (b) permutation, there is a matching type (c).

Since all permutations beginning with black cards are type (c), the number of "bad" permutations = 2 x the number beginning with black cards, and the probability of a permutation being bad = 2 x the probability of it beginning with a black card; in this case, 2 x 207/415 = 414/415.

The probability that a permutation has reds > blacks after every card is 1 - 414/415 = 1/415.

Now, to solve the original problem.

If the last card is red, then there will be 208 black and 207 red cards after card 415.

If the last card is black, this reduces the problem to the 415-card problem.

The probability of the last card being black = 1/2, so the probability of reds > blacks after every card from 1 to 415 is 1/2 x 1/415 = 1/830.

April 14th, 2016 at 4:56:58 PM
permalink

Quote:ThatDonGuyThe Solution

That is clever. This ring a bell from years ago. I think I've worked out a similar problem some 20 years ago but it was expressed as a drunken walk. Something like:

"A drunk takes a step to the left with 50% probability and right the other 50%. He takes 415 steps. What is the probability he never returns to the same spot he started at?"

Of course, this is a different problem because there is no effect of removal.

It's not whether you win or lose; it's whether or not you had a good bet.

April 14th, 2016 at 5:11:09 PM
permalink

There is no going back to where you were. Damn what the math says. 2F

Youuuuuu MIGHT be a 'rascal' if.......(nevermind ;-)...2F