'Dead' cards

[snoopdawg]

So let's say I take a deck and pull out Ace King of Spades. I then deal a lot of boards, shuffling the deck after every time (say i do this a few mill/bill times, enough for accurate probability). I figure the chance of getting a royal flush. Then i repeat the experiment, but instead of dealing from the 50 remaining cards, I put 2 jokers ('dead' cards) in the deck. Now, every time a joker comes up (has to be flop/turn/river, doesn't matter if it's a burn) during the hand, I immediatly stop dealing and cancel that hand, not counting it as a pass or a fail.

The probability of getting a royal would be higher in the 2nd situation, right? I figure because it takes 3 more cards to make a royal, which leaves 2 board cards left that can be jokers. If it's not going to be a royal, then I have 5 chances for a joker.

Is this logic flawed?

[Tom1975]

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Is this logic flawed?


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Yes. The odds would be the same in both cases.

[repulse]

Yes, the logic is flawed as Tom noted.

Here's a simple explanation that might help you see past what's tripping you up:

With the non-jokered 50-card deck, you deal it out eleventy billion times and count X times when you made the royal and Y times when you did not make the royal. So the probability of making the royal is X/(X+Y), the number of trials in which you hit the royal over the total number of trials.

With the deck with jokers added, you deal it out eleventy billion times and count X times when you made the royal, Y times when you did not make the royal, and Z times when you hit a joker and did not count it as a pass or fail. So since the times when you hit a joker are ignored, the probability of hitting the royal is just X/(X+Y) right? Since the times you hit jokers aren't counted as trials, the total number of trials is still X+Y.

Another perspective: your reasoning as to why the royal should be higher with the jokers is that one of the board cards is more likely to be a joker in the cases when a royal is not made. But whenever a joker hits, there is no distinction as to whether or not a royal was made, as the joker hitting makes the trial not count!

So while it seems that the probability of having a joker as one of the board cards when you have no royal is greater than the probability of having a joker as one of the board cards when you have a royal, both of these probabilities are 0. In the context of the experiment, it is impossible to have either "a royal" or "no royal" when there's a joker on the board, as the joker discounts the trial immediately.

I wonder if any of that made sense.

[AaronBrown]

Are you dealing 7 cards or 5?

If it's 5, you'll get three royal flushes on average every 2,118,760 deals the first way. When you add the jokers, you'll get three royal flushes every 2,598,960 deals, but you'll throw out 480,200 of those deals because you got a joker. None of those hands will be royal flushes. You'll be left with the same ratio, three royal flushes per 2,118,760 deals that don't have a joker.

To see that it makes no difference, shuffle the deck the first way, then use a second deck to make an exact copy of the first. Insert the two jokers at random locations in the second deck. This is the same as the second way, the manner of shuffling is different, but it has the same random card order.

If both jokers come after the fifth card, it will make no difference, both deals will count, and both will either be royal flushes or not. If one or more jokers comes in the first five cards, you'll throw the second deal away. The chance of that happening is the same whether you have a royal flush on the first deck or not. So you'll throw out 18% of the deals at random the second way, but that won't affect the long-term probabilities.