What are the odds of winning 13 hands in a row?

[jaygruden]

Was playing a 6D game, PA rules, with 3 players/spots at table, and won 13 hands in a row at the back end of a shoe. The streak started in a neg count but the count was increasing and went above the pivot point for 4-5 hands and was able to get some decent money out there (not to max bet though). The count fell to negative again so I dropped back to table min. but kept winning. I had nice take on the run (+58 units) but I feel like the odds of winning 13 hands in a row must be astronomical, especially when it's bookend'd by negative counts on both sides. How do you calculate the odds, if possible? PS: I was not a math major

Back in my "negative expectation days" (AKA-the Darkside) of using BS with a positive betting progression this streak would have made a killing:grin: (but of course I would have given it back rather quickly at that time as well).
Thank you in advance and, may the force of +var be with you.

 

[Sucker]

You'll win about 43% of your hands. This number really doesn't change much at all whether the count is neutral, positive, or negative. (BJs and double downs make up most of your profit in rich decks)

So the chance of winning 13 in a row is about .43 to the 13th power.

 

[iCountNTrack]

You'll win about 43% of your hands. This number really doesn't change much at all whether the count is neutral, positive, or negative. (BJs and double downs make up most of your profit in rich decks)

So the chance of winning 13 in a row is about .43 to the 13th power.

I just want to add something to clarify any misconceptions with gambler's fallacy, 0.43^13 is the probability of winning13 hands in a row BEFORE any hand is played, meaning before we have any information, however AFTER we play the first 12 hands and win them, the probability of winning is 1 because we now have the information, so the probability on the 13th hand is 1^12*0.43 = 0.43

 

[aslan]

Was playing a 6D game, PA rules, with 3 players/spots at table, and won 13 hands in a row at the back end of a shoe. The streak started in a neg count but the count was increasing and went above the pivot point for 4-5 hands and was able to get some decent money out there (not to max bet though). The count fell to negative again so I dropped back to table min. but kept winning. I had nice take on the run (+58 units) but I feel like the odds of winning 13 hands in a row must be astronomical, especially when it's bookend'd by negative counts on both sides. How do you calculate the odds, if possible? PS: I was not a math major

Back in my "negative expectation days" (AKA-the Darkside) of using BS with a positive betting progression this streak would have made a killing:grin: (but of course I would have given it back rather quickly at that time as well).
Thank you in advance and, may the force of +var be with you.

After a plus run with max bet out I never lower my bet until I lose one hand. This is not Hoyle, but consider when the count has just turned negative-- [begin ploppy thinking here lol] ... at a slightly negative count, the count can do three things-- (1) it can drop (meaning that good cards will continue falling which is to your advantage), (2) the count can rise (meaning that the hand will favor you in EV percentage terms, although more bad cards have to fall to make it so, one of which you are likely to catch), and (3) the count will remain the same (meaning that you are at a slight disadvantage). Besides, it's bad to slavishly adjust your bet as the count drops because it is telltale of card counting. Better to just leave than engrave an invitation for them to back you off. Just a thought.

 

[jaygruden]

After a plus run with max bet out I never lower my bet until I lose one hand. This is not Hoyle, but consider when the count has just turned negative-- [begin ploppy thinking here] ... at a slightly negative count, the count can do three things-- (1) it can drop (meaning that good cards will continue falling which is to your advantage), (2) the count can rise (meaning that the hand will favor you in EV percentage terms, although more bad cards have to fall to make it so, one of which you are likely to catch), and (3) the count will remain the same (meaning that you are at a slight disadvantage). Besides, it's bad to slavishly adjust your bet as the count drops because it is telltale of card counting. Better to just leave than engrave an invitation for them to back you off. Just a thought.

Good points. Thnx

 

[aslan]

Good points. Thnx

I was hoping some heavyweight mathies would check in (not that you're not one) to check me on my logic.

 

[Sucker]

Besides, it's bad to slavishly adjust your bet as the count drops because it is telltale of card counting.

This trumps anything a math geek might have to say. I've been picked off more for LOWERING my bet than I have for RAISING it.