# Odds on Streaks

Discussion in 'General' started by Mikeaber, Mar 29, 2005.

1. ### MikeaberWell-Known Member

Ken,

I once saw a chart on the odds of winning or loosing streaks of different lengths. This might prove beneficial to someone wanting to play a money management type game (betting strategy.)

My simulator will tell me what the length of the streaks are in a pre determined number of hands played with simulated shuffles of different numbers of decks of cards and using specified playing strategies. But it does not give any sort of analysis that would allow me to come up with the probabilities to use as a rule of thumb.

47% I believe is the percentage of winning 1 in a row (dealer on the average will win 53% of the hands played)

Would you figure that winning two in a row would be 1/2 that or 23.5%.
Three in a row would happen then 11.25% of the time and so on?

Formula

To figure the probability of a losing streak of n hands, given a probability p of losing a single hand (or roll, or coin-flip, or anything else), here's all you need:

p(Lose n in a row) = p^n

So, assuming p=0.53, n=10:
p(Lose 10 in a row) = 0.53^10 = 0.001749
That's once every (1/0.001749) tries, or roughly once every 572 tries.

3. ### rookie789Well-Known Member

^ ?

Ken,
For those of us that have not attended a math class in 20 years or so or are having memory math problems, can you explain the value Of ^, I believe it is an exponent but I am having a problem calculating the probabilities of winning 3 hands in a row instead of 10 in a row in your post with your " this is all you need to know" explanation.

4. ### MikeaberWell-Known Member

Howdee

Hi Rookie...
Yes, exponential. On the scientific view of your Microsoft Windows calculator, it is the x^y function. To get the inversion, use the 1/x function.

From a "winning" point of view using 47% as the frequency (the "Y" in the equation) you get

1 hand 1:2.12 or you'll win about 1 hand every 2 hands
2 hands 1:4.52 after winning 1 hand, you will win the next 1 our of every 4.5
3 hands 1:9.63
4 hands 1:20.49

A friend of mine gave me another list that could be off because of my assummed 47% / 53% split between dealer and any specific player:

2 ------ .2304 ------1 vs 3
3 ------ .1106 ----- 1 vs 8
4 ------ .0531 ----- 1 vs 18
5 ----- .0255 ----- 1 vs 38
6 ------ .0122 ----- 1 vs 81
7 ----- .0059 ------ 1 vs 168
8 ----- .0028 ------ 1 vs 356
9 ----- .0014 ------ 1 vs 713
10 ----- .0006 ------ 1 vs 1666
11 ----- .0003 ------ 1 vs 3332
12 ----- .0001 ------ 1 vs 9999

Keep in mind, these are for winning a given number of hands consecutively, not loosing as you would be interested in with Martingale negative progression. You loose more often than you win in Blackjack.

5. ### cheaney97Member

I'm not sure why you guys are using 47% as the probability for the player winning a hand and 53% as the probability for the dealer winning a hand. Isn't the correct probability for the player winning a hand 49.64% (Las Vegas 6-deck rules) and the probability for the dealer winning a hand 50.36%?

Here are the numbers

cheaney, It sounds like you are trying to back into the win/loss percentages by using the house edge. Blackjack is more complicated than that.

For a basic strategy player, if we count every hand with a net overall profit as a win, every hand with a breakeven outcome as a push, and every hand with a net loss as a loss, we get the following percentages for a typical 6-deck game:

Win: 43.3%
Push: 8.8%
Lose: 47.9%

If you look at just those numbers, the game looks very expensive. But, some of those wins are blackjacks paying 3:2, as well as multiple wins due to doubles and splits.

For purposes of figuring win or loss streaks, most players would choose to ignore pushes. If you eliminate pushes, that leaves 91.2% of hands. Of that, the ratio of wins to losses is 47.5% to 52.5%.

That's where the aforementioned 47% number comes from, though rounding could yield 48% here as well.

Here's another reason that the Martingale progression sucks, particularly for blackjack. This win percentage assumes you're willing to play correct basic strategy. What happens if you're in a big losing streak of your Martingale and your next bet is supposed to be \$500. If you get a string of 8s against a dealer ten followed up with 3s on each, are you willling to risk \$4000 on four hands plus doubles?

(The knowledgeable reader may note that splits in this example increase your percentage chance of winning, but the doubles actually reduce the win percentage but make more cash. Chill out, it's an example!)