Loss Streak Probabilities

I have a question about consecutive loss sequences. This question doesn’t apply to any game in particular. Ignoring any wins that occur after x losses, is the next sequence of losses more likely to be greater than or less than x? “x” can range from 1 or 2 losses to 13 or 26 losses in a row.

I suppose for an example, let’s say I lose 5 bets on red in roulette or 5 hands in blackjack in a row. Barring any wins that occur immediately after, is the immediate next sequence of losses more likely to be greater than or less than 5 losses in a row?

Thanks in advance for all help.

I should make clear that at least 1 win occurs after the loss sequence. So x losses occur, then at least 1 win. Is it more likely that the immediate next sequence of losses will be greater than or less than x? That should be a clarification. Thank you.

Depends on (a) the value of x, and (b) the probability of a win (this assumes a strictly win or lose case - ignore pushes).
For a game where the win probability is p, the probability of a string of n losses starting with the first loss after a win is (1 - p)^n-1 p.
Let q = 1 - p
The expected length = p + 2 qp + 3 q² p + 4 q³ p + ...
= p (1 + 2 q + 3 q² + 4 q³ + ...)
= p (1 + q + q² + q³ + ...)²
= p / (1 - q)² = 1 / p

So, for example, in double-zero roulette, the expected length of losses if you bet on double zero is 1 / (1 / 38) = 38.
If the last run of losses was 32, the next run is expected to be longer, and if the last run was 42, the next run is expected to be shorter.

Interesting, so is it possible to calculate the average loss streak for any game (whether 3 card poker, baccarat, blackjack, ultimate Texas holdem, etc.)? If the chances of wins/losses is known, is that all the information that is needed to know if n losses in a row is greater than or less than the expected loss streak? Thus, this information may offer a player some guidance for an approximate expectation of how long the next loss streak ought to be?

Quote: Rarush

Interesting, so is it possible to calculate the average loss streak for any game (whether 3 card poker, baccarat, blackjack, ultimate Texas holdem, etc.)? If the chances of wins/losses is known, is that all the information that is needed to know if n losses in a row is greater than or less than the expected loss streak? Thus, this information may offer a player some guidance for an approximate expectation of how long the next loss streak ought to be?

Depending on the complexity of the game, we can calculate the probability of the length of 'the next streak' after some trigger event, such as the clock striking the hour or 'the next time I win'. For some bets like Blackjack, it's easier. Probably the easiest is a fair coin toss, which is almost analogous to red/black on roulette.

So, taking the length of the next streak of heads after a coin toss lands tails.....
Probability of a streak of no 'Heads' (ie tails) is 50%
Probability of a streak of just one 'Head' is 25%
Probability of a streak of exactly two 'Heads' is 12.5%
Probability of a streak of exactly three 'Heads' is 6.25%
Probability of a streak of exactly four 'Heads' is 3.125%
Probability of a streak of exactly five 'Heads' is 1.5625%
Probability of a streak of exactly six 'Heads' is 0.78125%
etc...

add them up and you get
Probability of any possible set of outcomes approaches 100%

Quote: ThatDonGuy

So, for example, in double-zero roulette, the expected length of losses if you bet on double zero is 1 / (1 / 38) = 38.
If the last run of losses was 32, the next run is expected to be longer, and if the last run was 42, the next run is expected to be shorter.

Don't quite get what you're saying.
Are you actually saying that the previous run of losses is going to influence the numerical # of the next run of losses?
Huh? Please explain. Thanks.

Quote: lilredrooster

Don't quite get what you're saying.
Are you actually saying that the previous run of losses is going to influence the numerical # of the next run of losses?
Huh? Please explain. Thanks.

I initially read it like that, but how I now interpret it is... 'I observe a broken streak of events, such as 4 consecutive reds followed by a black at roulette... Can I expect the next streak of reds to be longer or shorter.' To which the answer was a calculation of the probability of the length of the next streak. which can, of course be estimated in isolation.
Thus if he said, I saw a streak of one red which was preceded and followed by a black, we can say 'yeah. the next streak of reds will probably be longer than one, when it eventually starts.'

A pretty pointless exercise, but hey oh.

Quote: ThatDonGuy

So, for example, in double-zero roulette, the expected length of losses if you bet on double zero is 1 / (1 / 38) = 38.

my simulations show the average length of run of losses to be 37, for your example
if we are talking about the same event
Sally

Quote: mustangsally

my simulations show the average length of run of losses to be 37, for your example
if we are talking about the same event
Sally

Is your simulation including streaks of length zero? The OP specified that the count should start at the trigger of the first loss, so the minimum streak would be one.

Quote: OnceDear

Depending on the complexity of the game, we can calculate the probability of the length of 'the next streak' after some trigger event, such as the clock striking the hour or 'the next time I win'. For some bets like Blackjack, it's easier. Probably the easiest is a fair coin toss, which is almost analogous to red/black on roulette.

Speaking of roulette, I think the win probability of hitting black/red or any of the 1:1 odds bets is 47.4% in double zero roulette. Given that information alone, can we calculate the average loss streak to a singular number of losses (i.e. 3 losses on average) or to a range of losses (i.e. 2-4 losses on average)?

I’m really fascinated by all of this.

Quote: mustangsally

my simulations show the average length of run of losses to be 37, for your example
if we are talking about the same event
Sally

I am counting the first loss as part of the streak. At least, that is how I am interpreting it, as the OP does not include streaks of zero losses.

From any particular starting point, the expected number of consecutive losses after that point before the next win is 1/p - 1, since, for example, there is probability p of the streak being zero losses instead of it being one loss.

Quote: ThatDonGuy

From any particular starting point, the expected number of consecutive losses after that point before the next win is 1/p - 1, since, for example, there is probability p of the streak being zero losses instead of it being one loss.

ok. agree. this is a simple geometric distribution

so for the OP post about 18/38 evens win
the average loss is 2 and 1/9 (2.11...)

with part of the distribution here

length	prob	cumulative
1	0.473684211	0.473684211
2	0.249307479	0.72299169
3	0.131214463	0.854206153
4	0.069060244	0.923266396
5	0.036347497	0.959613893
6	0.019130261	0.978744154
7	0.010068559	0.988812713
8	0.005299241	0.994111954
9	0.002789074	0.996901028
10	0.001467934	0.998368962
11	0.000772597	0.999141559
12	0.00040663	0.999548189
13	0.000214016	0.999762205
14	0.00011264	0.999874845
15	5.92841E-05	0.999934129
16	3.12022E-05	0.999965331
17	1.64222E-05	0.999981753
18	8.64326E-06	0.999990396
19	4.54909E-06	0.999994945
20	2.39426E-06	0.99999734
21	1.26013E-06	0.9999986
22	6.63229E-07	0.999999263
23	3.49068E-07	0.999999612
24	1.8372E-07	0.999999796
25	9.66947E-08	0.999999893
26	5.08919E-08	0.999999943
27	2.67852E-08	0.99999997
28	1.40975E-08	0.999999984
29	7.41973E-09	0.999999992
30	3.90512E-09	0.999999996