Gambler's ruin question

Just for info, I am not the formula guy

lets say I have a coin flip game and zero HE. each player will get 1000 chips and he needs to reach 100k chips with a minimum bet of 100 chips and maximum bet of 10k chips and every time he loses his 1000 chips he will get another 1000 chips to start again until he reaches the 100k chips.

how many buyin will he need till he reaches the win of 100k chips?

IMO no player will bet the 100 chips minimum under those conditions, he will always bet maximum to reach the 100k

tia

Well, the only way a player could win in the shortest time frame is to bet max and hope for 12 winners in a row (would only need to be 7 if there was no max bet) to parlay up the ladder.

Win 1: 2,000
Win 2: 4,000
Win 3: 8,000
Win 4: 10,000
Win 5: 10,000
Win 6: 10,000
Win 7: 10,000
Win 8: 10,000
Win 9: 10,000
Win 10: 10,000
Win 11: 10,000
Win 12: 10,000

Assuming an infinite bankroll as you established, to reach the 100k mark in the shortest amount of time the player would need to bet like above and win 12 times in a row. So we're looking for the probability of calling a coin flip correctly 12 times in a row...

P = .5*.5*.5*.5*.5*.5*.5*.5*.5*.5*.5*.5 = .000244, or .0244%

Given the first 3 bets are all max bets he could go bust on any 3 of them and need a new buy in. It would take a much more in depth analysis/simulator to find the "average" number of buy in's, but I thought I'd get the ball rolling and show you what it took to actually hit the 100k (after who knows how many buy ins).

Quote: Romes

Well, the only way a player could win in the shortest time frame is to bet max and hope for 12 winners in a row (would only need to be 7 if there was no max bet) to parlay up the ladder.

Win 1: 2,000
Win 2: 4,000
Win 3: 8,000
Win 4: 10,000
Win 5: 10,000
Win 6: 10,000
Win 7: 10,000
Win 8: 10,000
Win 9: 10,000
Win 10: 10,000
Win 11: 10,000
Win 12: 10,000

Assuming an infinite bankroll as you established, to reach the 100k mark in the shortest amount of time the player would need to bet like above and win 12 times in a row. So we're looking for the probability of calling a coin flip correctly 12 times in a row...

P = .5*.5*.5*.5*.5*.5*.5*.5*.5*.5*.5*.5 = .000244, or .0244%

Given the first 3 bets are all max bets he could go bust on any 3 of them and need a new buy in. It would take a much more in depth analysis/simulator to find the "average" number of buy in's, but I thought I'd get the ball rolling and show you what it took to actually hit the 100k (after who knows how many buy ins).

thanks for get the ball rollin :)

now lets see who will have the answer for how many buy ins are needed

cheers

Quote: seven

now lets see who will have the answer for how many buy ins are needed

The problem is, it looks like it's a 43-state problem.
Let each bankroll amount (in thousands) be a state number; the starting state is 1
State 0 is a terminating condition
State 1 becomes 0 or 2
State 2 becomes 0 or 4
State 4 becomes 0 or 8
State 6 becomes 0 or 12
State 8 becomes 0 or 16

State 12 becomes 2 or 22
State 14 becomes 4 or 24
State 16 becomes 6 or 26
State 18 becomes 8 or 28

State 22 becomes 12 or 32
...
State 88 becomes 78 or 98

State 92 becomes 84 or 100
State 94 becomes 88 or 100
State 96 becomes 92 or 100
State 98 becomes 96 or 100
State 100 is a terminating condition

Quote: ThatDonGuy

The problem is, it looks like it's a 43-state problem.
Let each bankroll amount (in thousands) be a state number; the starting state is 1
State 0 is a terminating condition
State 1 becomes 0 or 2
State 2 becomes 0 or 4
State 4 becomes 0 or 8
State 6 becomes 0 or 12
State 8 becomes 0 or 16

State 12 becomes 2 or 22
State 14 becomes 4 or 24
State 16 becomes 6 or 26
State 18 becomes 8 or 28

State 22 becomes 12 or 32
...
State 88 becomes 78 or 98

State 92 becomes 84 or 100
State 94 becomes 88 or 100
State 96 becomes 92 or 100
State 98 becomes 96 or 100
State 100 is a terminating condition

uh oh......now I am lost, is there a link where I can read what the 43-state problem is? thanks

Quote: seven

uh oh......now I am lost, is there a link where I can read what the 43-state problem is? thanks

Markov Chain is what ThatDonGuy is getting at (at least I’m assuming). There are 100 states as he described (and technically state 0 will become state 1 when you rebuy). You end up with a pretty large matrix that has to be taken to pretty large powers. Which a computer can solve.

A computer can also be used to simply simulate the problem. Which I did. The results of 500,000 simulated games are as follows:

Average number of Buyins: 99.0
Average number of Coin Flips: 280.5


Edit: Average number of coin flips should read 280.5 not 208.5

Quote: PeeMcGee

Markov Chain is what ThatDonGuy is getting at (at least I’m assuming). There are 100 states as he described (and technically state 0 will become state 1 when you rebuy). You end up with a pretty large matrix that has to be taken to pretty large powers. Which a computer can solve.

A computer can also be used to simply simulate the problem. Which I did. The results of 500,000 simulated games are as follows:

Average number of Buyins: 99.0
Average number of Coin Flips: 208.5

first of all many thanks again for trying to help :)

let me see if I understand Your explanation

after 99 buyins the player would hit 100k target?
and
after 208.5 coin flips the player hits the 100k target? 208 coin flips ? sounds low to me but it looks that I missed it again :(

Here is my way of looking at it:

Each coin flip has zero house edge. Therefore, the conglomeration of all coin flips has zero house edge.

The overall game has two possible results. Either 1k chips are lost or 99k chips are won. In order for this game to have zero house edge, the likelihood of winning must be 1 in 100. So, on average, 99 buy ins will be required.

This assessment is independent of the detailed rules of the game. Each wager must be one chip. Each wager must be twice that of the previous wager. Maximum wager is 10k chips. Doesn't matter.

Am I doing it wrong?

Quote: PapaChubby

Here is my way of looking at it:

Each coin flip has zero house edge. Therefore, the conglomeration of all coin flips has zero house edge.

The overall game has two possible results. Either 1k chips are lost or 99k chips are won. In order for this game to have zero house edge, the likelihood of winning must be 1 in 100. So, on average, 99 buy ins will be required.

This assessment is independent of the detailed rules of the game. Each wager must be one chip. Each wager must be twice that of the previous wager. Maximum wager is 10k chips. Doesn't matter.

Am I doing it wrong?

I dont know if You are right or wrong, but I am sure that the maximum of allowed bet and the target is important and should be considered in the math
10k max and target 100k

99 buyins required with what betting strategy?

thanks

Quote: PeeMcGee

Markov Chain is what ThatDonGuy is getting at (at least I’m assuming). There are 100 states as he described (and technically state 0 will become state 1 when you rebuy). You end up with a pretty large matrix that has to be taken to pretty large powers.

You assume correctly - it is a Markov chain. The chain determines the probability of a buy-in winning; if it is w, then the expected number of buy-ins = (1 x w) + (2 x (1-w) x w) + (3 x (1-w)² x w) + (4 x (1-w)³ x w) + ...
= w x (1 + 2 (1-w) + 3 (1-w)² + 4 (1-w)³ + ...)
= w x (1 + (1-w) + (1-w)² + (1-w)³ + ...)²
Since 0 < (1-w) < 1, this equals w x (1 / (1-(1-w)))² = w / w² = 1 / w.

Actually, by your count it would be 101 states (both 0 and 100 are states), but the only possible odd-numbered state is 1, and the only states that are multiples of 10 are 0 and 100, which leaves 43.

I don't know where powers come into it, although if you are looking for an "exact" value, the result's denominator is probably rather large - it's 43 simultaneous equations in 43 unknowns, isn't it?
P(0) = 0
P(1) = 1/2 (P(0) + P(2))
P(2) = 1/2 (P(0) + P(4))
P(4) = 1/2 (P(0) + P(8))
P(6) = 1/2 (P(0) + P(12))
P(8) = 1/2 (P(0) + P(16))
For all integers N from 1 to 8 inclusive:
P(10N+2) = 1/2 (P(10N-8) + P(10N+12))
P(10N+4) = 1/2 (P(10N-4) + P(10N+14))
P(10N+6) = 1/2 (P(10N-6) + P(10N+16))
P(10N+8) = 1/2 (P(10N-2) + P(10N+18))
Finally:
P(92) = 1/2 (P(84) + P(100))
P(94) = 1/2 (P(88) + P(100))
P(96) = 1/2 (P(92) + P(100))
P(98) = 1/2 (P(96) + P(100))
P(100) = 1

Quote: ThatDonGuy

Quote: PeeMcGee

Markov Chain is what ThatDonGuy is getting at (at least I’m assuming). There are 100 states as he described (and technically state 0 will become state 1 when you rebuy). You end up with a pretty large matrix that has to be taken to pretty large powers.

You assume correctly - it is a Markov chain. The chain determines the probability of a buy-in winning; if it is w, then the expected number of buy-ins = (1 x w) + (2 x (1-w) x w) + (3 x (1-w)² x w) + (4 x (1-w)³ x w) + ...
= w x (1 + 2 (1-w) + 3 (1-w)² + 4 (1-w)³ + ...)
= w x (1 + (1-w) + (1-w)² + (1-w)³ + ...)²
Since 0 < (1-w) < 1, this equals w x (1 / (1-(1-w)))² = w / w² = 1 / w.

Actually, by your count it would be 101 states (both 0 and 100 are states), but the only possible odd-numbered state is 1, and the only states that are multiples of 10 are 0 and 100, which leaves 43.

I don't know where powers come into it, although if you are looking for an "exact" value, the result's denominator is probably rather large - it's 43 simultaneous equations in 43 unknowns, isn't it?
P(0) = 0
P(1) = 1/2 (P(0) + P(2))
P(2) = 1/2 (P(0) + P(4))
P(4) = 1/2 (P(0) + P(8))
P(6) = 1/2 (P(0) + P(12))
P(8) = 1/2 (P(0) + P(16))
For all integers N from 1 to 8 inclusive:
P(10N+2) = 1/2 (P(10N-8) + P(10N+12))
P(10N+4) = 1/2 (P(10N-4) + P(10N+14))
P(10N+6) = 1/2 (P(10N-6) + P(10N+16))
P(10N+8) = 1/2 (P(10N-2) + P(10N+18))
Finally:
P(92) = 1/2 (P(84) + P(100))
P(94) = 1/2 (P(88) + P(100))
P(96) = 1/2 (P(92) + P(100))
P(98) = 1/2 (P(96) + P(100))
P(100) = 1

I am really glad to get some answers but I mentioned in my first posting that I am not the formula guy :( do You guys have
a ELI5 explanation for me?

would one need 99 rebuys of 1k to hit the 100k target? and with what betting strategy ?
and would he hit the 100k target after an average of 208.5 coin flips?

how goes this together with the 99 buyins?




thanks

Quote: seven

I am really glad to get some answers but I mentioned in my first posting that I am not the formula guy :( do You guys have
a ELI5 explanation for me?

would one need 99 rebuys of 1k to hit the 100k target? and with what betting strategy ?
and would he hit the 100k target after an average of 208.5 coin flips?

how goes this together with the 99 buyins?

First of all, understand that you can't "guarantee" that you will ever reach the 100K target. What we are trying to figure out is the expected, or "average", number of buyins necessary. If you include the initial 1000 as a buyin, I calculate this as exactly 100.

The strategy I am using is:
If your bankroll is 10K or less, bet everything
If it is between 10K and 90K, bet 10K
If it is more than 90K, bet just enough to reach 100K if you win (so if you have 92K, you bet 8K).
However, I have a feeling that the betting strategy does not matter, and that the expected number of buyins will always be 100, in line with PapaChubby's reasoning earlier in this thread.

Quote: ThatDonGuy

First of all, understand that you can't "guarantee" that you will ever reach the 100K target. What we are trying to figure out is the expected, or "average", number of buyins necessary. If you include the initial 1000 as a buyin, I calculate this as exactly 100.

The strategy I am using is:
If your bankroll is 10K or less, bet everything
If it is between 10K and 90K, bet 10K
If it is more than 90K, bet just enough to reach 100K if you win (so if you have 92K, you bet 8K).
However, I have a feeling that the betting strategy does not matter, and that the expected number of buyins will always be 100, in line with PapaChubby's reasoning earlier in this thread.

You mean that the 100k should hit in an average of 100 buyins sure thing that it can hit immediately or after 500 buyins but the average is 100 buyins.
I hope this is right :)

players betting strategy will and should be as You said with the given max bet and 100k target.

but if average hit number is after 100 buyins what if there are 100 players playing the same time? so one should hit the target, right?

if yes then we need to know when? after how many (average of) coin flips? or am I wrong?

thanks again

Quote: ThatDonGuy

Quote: PeeMcGee

Markov Chain is what ThatDonGuy is getting at (at least I’m assuming). There are 100 states as he described (and technically state 0 will become state 1 when you rebuy). You end up with a pretty large matrix that has to be taken to pretty large powers.

You assume correctly - it is a Markov chain. The chain determines the probability of a buy-in winning; if it is w, then the expected number of buy-ins = (1 x w) + (2 x (1-w) x w) + (3 x (1-w)² x w) + (4 x (1-w)³ x w) + ...
= w x (1 + 2 (1-w) + 3 (1-w)² + 4 (1-w)³ + ...)
= w x (1 + (1-w) + (1-w)² + (1-w)³ + ...)²
Since 0 < (1-w) < 1, this equals w x (1 / (1-(1-w)))² = w / w² = 1 / w.

Actually, by your count it would be 101 states (both 0 and 100 are states), but the only possible odd-numbered state is 1, and the only states that are multiples of 10 are 0 and 100, which leaves 43.

I don't know where powers come into it, although if you are looking for an "exact" value, the result's denominator is probably rather large - it's 43 simultaneous equations in 43 unknowns, isn't it?
P(0) = 0
P(1) = 1/2 (P(0) + P(2))
P(2) = 1/2 (P(0) + P(4))
P(4) = 1/2 (P(0) + P(8))
P(6) = 1/2 (P(0) + P(12))
P(8) = 1/2 (P(0) + P(16))
For all integers N from 1 to 8 inclusive:
P(10N+2) = 1/2 (P(10N-8) + P(10N+12))
P(10N+4) = 1/2 (P(10N-4) + P(10N+14))
P(10N+6) = 1/2 (P(10N-6) + P(10N+16))
P(10N+8) = 1/2 (P(10N-2) + P(10N+18))
Finally:
P(92) = 1/2 (P(84) + P(100))
P(94) = 1/2 (P(88) + P(100))
P(96) = 1/2 (P(92) + P(100))
P(98) = 1/2 (P(96) + P(100))
P(100) = 1

Yea, this is correct. And you do get a solution of P(1) = 1/99.

Quote: seven

Quote: ThatDonGuy

First of all, understand that you can't "guarantee" that you will ever reach the 100K target. What we are trying to figure out is the expected, or "average", number of buyins necessary. If you include the initial 1000 as a buyin, I calculate this as exactly 100.

The strategy I am using is:
If your bankroll is 10K or less, bet everything
If it is between 10K and 90K, bet 10K
If it is more than 90K, bet just enough to reach 100K if you win (so if you have 92K, you bet 8K).
However, I have a feeling that the betting strategy does not matter, and that the expected number of buyins will always be 100, in line with PapaChubby's reasoning earlier in this thread.

You mean that the 100k should hit in an average of 100 buyins sure thing that it can hit immediately or after 500 buyins but the average is 100 buyins.
I hope this is right :)

players betting strategy will and should be as You said with the given max bet and 100k target.

but if average hit number is after 100 buyins what if there are 100 players playing the same time? so one should hit the target, right?

if yes then we need to know when? after how many (average of) coin flips? or am I wrong?

thanks again

Just want to make sure you caught that the 100 buyins number does include the initial 1000 chips.

Also, I did typo on the average number of coin flips that I got…it should had read 280.5 not 208.5. A little bit dyslexic I guess, sorry.

ok so please let me ask again.

100 buyins of 1000 chips and one of those buyins of the player should hit the target of 100k?

the target of 100k should be hit in an average of 280.5 coin flips?

so lets say we have 100 players (equal to 100 buyins of one player) at the coin flip table playin the same time, so one of them should hit the target during the 280.5 coin flips?

thanks again for trying to help

Quote:

how many buyin will he need till he reaches the win of 100k chips?

Guys, it is a zero-HE game. It makes all calculations very simple.

For any reasonable strategy (i.e. a strategy where you won't overshoot the target), the chance p of hitting your target (100k) from a start (1k) is exactly 1%.

Why ? If it were p>1%, you could buy in for 1k, aim for the 100k target, and would hit it with probability >1%, your expected value is thus >1k, i.e. larger than your buy-in. You would consistently profit from a zero-HE game, which is impossible.

Same is true for p<1%. You could offer that game to someone else and consistently make profit from a zero-HE game. Which is also impossible.

So it must be p=1% exactly. Then the average number of buy-ins is <n> = {sum(n=1 to infintiy) n*p*(1-p)^(n-1)} / {sum(n=1 to infintiy) p^n} = 1 / p = 100.

Quote: MangoJ

Quote:

how many buyin will he need till he reaches the win of 100k chips?

Guys, it is a zero-HE game. It makes all calculations very simple.

For any reasonable strategy (i.e. a strategy where you won't overshoot the target), the chance p of hitting your target (100k) from a start (1k) is exactly 1%.

Why ? If it were p>1%, you could buy in for 1k, aim for the 100k target, and would hit it with probability >1%, your expected value is thus >1k, i.e. larger than your buy-in. You would consistently profit from a zero-HE game, which is impossible.

Same is true for p<1%. You could offer that game to someone else and consistently make profit from a zero-HE game. Which is also impossible.

So it must be p=1% exactly. Then the average number of buy-ins is <n> = {sum(n=1 to infintiy) n*p*(1-p)^(n-1)} / {sum(n=1 to infintiy) p^n} = 1 / p = 100.

thanks MangoJ

I understood now that the average number of buy-ins to hit the 100k target is 100 buy-ins. is this right? so even I will up the target to 200k the average number of buy-ins to hit the 200k target stays at 100 buy-ins?

so if there are 100 players at the site/table the same day, one should hit the target?

how about the coin flips? how about the number of 280.5 coin flips (mentioned before by PeeMcGee) that would be needed to hit the target?

Quote: seven

I understood now that the average number of buy-ins to hit the 100k target is 100 buy-ins. is this right?

That is right, yes.

Quote:

so even I will up the target to 200k the average number of buy-ins to hit the 200k target stays at 100 buy-ins?

It depends on the buy-in. If the target is upped to 200k and the buy-in is still 1k, one of 200 buy-ins will hit the target.
If the buy-in is also upped to 2k, then one of 100 buy-ins will hit the target.

Quote:

how about the coin flips? how about the number of 280.5 coin flips (mentioned before by PeeMcGee) that would be needed to hit the target?

The number of coin flips is a different story, which is way more complicated to calculate exactly. It depends alot on the strategy employed. In general, if you bet big you need less coin flips. The optimal strategy for minimum coin flips is to max-bet (even if it is all you have), with the only exception of an overbet (when you would overshoot your target if the bet is won).

For some specific rules the expected number of coin flips for the minimum coin flip can be calculated. Most simply when there is no cap on the bet (or the cap is larger than half the target). Another scenario (more complicated) is when the cap is way smaller than the buy-in. For some intermediate caps and antes, a simulation will give you the most precise answer in a reasonable time (like the 280.5).

thanks MangoJ

now another important point is cleared now = with 1k chips buy-ins the number of the target is the average number of buy-ins needed to hit the target.

I am sure that each player will always try to bet maximum beside the last bet to reach target. so in case the 280.5 coin flips are correct
it would mean that a player will hit the target either after 280.5 coin flips or 100 buy-ins (in case of 100k target)

could You please answer this question: so if there are 100 players at the site/table the same day, one should hit the target? (in case 100k is the target with
1k chips at the start)

Quote: seven

could You please answer this question: so if there are 100 players at the site/table the same day, one should hit the target? (in case 100k is the target with
1k chips at the start)

On average, yes. Of course you can have days without a winner, or days with multiple winners....