joue
joue
  • Threads: 1
  • Posts: 3
Joined: Feb 2, 2017
February 2nd, 2017 at 10:57:54 AM permalink
Hello,

I have been trying to find the answer to that question for quite some time now and eventually thought the best would be to ask here!

Let's say someone offers to play a totally fair heads or tails game with you. The rules are that:
- he will always bet heads
- he will start betting with 1 unit
- he will double after every loss until he recovers or looses everything (martingale).
- you will receive every bet and must pay every bet he wins when he does.
- you can both play for years

Can you win? If so:
- How much funds do you need?
- With what probability will he loose everything in x bets?

In that case, aren't you like a casino or bookmaker? (minus their edge of course)

Thanks a lot!!
ChesterDog
ChesterDog 
  • Threads: 9
  • Posts: 1732
Joined: Jul 26, 2010
Thanked by
Romes
February 2nd, 2017 at 11:27:33 AM permalink
This is the "gambler's ruin problem." Here is the Wikipedia article. Reading the section called "Fair coin flipping," I see from the two equations that when I act as the casino, I cannot guarantee a win for myself if I limit myself to some fixed, albeit large, initial bankroll.

edit: Sorry; disregard the above because I forgot to take into account the player's Martingale on his losing bets.
Last edited by: ChesterDog on Feb 2, 2017
joue
joue
  • Threads: 1
  • Posts: 3
Joined: Feb 2, 2017
February 3rd, 2017 at 6:14:37 AM permalink
Thanks for the quick reply, the article was still interesting even though not including the martingale aspect!
I do understand that without martingale then there is no question to be asked here. But I thought maybe there is something if the other player chases losses...
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6743
Joined: Jun 22, 2011
February 3rd, 2017 at 9:00:57 AM permalink
The only way you can win in this situation is if the other player decides to stop after a significant number of consecutive losses.

How much money do you need? It depends on the "stop condition" - the point at which the game ends - but the best answer I can come up with off the top of my head is, "More than the other guy."

Systems like Martingale and D'Alembert work if either (a) you have unlimited time and unlimited budget, neither of which are true, or (b) the probability of winning each individual bet is greater than 1/2.

I am a little confused as to the question, "With what probability will he lose everything in X bets?"; this also depends on his initial budget.
joue
joue
  • Threads: 1
  • Posts: 3
Joined: Feb 2, 2017
February 12th, 2017 at 5:03:48 AM permalink
Thanks a lot!

To clarify further:

He (the other guy) doesn't have unlimited funds
He will stop only when he has lost everything or cannot place the next bet.
He will always start back at 1 unit after recouping all losses, i.e. never increase the base stake even when his balance increases.

With those real life conditions the probability of him losing everything eventually is 1. But only if I have enough funds to continue paying him until that happens. Before that he could easily double, triple or whatnot his funds before finally losing everything.
I can find out what is the probability of him losing everything for X bets but my question is if mathematics can clearly say if it's possible and how to ensure my survival until he loses everything.
  • Jump to: