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I was just thinking about something to do with gambling, and it occurred to me that surely you could guarantee success at a gambling game in the long run if you set aside profits.

Allow me to explain. Say you had 4 betting units.

If you were to flat bet one 1 unit each round, but set aside any profit that you made on top of your original 4 units. When you reach 4 units in profit or loss, (disregarding any previous winnings), you quit.

Therefore, if we assume there is a 50% chance of either winning or losing your 4 units, surely there is also a chance you will near your 4 unit profit goal, but then lose your original 4, leaving you with 3 units. Therefore, your loss for that session is -1.

Therefore, there are 4 different ways one can lose. Having lost 4 units but gained 3, lost 4 gained 2, lost 4 gained 1, or lost all 4 straight away.

If it is 50/50 that you will win or lose 4 units, then surely by safeguarding profits, then the possible outcomes would be, +4, -4, -3, -2, -1. Because of this surely the 50% that you lose must be divided up again to consider losing only 3 or 2 or 1 unit.

Surely then that means the probability of each outcome is: +4 : 1/2 -1 : 1/8 (1/2 divided by all negative outcomes.) -2 : 1/8 -3 : 1/8 -4 : 1/8

All the probabilities add to 1 as should be expected. But when calculating expected value we get:

+4 * 1/2 + -1 * 1/8 + -2 * 1/8 + -3 * 1/8 + -4 * 1/8 = 3/4

This shows a positive expectation, which should not be possible.

Can anyone tell me the flaw in my reasoning or workings, as this is puzzling me greatly.

Roskiller
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  • Please clarify the strategy : Are you playing until you lost $4$ times ? – Peter May 08 '18 at 15:20
  • Suppose you start with a loss and then a win. You have four coins, but is the second a winning that cannot be risked? As Peter says, it sounds like you play until you either win 4 or have 4 losses. Is that correct? – Ross Millikan May 08 '18 at 15:23
  • only profit on top of the original 4 is set aside. If i lose and then win I do not set aside, but if i win and then lose i set aside 1 in profit – Roskiller May 08 '18 at 15:30
  • So if you go $WLLWWL$ have you set aside two and $WLLWWLLL$ causes you to stop? I think you can stop ahead less than $4$, say from $WLWLWWLWL$ where you are $+1$. You need to carefully define the strategy. I guarantee it will be break even. – Ross Millikan May 08 '18 at 15:43

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Your chance of ending at $+4$ is not $50\%$. The symmetry is broken when you refuse to bet your winnings. If you use the stopping criterion of stopping when your net is $+4$ or $-4$ your chance of ending at $+4$ is indeed $50\%$, but in your system a series of $WWLLLL$ would cause you to stop and you could never reach $+4$ that way. That reduces the chance you will reach $+4$.

Ross Millikan
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  • I apologise for asking, but what would be the probability of reaching +4 with the above criterion. – Roskiller May 08 '18 at 15:34
  • I haven't understood the strategy well enough to answer. – Ross Millikan May 08 '18 at 15:42
  • The game ends under 3 conditions, +4, -4, or whatever the situation is on getting LLLL. Intuitively, symmetry says that finishing with a profit or a loss have equal probability. That is, playing against someone using the same strategy, you both can't win. – Phil H May 08 '18 at 17:05
  • It doesn't seem the four losses have to be sequential. That is what I mean by saying the strategy is not well defined. There is an asymmetry because of your set asides. The game will still have expectation $0$, but you may have a greater than $50%$ chance to lose money, balanced by a greater win when you do win. – Ross Millikan May 08 '18 at 17:51