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## Question

Assume you start off with a bankroll of \$1,000,000. Consider the following 3 games:

Game 1: Flip a biased coin which lands Heads 45% of the time, and Tails the rest of the time.

If you play this game, you get a \$1 to add to your bankroll if you flip Heads and you lose

\$1 from your bankroll if you flip Tails.

Game 2: There are 2 cases:

_______Case 1: If your bankroll modulo 3 is 0, you flip a coin which lands Heads 1% of the

time, and Tails 99% of the time. You get \$1 to add to your bankroll if you flip Heads,

Otherwise you lose \$1 from your bankroll.

_______Case 2: If your bankroll modulo 3 is 1 or 2, you flip a coin which lands Heads 90%

of the time, and Tails 10% of the time. You get \$1 to add to your bankroll if you flip Heads,

Otherwise you lose \$1 from your bankroll.

Game 3: Flip and unbiased coin (50% of the time Heads, 50% of the time Tails). If the coin

lands Heads, play Game 1; if it land Tails, play game 2.

After 1,000,000 plays of each game (always starting with a bankroll of \$1,000,000), which of

these games should increase your bankroll and which should decrease your bankroll?

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Quick answer without doing the math

Case 1, decrease since odds are against you

Case 2, increase since 2/3rds of the time significant probability of winning.

Case 3.increase, quick estimate of odds in case 2 > 60%

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Quick answer without doing the math

Case 1, decrease since odds are against you

Case 2, increase since 2/3rds of the time significant probability of winning.

Case 3.increase, quick estimate of odds in case 2 > 60%

If you simulate this, you're in for some surprises!

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I am never going to this casino

Obviously game 1 isn't going to work out in your favor. You will on average lose 100k and the chances of you breaking even or gaining money is about 10%

Game 2 is interesting. You will most likely start off losing \$1 to start off and since the increment is \$1 every time are you will always pull back to that 1 million you started out with at best and since that has a modulo of zero you have to lose money again. Since it is 10x more likely that you will lose money on module 1 or 2 versus gaining money on modulo 0 then you will probably end up losing money here too. Again not in a probability running mood, but I would assume that you would lose \$10,000 on average through 1 million plays of game 2 but your chances of breaking even on this game are probably less than in Game 1.

Well..this one is obvious. If both games involve you losing money than letting a coin decide which game you lose on isn't going to net you any cash. However I am pretty sure you asked this to see that maybe by going between the two games, one where you lose more money overall but have a better shot of breaking even/gaining money and the other where you have less chance of breaking even but would lose less money in general, you could gain something. I don't see the point as far as figuring out this as my logic tells me that I would lose money so as far as answering the question goes...don't gamble

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They all (or technically both [in reference to Games 1 & 2 considering 3 merely redirects you to one of the aforementioned]) have the capability to increase and/or decrease your bankroll. You can't really ascertain probabilistic circumstances; you could land Heads one million times, you could land Tails one million times, or you could land a mixture of both Heads and Tails with either one outweighing the other -- how do we know which one specifically? It's impossible to tell.

Edited by Omniscience
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Got a question:

After 1,000,000 plays of each game (always starting with a bankroll of \$1,000,000), which of

these games should increase your bankroll and which should decrease your bankroll?

Does this mean:

(a) Start with \$1M, make 1M plays of game 1; start with \$1M, make 1M plays of game 2, start with \$1M, make 1M plays of game 3, or

(b) (1 M times: (start with \$1M, play game 1)), (1M times: (start with \$1M, play game 2)), (1M times (start with \$1M, play game3))

(b)

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i'd pick game 2.

on your first play, you're at mod 3 = 1. So 90% to win and increase \$1. Then mod 3 = 2. Again 90% to win and increase \$1. Now mod 3 = 0 so 99% to lose and decrease \$1.

Working it out, the chances to lose 2 games in a row when mod 3 = 1 or 2 is 1% which is the same chance to win 1 game when mod 3 = 0. So it seems like you're likely to bounce around between \$1,000,002 and \$1,000,000. That doesn't seem too bad to me.

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Game 1 is pretty obviously a losing game.

For game 2, this isn't a totally legit formal calculation, but could help provide insight.

Suppose you start of with N dollars and want to find out what the odds are that after three plays you would have N+3 dollars or N-3 dollars, thus putting yourself at the same spot modulo 3 but with a net gain or loss of \$3.

PW3 = 0.9 * 0.9 * 0.01 = 0.0081

PL3 = 0.1 * 0.1 * 0.99 = 0.0099

So game 2 is also a losing game.

For game 3, if your bankroll is 0 mod 3 then your probability of winning the next game is

PW = 0.5 * 0.45 + 0.5 * 0.01 = 0.23

If your bankroll is 1 or 2 mod 3 then your probability of winning the next game is

PW = 0.5 * 0.45 + 0.5 * 0.9 = 0.675

Since the probability of winning is dependent on your cash in mod 3, using a similar approach as for game 2 of calculating the probability of winning or losing three consecutive games gives

PW3 = 0.675 * 0.675 * 0.23 = 0.1048

PL3 = 0.325 * 0.325 * 0.77 = 0.0813

And this is a winning game.

I realize that my math isn't totally legit since I'm only considering the probability of getting from \$N to \$N+3 or \$N-3 by winning three consecutive games, and it doesn't consider more circuitous routes. If you really want to, you can make a spreadsheet to get exact probabilities after each round.

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Got a question:

Does this mean:

(a) Start with \$1M, make 1M plays of game 1; start with \$1M, make 1M plays of game 2, start with \$1M, make 1M plays of game 3, or

(b) (1 M times: (start with \$1M, play game 1)), (1M times: (start with \$1M, play game 2)), (1M times (start with \$1M, play game3))

(b)

I meant (a). I said 1M trials, but what I really want is how these compare for a lot of trials -- the long run, if you will.

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Game 1 is pretty obviously a losing game.

For game 2, this isn't a totally legit formal calculation, but could help provide insight.

Suppose you start of with N dollars and want to find out what the odds are that after three plays you would have N+3 dollars or N-3 dollars, thus putting yourself at the same spot modulo 3 but with a net gain or loss of \$3.

PW3 = 0.9 * 0.9 * 0.01 = 0.0081

PL3 = 0.1 * 0.1 * 0.99 = 0.0099

So game 2 is also a losing game.

For game 3, if your bankroll is 0 mod 3 then your probability of winning the next game is

PW = 0.5 * 0.45 + 0.5 * 0.01 = 0.23

If your bankroll is 1 or 2 mod 3 then your probability of winning the next game is

PW = 0.5 * 0.45 + 0.5 * 0.9 = 0.675

Since the probability of winning is dependent on your cash in mod 3, using a similar approach as for game 2 of calculating the probability of winning or losing three consecutive games gives

PW3 = 0.675 * 0.675 * 0.23 = 0.1048

PL3 = 0.325 * 0.325 * 0.77 = 0.0813

And this is a winning game.

I realize that my math isn't totally legit since I'm only considering the probability of getting from \$N to \$N+3 or \$N-3 by winning three consecutive games, and it doesn't consider more circuitous routes. If you really want to, you can make a spreadsheet to get exact probabilities after each round.

I agree that this has got to be it. I figured there was something to the combination of the two games that would make everything work. I think that a good follow-up to this problem would be if you had the choice to switch freely between each game and if so what is the best strategy to make money on these 2 losing games

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Game 1 is simple enough, you have a 45% chance of winning and 55% chance of losing.

For game 2, let P0, P1, and P2 be defined as probabilities of having a bankroll congruent to 0, 1, and 2 respectively (modulo 3) after already having played a large number of games. Doing the math, which is a bit tedious, we get P0 = 910/2018, P1 = 109/2018, and P2 = 999/2018. So the probability of winning your next game is P0*1/100 + P1*9/10 + P2*9/10 = 10063/20180, which is roughly 0,499. In the long run, the actual probability of winning a game will converge to this number, so game 2 is a slightly losing game. It is deceptively close to even money, but the fact remains: it will decrease your bankroll in the long run.

For game 3, define W0 and W12 as the probability of winning a game if your bankroll is congruent to 0 and not 0 respectively (modulo 3). We get W0 = 23% and W12 = 67,5% = 27/40. Now define P0, P1, and P2 as we did for game 2. They will be different for this game. We get P0 = 3842/17489, P1 = 6245/17489, and P2 = 7402/17489. The probability of winning the next game is P0*23/100 + P1*27/40 + P2*27/40, which is roughly 0,577. In the long run, the actual probability of winning a game will converge to this number, so game 3 is a winning game.

Edited by shakingdavid
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game 1: clearly a losing game by the expected value principle

game 2: after each 3 rounds, you will gain \$3 with .81% chance and lose \$3 with .99% chance. if neither of these happen, then you just gain or lose one, remaining in the same set of 3. Therefore, it is also a losing game.

Game 3: you are given an equal chance of playing one of two losing games. therefore, it is also a losing game.

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game 1: clearly a losing game by the expected value principle

game 2: after each 3 rounds, you will gain \$3 with .81% chance and lose \$3 with .99% chance. if neither of these happen, then you just gain or lose one, remaining in the same set of 3. Therefore, it is also a losing game.

Game 3: you are given an equal chance of playing one of two losing games. therefore, it is also a losing game.

You are correct for game 1 of course, but your logic is too hasty on game 2 and 3. Case in point, what if we copy the logic you use for game 2 and apply it to game 3?

Clearly if your bankroll is congruent to 0, you have a winning chance of (45%+1%)/2 = 23%

If your bankroll is congruent to 1 or 2, you have a winning chance of (45%+90%)/2 = 67.5%

So after each 3 rounds you will gain \$3 with ~10% chance (.23*.675*.675) and lose \$3 with ~8% chance (.77*.325*.325). This would make it a winning game according to your logic for game 2. (It actually is a winning game, but it's not so easy to conclude.)

Edited by shakingdavid
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I realize that the credibility of my earlier solution can be doubted since I omitted the math part, so here we go with lenghty math:

Game 2: P, Q, and R (previously P0 through P2) are defined as the probabilities of having a bankroll congruent to 0, 1, and 2 respectively (modulo 3) after already having played a large number of games. Let #n mean "congruent to n mod 3". We know that in order to reach a bankroll #0, you must either lose from a bankroll #1 or win from a bankroll #2. So we get

(1) P = Q*.1 + R*.9, and similarly we get

(2) Q = P*.01 + R*.1, and

(3) R = P*.99 + Q*.9

Using (2) to substitute Q with P*.01 + R*.1 in (1), we get

(4) P = (P*.01 + R*.1)*.1 + R*.9, which can be simplified to

(5) P*.999 = R*.91

Using (3) to substitute R with P*.99 + Q*.9 in (2), we get

(6) Q = P*.01 + (P*.99 + Q*.9)*.1, which can be simplified to

(7) Q*.91 = P*.109

(5) and (7) combine to give us the ratio P to Q to R = .91 to .109 to .999. Since we also know that P+Q+R=1, we merely add .91+.109+.999=2.018 and conclude that P = 910/2018, Q = 109/2018, and R = 999/2018.

Game 3 is done the same way, but with different probabilities (23% to win from #0, 77% to lose. 67.5% to win from #1 or #2, 32.5% to lose)

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Plasmid and shakingdavid got it correct. Magician would have been correct had Games 1 and

2 been independent. But the fact that Game 1 influences the bankroll used in Game 2 makes

then dependent (albeit, rather weakly). This type of situation is called a Parrondo Paradox,

after Spanish physicist Juan Parrondo, who discovered the paradox in 1996. Of course, it

is not a paradox at all, but can be explained by the dependencies between the games.

By the way, based on a simulation of 100,000,000 of each game, the Approximate expected \$ win

in each game:

Game 1: -\$0.10

Game 2: -\$0.00269

Game 3: +\$0.00323

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