# Math wonks: A question about probability in a 'coinflip' poker tournament **URL:** \ **Category:** Factual Questions\ **Created:** [October 13, 2015, 3:04am UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512 "2015-10-13T03:04:22Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![BigT](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bigt/32/12044_2.png) [@BigT](https://boards.straightdope.com/u/BigT)\ **Post date:** [October 14, 2015, 11:40am UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/21 "2015-10-14T11:40:57Z")
> [@septimus](#): > > At American roulette, betting $18 on Red is _exactly_ the same as betting $1 on each of the 18 Red numbers. (I’m careful to specify _American_ roulette since European casinos often have an _en prise_ rule, improving the odds for even-money bets.) > > Your expected loss as a percentage of your wager is constant regardless of how you bet at American roulette, _but see my preceding post for why it matters how you bet_. Did you understand and agree with the claim I made there? I don’t want to quibble about the semantics of “betting against yourself” but IMO this is an informative way to understand the underlying point. I sure don’t. I will ultimately be wagering the same amount in either scenario, whether it is 18 bets on one number or one bet on 18 numbers. So if my expected loss is constant no matter how I bet, then it should be the same in both scenarios. It only makes sense to me if you are expecting me to choose to stop betting should I reach a certain return, meaning I risk less money if I win early on. If that’s the case, then obviously many smaller bets is better. But only because there’s now a chance I will not bet the entire amount. To use the coins that **brickbacon** attempted to use to simplify the problem: if I have $2 to spend on two coin flips, it’s better for me to bet $1 even money on heads for the first flip, and then, if I win, walk away now with $3, than it is for me to bet $0.50 even money on both sides for both flips, where I will just get a guaranteed $2. In scenario 1, my chance of winning more than I started with is 1:2. In scenario 2, there is no chance at all. ---
**Author:** ![Snarky\_Kong](https://avatars.discourse-cdn.com/v4/letter/s/a183cd/32.png) [@Snarky\_Kong](https://boards.straightdope.com/u/Snarky_Kong)\ **Post date:** [October 14, 2015, 11:54am UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/22 "2015-10-14T11:54:07Z")
> [@septimus](#): > > The simplest way to understand the point is to imagine that you have $A and are willing to risk it all for a chance to end up with $B. Your goal is to construct an optimal, possibly compound, bet from a set of permitted single bets. > > As just one example, suppose that B = 2A and your only options are those available at an American roulette table. One option is to bet the entire $A on Red. Another option is to bet A/35 on a single number, and continue appropriately if the bet loses. The latter strategy is superior. This can be demonstrated mathematically, or by simulation, or by the following argument: if you bet $A on Red, you’ve subjected all $A to the constant house vigorish; if you bet $A/35 on a single number, as little as $A/35 is subjected to the vigorish. > > B = 2A is an obvious simple case, but the same principle applies for any B \> A. If you really want to review the prior thread(s), “roulette septimus” should find them using the SDMB search facility. Assume there is a bystander that is willing to pay you C to not play roulette. C = B-A-epsilon where epsilon is some arbitrarily small value. Using your definition of better bet, it is always better to walk away from free money and to make a wager with negative expected value. Looking at it from an expected utility standpoint: assume log-utility, although any decreasing marginal utility works, it’s better to guarantee a loss of $2 rather than probably a loss of your entire bet. This, to me, seems to be a better way to frame the question than to assume you need an arbitrarily larger amount of money or else some nebulous _bad thing_ will happen. (What about “I need to leave Vegas with at least $D where D is less than A, but enough for a bus ticket and for the old lady to not throw me out”, a much more common situation) Regardless, there are multiple subjective ways of looking at it. All valid for different needs, so claiming that it’s proven to be better to bet with higher variance is odd. ---
**Author:** ![Your\_Great\_Darsh\_Face](https://avatars.discourse-cdn.com/v4/letter/y/ac8455/32.png) [@Your\_Great\_Darsh\_Face](https://boards.straightdope.com/u/Your_Great_Darsh_Face)\ **Post date:** [October 14, 2015, 12:15pm UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/23 "2015-10-14T12:15:07Z")
The situation given is not the same as the roulette situation. Playing roulette, your chance that a given number will come up is 36/38 so if you are wagering dollar stakes, you will get paid an average of $36/38 per turn (on an American wheel. In Monte Carlo they recognise that a mere 3% house edge is more than enough to get rich on, and they don’t need to be so greedy). If you bet on one number, it costs you $1 to win your $36/38. If you bet on two numbers, it costs you $2 to win your $36/38. Etc. More later. ---
**Author:** ![brickbacon](https://avatars.discourse-cdn.com/v4/letter/b/898d66/32.png) [@brickbacon](https://boards.straightdope.com/u/brickbacon)\ **Post date:** [October 14, 2015, 12:26pm UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/24 "2015-10-14T12:26:12Z")
> [@septimus](#): > > At American roulette, betting $18 on Red is _exactly_ the same as betting $1 on each of the 18 Red numbers. (I’m careful to specify _American_ roulette since European casinos often have an _en prise_ rule, improving the odds for even-money bets.) First, you said 36, not 18. Second, if you think betting $18 on one number is the same as $1 on 18 numbers, then what are we arguing about and how do you defend your initial claim? > [@septimus](#): > > Your expected loss as a percentage of your wager is constant regardless of how you bet at American roulette, _but see my preceding post for why it matters how you bet_. Did you understand and agree with the claim I made there? I don’t want to quibble about the semantics of “betting against yourself” but IMO this is an informative way to understand the underlying point. No, I don’t understand the point you were trying to make. Why is it better to bet $2 on one number than $1 on two numbers? ---
**Author:** ![Turble](https://avatars.discourse-cdn.com/v4/letter/t/8dc957/32.png) [@Turble](https://boards.straightdope.com/u/Turble)\ **Post date:** [October 14, 2015, 3:06pm UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/25 "2015-10-14T15:06:33Z")
> [@borschevsky](#): > > There may be a slight advantage to splitting your entries. The expected value of an entry is: > > Prize Money / Number of Entries > > Say that there are 9,999 entries in this week’s tournament and 9,999 in next week’s. If you put one entry in each, then your expected return is: > > $5000 / 10000 + $5000 / 10000 = $1 > > If you put both in next week’s, then your expected return is: 2 \* $5000 / 10001 = $0.99990001. Your first formula is correct but the second formula ignores the fact that you cannot win 1st place twice in one week. It should account for the fact that you have a chance to win both 1st and 2nd place. So the answer to the OP is that it is better to enter once each week than to enter twice every other week because it is better to have two chances to win 1st place than it is to have one chance to win both 1st and 2nd place. ---
**Author:** ![Your\_Great\_Darsh\_Face](https://avatars.discourse-cdn.com/v4/letter/y/ac8455/32.png) [@Your\_Great\_Darsh\_Face](https://boards.straightdope.com/u/Your_Great_Darsh_Face)\ **Post date:** [October 14, 2015, 4:36pm UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/26 "2015-10-14T16:36:17Z")
> [@Your\_Great\_Darsh\_Face](#): > > The situation given is not the same as the roulette situation. Playing roulette, your chance that a given number will come up is 36/38 so if you are wagering dollar stakes, you will get paid an average of $36/38 per turn (on an American wheel. In Monte Carlo they recognise that a mere 3% house edge is more than enough to get rich on, and they don’t need to be so greedy). > > If you bet on one number, it costs you $1 to win your $36/38. > If you bet on two numbers, it costs you $2 to win your $36/38. Etc. > More later. Note: I was a little rushed posting this. I suspect it’s complete tosh. More later… maybe. Meanwhile, the situation in the OP breaks down quite simply: If you enter twice in the same week there is a small chance that one of your entries competes with the other, knocking one of them down to a lower place. If you enter once per week that won’t happen. The probability may be small, but there’s no compensating advantage, so it’s a marginally losing option. ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [October 14, 2015, 8:10pm UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/27 "2015-10-14T20:10:56Z")
**Here’s the simplest scenario to clarify the essential truth:** \*Mr. K bets all $1000 on a group of 9 numbers. He walks away $3000 winner 23.6842% of the time (9/38). Mr. J, OTOH, bets $1000 on a group of 18 numbers, e.g. Red. If the bet wins, he lets it ride. Mr. J walks away $3000 winner 22.4377% of the time (18/38 \* 18/38). Mr. K’s approach was clearly better than Mr. J’s (though not nearly as good as if he’d started with an $85 single-number bet and gone on from there. \* The claims that I made in #14 were very specific and were either true or false. In fact, they are true. Their truth can be demonstrated arithmetically, with simulation, or by an argument which appeals to intuition (but can be made rigorous). The simple scenario at the beginning of this post should point the way. At least two respected SDMB mathematicians have already posted in this thread; I will be interested to see if they return and address #14 specifically. If Mr. J and Mr. K each end up betting a total of exactly $1000 each at American roulette, then each will end up losing $52.63 on average. That much we all agree on. And we also agree it makes a difference _how_ they bet. I’ve already shown a “strategy” for Mr. J to arrange that he always loses exactly $52.63 ! Mr. K, OTOH, can make a single $1000 bet on Number 17 and have a chance of walking away with $35,000 profit. But in the scenario relevant to my #14, the total wagering is variable. And in the scenario at the top of this post, Mr. K wagers exactly $1000, while Mr. J wagers on average $1947.37. That’s one way to understand why Mr. J’s strategy is inferior! The underlying point, which leads to the correct answers given to OP’s original question by #2, #6, #7, most clearly #8, is obviously well known, but its application to specific scenarios like my roulette examples surprises most people (even good mathematicians) when they first learn of it. If you refer to the earlier thread you will see that nobody accepted the claim at first but eventually most came around. Lance Turbo, who presumably has his PhD in mathematics by now, took my side … though the argument went on and on. (If you do review that 5-year old thread, as I did now, you’ll see that **septimus** ’ irritation might have become the most salient detail! Let’s hope that doesn’t happen again. :o ) > [@brickbacon](#): > > First, you said 36, not 18… > > Why is it better to bet $2 on one number than $1 on two numbers? The number of times I’ve typed the number you mention (between "First you said " and “not 18”) is ZERO, Nada, Zip. I’ve never typed that number in this thread. I have mentioned 38 several times; are you referring to one of those mentions? In the scenario which heads this post, I demonstrate that betting on 9 numbers is better than 18 numbers. The same principle applies to one number versus two numbers, but to demonstrate it with detailed arithmetic would be far more tedious. Does this help? ---
**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\ **Post date:** [October 14, 2015, 8:39pm UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/28 "2015-10-14T20:39:45Z")
> [@Crazyhorse](#): > > There are usually 15k to 25k players per tournament. I think the best strategy is to save your entries and wait for weeks where there are smaller number entries. ---
**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\ **Post date:** [October 14, 2015, 8:58pm UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/29 "2015-10-14T20:58:42Z")
> [@septimus](#): > > Lance Turbo, who presumably has his PhD in mathematics by now, took my side … though the argument went on and on. Thanks for remembering. I do, in fact, have my PhD in mathematics now. I have since learned that your strategy is called ‘bold play’ and it [has been proven](http://www.ncbi.nlm.nih.gov/pmc/articles/PMC223086/pdf/pnas00211-0067.pdf) to be optimal. Man that earlier thread was frustrating. ---
**Author:** ![borschevsky](https://avatars.discourse-cdn.com/v4/letter/b/97f17d/32.png) [@borschevsky](https://boards.straightdope.com/u/borschevsky)\ **Post date:** [October 14, 2015, 9:41pm UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/30 "2015-10-14T21:41:26Z")
> [@Turble](#): > > Your first formula is correct but the second formula ignores the fact that you cannot win 1st place twice in one week. It should account for the fact that you have a chance to win both 1st and 2nd place. I don’t think that matters, at least as far as your expectation is concerned. If you split your entries, it’s true that you have the chance to get first place in both draws. But you also have the chance to get 3rd in both draws, or 4th in both, etc. Take an example. Say there are two prizes, $4000 for first place and $1000 for second, and say there are 10,000 total entries in all cases. If you put one entry in each draw, then your expectation is: $4000 \* 1/10000 + $1000 \* 1/10000 + $4000 \* 1/10000 + $1000 \* 1/10000 = $1. If you put both in a single draw, then your expectation is: 2/10000 \* 1/9999 \* $5000 + 2/10000 \* 9998/9999 \* $4000 + 9998/10000 \* 2/9999 \* $1000 = $1 In the first case, your chances of winning nothing are 9998 / 10000 \* 9998 / 10000. In the second case, your chances of winning nothing are 9998 / 10000 \* 9997 / 9999 - slightly less. ---
**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\ **Post date:** [October 14, 2015, 10:13pm UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/31 "2015-10-14T22:13:08Z")
To expand on my post #28. If the prize pool is fixed at $5000 the expectation of an entry is $5000/N where N is the number of players. If one week there are 25k players and the next week there are 15k players, playing one entry each week has an expectation of $0.53. Playing two entries the second week has an expectation of $0.67. For completeness, playing two entries the first week has an expectation of $0.40, but this is obviously a bad move. ---
**Author:** ![brickbacon](https://avatars.discourse-cdn.com/v4/letter/b/898d66/32.png) [@brickbacon](https://boards.straightdope.com/u/brickbacon)\ **Post date:** [October 15, 2015, 1:16am UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/32 "2015-10-15T01:16:33Z")
> [@septimus](#): > > **Here’s the simplest scenario to clarify the essential truth:** > > \*Mr. K bets all $1000 on a group of 9 numbers. He walks away $3000 winner 23.6842% of the time (9/38). Mr. J, OTOH, bets $1000 on a group of 18 numbers, e.g. Red. If the bet wins, he lets it ride. Mr. J walks away $3000 winner 22.4377% of the time (18/38 \* 18/38). You are changing the scenario by letting it ride in the second scenario, and changing the amounts that were bet. The issue is simple. Use a die for simplicity. Why would you be better of betting $100 on one number once than $50 on each of two numbers? In the former, you have a 1/6 chance of winning. In the latter you have 2/6 chance or winning half as much money. Why is the expected return different? Is there some roulette rule that changes the odds? Please point out where the difference lies? > [@septimus](#): > > At least two respected SDMB mathematicians have already posted in this thread; I will be interested to see if they return and address #14 specifically. Who are you talking about? You explicitly mentioned Chronos before, and he said: > [@](#): > > Which is better in the roulette example depends on whether you put a positive or a negative value on variance. That’s basically a matter of taste… but if you put a zero or negative value on variance, then the winning move is not to play, so it’s probably safe to assume that everyone at the table puts a positive value on it. That doesn’t seem to back your understanding of the issue. Perhaps he’ll be back to clarify. > [@septimus](#): > > The number of times I’ve typed the number you mention (between "First you said " and “not 18”) is ZERO, Nada, Zip. I’ve never typed that number in this thread. I have mentioned 38 several times; are you referring to one of those mentions? Sorry, I meant 35. ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [October 15, 2015, 1:34am UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/33 "2015-10-15T01:34:23Z")
> [@brickbacon](#): > > Why is the expected return different? The expected return is NOT different, _if your total wager is the same_. (But expected return _does not tell the full story._) Debate, if you wish, the claim I made in post #14, not some claim you imagined I made. ---
**Author:** ![brickbacon](https://avatars.discourse-cdn.com/v4/letter/b/898d66/32.png) [@brickbacon](https://boards.straightdope.com/u/brickbacon)\ **Post date:** [October 15, 2015, 1:40am UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/34 "2015-10-15T01:40:31Z")
> [@septimus](#): > > The expected return is NOT different, _if your total wager is the same_. (But expected return _does not tell the full story._) Debate, if you wish, the claim I made in post #14, not some claim you imagined I made. But that is what you said, no? To quote you: > [@](#): > > As a trivial example, at the roulette table it is better to bet $2 on a single number than to bet $1 on each of two numbers. Are you now disavowing this? Please defend the above? ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [October 15, 2015, 1:51am UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/35 "2015-10-15T01:51:25Z")
> [@brickbacon](#): > > But that is what you said, no? To quote you: > > > [@](#): > > > > As a trivial example, at the roulette table it is better to bet $2 on a single number than to bet $1 on each of two numbers. > > Are you now disavowing this? Please defend the above? I explained and proved explicitly in what sense it was better to bet 9 numbers than to bet 18 numbers. Did you understand that discussion? I suggested that a similar argument would apply when {9; 18} are replaced with {1; 2} (but the detailed arithmetic proof might be more tedious). If you understand the {9;18} discussion I ask you what your intuition tells you about {1; 2}. ---
**Author:** ![brickbacon](https://avatars.discourse-cdn.com/v4/letter/b/898d66/32.png) [@brickbacon](https://boards.straightdope.com/u/brickbacon)\ **Post date:** [October 15, 2015, 2:08am UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/36 "2015-10-15T02:08:02Z")
> [@septimus](#): > > I explained and proved explicitly in what sense it was better to bet 9 numbers than to bet 18 numbers. Did you understand that discussion? > > I suggested that a similar argument would apply when {9; 18} are replaced with {1; 2} (but the detailed arithmetic proof might be more tedious). If you understand the {9;18} discussion I ask you what your intuition tells you about {1; 2}. I don’t think you have proved that. Where is this proof that doesn’t change the scenario. You also said: > [@](#): > > But, as is so often the case, one can guess the correct answer by considering extreme cases; betting on three numbers is worse than two, betting four worse still and so on. > **And if you bet $1 each on all 38 numbers you have no chance to win; you’re guaranteed to lose $2!** To which MikeS correctly responded: > [@](#): > > I don’t think this example illustrates the principle you’re saying. In both cases, your expected winnings are $2 times the odds on a single number. And if (as you suggest in your later post) you bet $1 on all 38 numbers, you’ll only get $36 back each time; but if you bet $38 on a single number, you’ll get $1368 back one time out of 38, and $0 back the other 37 times. So on average, you get $36 back per time you play, for a net loss of $2, regardless of strategy. Why are we both wrong? And please don’t add in something about letting your winnings ride, betting different aggregate amounts, or whatever. Please address the original claim. ---
**Author:** ![Turble](https://avatars.discourse-cdn.com/v4/letter/t/8dc957/32.png) [@Turble](https://boards.straightdope.com/u/Turble)\ **Post date:** [October 15, 2015, 2:34am UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/37 "2015-10-15T02:34:54Z")
> [@borschevsky](#): > > If you put both in a single draw, then your expectation is: > > 2/10000 \* 1/9999 \* $5000 + > 2/10000 \* 9998/9999 \* $4000 + > 9998/10000 \* 2/9999 \* $1000 > = $1 > > In the first case, your chances of winning nothing are 9998 / 10000 \* 9998 / 10000. In the second case, your chances of winning nothing are 9998 / 10000 \* 9997 / 9999 - slightly less. No. You don’t have two chances to win 1st AND two chances to win 2nd. The times you do win 1st place you no longer have two chances to win 2nd place; you then only have one chance to win 2nd. ---
**Author:** ![borschevsky](https://avatars.discourse-cdn.com/v4/letter/b/97f17d/32.png) [@borschevsky](https://boards.straightdope.com/u/borschevsky)\ **Post date:** [October 15, 2015, 2:58am UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/38 "2015-10-15T02:58:00Z")
> [@Turble](#): > > The times you do win 1st place you no longer have two chances to win 2nd place; you then only have one chance to win 2nd. Right, that’s where the first line in that calculation comes from. 2/10000 of the time, you win the first prize. When that happens, **1** /9999 of the time you also win second prize, for $5000 total; the other 9998/9999 of the time you don’t win second and only get $4000. In the 9998/10000 cases where you don’t win first prize, 2/9999 of the time you win second prize and $1000. ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [October 15, 2015, 5:18am UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/39 "2015-10-15T05:18:29Z")
There is one argument which _might_ help the intuition to grasp the apparently paradoxical-looking observation I made about roulette bets. Your expected loss is exactly 5.26% of _your_ wager no matter how you bet. But let’s turn it around and consider it from the casino’s point-of-view. On an even-money bet, their expected winning is exactly 5.26% of the money _the casino wagers._ **But, if a customer bets $1 on a single number, _the casino is wagering $35 to win $1_. If you calculate the house advantage on the $35 it wagers, you will find it to be 0.15% – much much less than 5.26%.** For some, seeing how much different the bets are _from the casino’s standpoint_ may lead to an Aha! moment and to enlightenment! (And it will help explain why high-odds games like Keno have higher vigorish than even-money bets.) If not … forget it. As I said, we discussed this to the point of irritation and exhaustion in the earlier thread. Let’s not let s **eptimus** get frustrated! I am certainly correct; I do not think any mathematicians will appear to say otherwise. My arguments need to stand on their own merits, but I suspect you’d be more eager to learn rather than teach if you knew my credentials. I’m just another asshole on a message board here, but on some subjects I do know what I’m talking about. 😃 > [@brickbacon](#): > > I don’t think you have proved that. Where is this proof that doesn’t change the scenario. You also said: > > To which MikeS correctly responded: > > Why are we both wrong? And please don’t add in something about letting your winnings ride, betting different aggregate amounts, or whatever. Please address the original claim. **brickbacon** , did you understand and agree with the example I present at the top of post #27? I show that if you want to multiply your bankroll by 4, a 9-number bet is better than is an 18-number bet. Please decide Yes-or-No, whether you agree with this before proceeding. This claim is about three numbers, \< 4, 9, 18 \>. _ **Let’s consider the same claim, but the general case \.** _ I assert that the b-number bet will be better in any of these cases than the c-number bet, when b \< c. If you’re still with me, do you think that this statment is also true? _I have tried to be careful in the phrasing of my posts in this thread, but perhaps you’ve misinterpreted something I wrote._ Obviously I don’t want to defend a claim that I never made. If you think you’ve got some Gotcha, where I phrased a claim poorly, sorry. We’ve had one PhD in mathematics show up and agree with me. If I get 97% consensus from mathematicians will you acquiesce, or will this become like the climate change debate? 😃 ---
**Author:** ![Pasta](https://avatars.discourse-cdn.com/v4/letter/p/ecccb3/32.png) [@Pasta](https://boards.straightdope.com/u/Pasta)\ **Post date:** [October 15, 2015, 6:59am UTC](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512/40 "2015-10-15T06:59:30Z")
Folks are discussing different scenarios. The key in **septimus** ’s scenario is that the player is trying to maximize the probability of reaching a certain gain (a “stop win”). The optimal strategy at a roulette table is indeed a series of small bets for that scenario. Anyone not overtly taking the stop-win condition into consideration is examining a scenario other than the one **septimus** is outlining. > [@brickbacon](#): > > Your scenario, in my mind, seems closer to saying it makes more sense to bet two dollars on heads than one dollar on both heads and tails. I think the math says you are in the same boat given enough coin flips. What am I missing here? You are in the same boat if the goal is to maximize your expected return over some large number of flips. That isn’t the goal in **septimus** ’s scenario. The goal is to maximize the probability of reaching some pre-determined total winnings. If you bet $1 on heads and $1 on tails everytime, your expected winnings are zero and you have zero chance of ever earning (say) $5. If you bet $2 on heads and $0 on tails everytime, your expected winnings are still zero but now you have _some finite_ chance of earning (say) $5. If your goal is to earn $5, the second strategy is clearly better. This concept extends into the roulette case. [Previous page](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512.md?page=1) [Next page](https://boards.straightdope.com/t/math-wonks-a-question-about-probability-in-a-coinflip-poker-tournament/734512.md?page=3)