An online poker room offers a reward to qualifying players in the form of one entry per week to a weekly tournament with a $5,000.00 prize pool.
It’s a ‘coinflip’ tournament, meaning that no poker skill is involved. In fact you don’t even need to be online to win. Every hand, every player is put ‘all in’ until only one remains. There are usually 15k to 25k players per tournament. The top 10% or so of the players make at least some money, and the payout gets higher the longer you survive.
It’s also a ‘multi-entry’ tournament, meaning that you can have two entries to the same tournament, doubling your chances. But since only one entry is awarded each week that means if you want to play two entries you can only play every 2 weeks.
When the tournament begins every player goes all in with whatever cards they’re dealt, and after one hand 5/6 of the field has been eliminated.(There are 6 seats per table. Sometimes there could be a tie between two or more players at a table but that is pretty rare). The same thing happens next hand, and the next. You only need to survive through a few of these to make at least some money and the longer you survive the higher the payout.
Logic tells me that the long term odds of playing one entry every week or two entries every two weeks should be exactly the same. But a gut feeling keeps nagging at me that somehow the chances of getting to the money will be better playing 2 entries every two weeks.
The most important factor of course is how many other players are entered, which varies from week to week. But this can’t be predicted, so for the purposes of the question we can consider the number of entries to be equal every week.
Is there any advantage to playing either one entry per week or two entries every two weeks?
It doesn’t matter, there is no real benefit to collecting tickets for use 2,3,4 at a time…
There is a very very small draw back, you do rule out winning two (etc) weeks in a row… but its very slim difference in expected outcomes.
If the game was limitted to a small N (eg 20) players, and your entry blocked other players, you could collect N entries to totally ensure you win once… this reduces the maximum you can you win, you could have won N times (at extremely remote chance…), but you traded off the win total by improving the chance you win . That means the expected outcome is the same no matter what…
Only in the sense that luck determines the outcome. But otherwise it’s different in several ways. Instead of being a single 1 in 25,000 chance (or 1 in 12,500 chance with two entries) it’s a series of several 1 in 6 chances, and you only need to survive a couple of those to win at least something instead of nothing.
You can buy 2 entries to a lotto every week and double your chances of winning over the long term. But as noted in this case if you use two entries one week, you have to skip the tournament the next week. Thus my question.
Thanks. This is what I keep telling myself. But, a couple of points:
But remember also that with two entries in the same tournament you could win, for example, both first and second place in the same tournament. Or 60th and 250th place, etc. So unlike the lottery where you either win or tie no matter how many entries you have, multiple entries in this tournament could each win in different positions.
This is what I think too but what keeps me wondering about whether or not I may be wrong is that again it is not just a question of winning 1st place or losing, but also a matter of getting at least something if you manage to survive the massive cull of the first couple of hands.
On the first hand of the game the odds are 1 in 25,000 of making first place. But after surviving just one hand, the odds are 1 in 4100 of making first place. And after surviving just 2 hands, the odds are 1 in 700, etc. The field gets small very quickly. This is why I keep going back to the idea that having 2 entries, and so twice the chance of surviving that first massive cull of players in any one tournament still somehow seems like it could be an edge over having 2 entries spread across two different tournaments, both in terms of getting paid any money at all and also in the chance of making 1st place.
I’m not sure it’s useful to worry about the details of the tournament process like this. The tournament is just a random process that creates an ordered list of players; all the stuff about poker hands is just window dressing. So I think your expected value is the same either way, and your variance will be slightly less when you put both your entries in the same game.
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.
This is actually a slight advantage in favor of only playing once per week. You can win different places on separate weeks as well with single bets. However, if you play more than once in the same week, you ruled out that you will win multiple first places (or any other specific place) which is possible if you play once on different weeks. In effect, you are competing with yourself for prizes and that isn’t good.
It isn’t as bad as buying 10,000 lottery tickets with the exact same numbers versus just one in the same week but still offers slightly worse odds overall.
On the other hand, while the expectation is definitely higher for one entry at a time, it’s higher by such a small amount that it’s not worth worrying about. In practice, you should just choose whichever option seems like more fun to you, since the difference in fun will almost certainly be greater than the difference in monetary value.
As Chronos points out, the difference between the two approaches in this problem is insignificant. However the general principle is worth remembering: Don’t bet against yourself. 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.
As with many “theorems” of gambling, the best policy reverses if bets have positive expectations. In the stock market, hedging plays (“betting against yourself”) can be wise as they allow larger “bets” for the same net risk.
You can search for old threads on this topic: the point has been discussed to the point of exhaustion. :eek:
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!
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.
Or am I missing something? I can’t recall any of the previous threads that discussed this, so if you have any links I’d be happy to peruse them.
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.
But you are changing the odds since “red vs one number” is not the same as “one number vs two numbers” since there is 0/00 eating into odds in the former case resulting in a less than 50% chance of winning. Plus, I don’t think there are typically 35 red numbers, so your numbers don’t make sense.
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?
The difference between the roulette wheel and the original bet is that with the roulette wheel you are not betting against yourself. Both bets are against the house; the number of bets placed on a single spin does not change the expected payout to any single bet (assuming the bank can’t go broke).
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.
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.
It’s usual to assume one’s goal is to maximize the Expectation of B, where B is your resultant wealth, but it is better to maximize Expectation of u(B), where u(.) assigns utility to a given wealth.
It is usually reasonable to assume that u(B) has a negative 2nd derivative, in which case (assuming other plausible conditions), no bet with positive vigorish is advisable (and large bets are ill-advised even with negative vigorish). To make gambling puzzles meaningful, it is simplest to frame the problem as I did upthread: you’re willing to risk $A to acquire $B. This is not far-fetched; $B may be an important goal if it’s the cost of [del]the heroin fix for which you’re desperate[/del] a life-saving medication you need.
But being entered in every week means that you have a chance of winning the top prize each week. With two entries in the same tournament you could, at best, win a first and second place.
This is what I meant when I said that it’s just a lottery. If the winners are found by some random process, that is what it is - there is no skill involved, you are just buying a chance to win. Naturally, the organisers are the real winners.
:eek: