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This is a theoretical question, following what I have seen in some low blinds cash games. You generally cannot buy-in more than 100BB.

Say at some point, you reached 10,000 BB. All you adversaries have about 100BB: how good is to go all-in on each hand? My guess is that it's bad, but how bad?

I would be interested in seeing the probability study behind this. Maybe this can be found somewhere? It should depend on the number of opponents, the hand odd, etc. For instance, with a stack K times bigger than the average stack of your opponents, what it is the expected number of hands you can play all-in preflop before your stack hits 0?

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3 Answers 3

My short answer is this: as long as your opponents have 100BBs, it doesn't matter how many BBs you have (as long as you have them covered). You will only play for their 100 BBs.

Shoving every hand when you have 10000 BBs is as bad as shoving every hand when you have 100 BBs. The difference is that in former case you don't have to re-buy every time after you lose.

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Agreed this would be identical as long as he rebuys for the same as the largest stack at the table. Then the question is how many hands before he burns through 10000 BBs. –  TTT Apr 12 '13 at 23:59
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Poker is a zero-sum game so, in the long run, he might not reach the point where he burns his 10k BBs, because he-s mega-deep stacked. –  Bogdan Apr 13 '13 at 6:00

This is actually an interesting question and not because it might relate to good strategy when considering hand equity match-ups or Expected Value, (let's face it, random All-In shoves every hand don't fair well in terms of overall strategies) but because it's a good example of bankroll management, and how people misapply it.

The key concept here is how much of that 10,000bb's can you afford to wager each time given the knowledge of your average equity.

This is the difference between Expected Value (EV) and Expected Growth (EG). Most people understand the importance of EV but few take into account the importance of EG, and to be clear, the two are INEXTRICABLY linked to correct long-term bankroll management.

If you're shoving All-In each time, every time, then running a "random hand" against what you assume is your opponents calling range will give you this EV information. To know whether this is the right choice will require you to calculate your EG as well. Just consider for a moment that the 100bb's your opponent has is the ONLY money he has left in the world. Does it make sense for him to call? Even knowing that he probably has the best possible matchup, for example AA vs 72o? Well, EV-wise, easy call. EG-wise, it depends but in this case almost certainly not.

What matters here, is whether the players have the means to be making huge bets,or calling them. In this situation, the 10,000bb player can repeatedly cover his losses, assuming that the opponents can't continuously reload their stacks. The important thing is not to destroy your bankroll, leaving you incapable of playing anymore. Hero obviously doesn't have to worry about his risk too much here.

I won't go into the deep intricacies of it, but I'll scratch the surface... This relates to the Kelly Criterion. Here's the formula for EG...

  • E(G) = (1 + (O-1) * X)p * (1 - X)1-p - 1

Where X = the percentage of bankroll wagered on the given bet; O = the decimal odds being offered.

In terms of win probability (for win probability * Odds ≥ 1)...

  • Stake as a percentage of bankroll = (Prob * Odds – 1) / (Odds – 1)

Without breaking it down, intuitively I can see that the probability of the 10,000bb player going broke at the assumed win probability (40%) is very, very small. It would take a run of incredibly poor luck. I'd hazard to suggest that even in a worst case scenario every time, you'd probably win in the long run as well just by virtue of time, although I'm not certain.

As it stands, its probably best to go All-In every time given these conditions!

n.b. I'm no expert on these things, but I believe I've done a lot more work on this than most people making me a "relative" expert. Take what you read with a grain of doubt and research it further if it peaks your interest, like it did me.

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Intuitively to me I thought the table could devise a strategy to win, and you seem to feel the opposite. But your thought might be based on a guess of 40% win probability. Suppose 9 other players all agreed to call every hand and split up hero's winnings later if they break him. Might that tip the scales in their favor? Hmmmm.... maybe not- even if hero wins 1/50 hands instead of 1/10, he might still clear the table each time it happens. –  TTT Apr 13 '13 at 0:25
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@TTT I don't think it matters because even as Hero's odd's of winning decrease, his earnings from each win still cover his losses (+EV). Hero's losses still only equate to 100bbs no matter how many opponents are in the hand. I think you can see that too :) –  Toby Booth Apr 13 '13 at 0:41
    
Here's what I'm thinking: suppose a 10 person table and everyone goes all in every hand, and a villian wins 3 in a row. After the first hand they have 1000bbs, hero has 9900, and everyone else rebuys for 100. After hand 2, villian has 2800, here has 8900, and everyone else rebuys for 100. After hand 3, villian has 6400, hero has 6100, and everyone else buys in for another 100. So someone has to win 3 hands in a row to but hero into second place. The chances of the villain winning 3 in a row are 1/1000, but, the reality is that it doesn't have to be in a row, –  TTT Apr 13 '13 at 3:36
    
cont... if they lose some hands to a 100bbs player they only lose 100bbs. The chances of any one player winning 3 in a row is just 1/100 (since anyone can win the first one to reset the count). Of course, hero's stack can grow well beyond 10000bbs every time he clears the table, and every else is reset, so the stars may need to align a bit in the beginning for hero to start worrying. –  TTT Apr 13 '13 at 3:40
    
@TTT ...you're also assuming that villain still calls shoves now he has a 10x buy-in stack! I think we've past the realms of reasonable assumptions ;)! –  Toby Booth Apr 13 '13 at 13:18

This question is actually very similar to the "double your bet" strategy for baccarat, or any other close to even odds game. Casinos typically will always have a table max bet for these games to prevent some billionaire from coming in and keep doubling their bet until win back all of their loses. The number of times you can lose before you hit the limit is the breaking point. In your scenario, the breaking point is 7, calculated this way:

100 * (2^x) > 10,000

The lowest X where the above equation is true is 7. So the worst case scenario for you is if you are called and lose 7 times in a row. If you were heads up in a tournament playing this way, the chances of you losing the tournament if your opponent just called every hand are 1/128. However, given that the person will not call every hand, and instead only call with better than average hands, you're less likely to win each hand that you are played against. Presumably the person you are playing against may be able to use a strategy that gives him an edge, I would guess around 20% or so, making the position worse, perhaps making your chances of winning in the range of 1/20-1/40. (I'm too tired to do the math on this right now, and I'm not 100% sure I'd get it right even if I tried.)

In your situation, a cash game, with multiple people at the table, on the positive side, you're less likely to continuously double up the same person 7 times in a row, but on the negative side, you're also less likely to win against multiple people, especially since you'll be against an even stronger than average hand that calls you. If the table got together and devised a strategy, surely they could break you. I don't know what that strategy would be exactly, and I believe it would depend on the number of opponents at the table, but it seems like it's easy to imagine a scenario where X people call with Y or better, and you would get called often enough to lose more than you pick up in blinds when everyone folds.

As for a mathematical answer to your question, I believe the number of opponents (P) you have will need to be figured into the equation.


Update: After further thought, I believe we can solve this. If we reduce your scenario to heads up, and make the assumption that villain calls all in every hand, (I assume that isn't the best strategy for villain, but it's one strategy, and one I believe we can quantify), and continues to rebuy at 100bb every time he busts out, then I believe the following:

  1. Villain will always win eventually.
  2. The formula for calculating the expected number of hands hero will last has an e in it (Eulers number).
  3. This is just pure speculation, but I believe the expected number of hands hero will last is less than 1 order of magnitude of K (less than 10*K).

My thinking is this (for simplicty I'm going to assume hero has 128 times villain's stack rather than 100 times:

At 128 to 1, villain has to win 7 in a row to bust hero. Villain has 128 tries to achieve this. If Villain fails, he then has 256 tries to hit 8 in a row, which is 1 in 256. If he still fails to do that, he then has 512 tries to hit a 1 in 512 shot, etc. The probability of hitting a 1 in X shot with X tries, as X goes to infinity, is related to the number Phi. (I forgot what the actual formula is, but I know where to go to find out.)

That was slightly over simplifying, since once villain's stack is greater than hero, there is some opportunity for them to toggle back and forth a few times before one busts out the other, but I'm assumming that will have a negligible effect on the total answer.

That's where I'm at, I'll update again when I get the actual formula...

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I think the "double bet" strategy you mention is the Martingale system. I could be wrong! –  Toby Booth Apr 13 '13 at 0:11
    
@TobyBooth - yep- you're correct! –  TTT Apr 13 '13 at 0:17

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