What is the difference between house edge and the hold percentage in gambling?

Dec 2012
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0
A single-zero roulette has 37 slots, they pay 35+1
House edge is 2,7%, you lose it in every bet, in a long run.
But, there is a "hold percentage" which is supposed the earning of a casino in a certain time. I read Las Vegas strip casinos have 11% earnings from the hold.
I believe, it is due to their unlimited bankroll, and player's finite bankroll.
How do they calculate it?
How could they earn more than their house edge?

Any help is welcome.
Regards
ybot
 

skipjack

Forum Staff
Dec 2006
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I doubt that any casino operates only roulette. I've seen suggestions that the hold percentage is much more than the house edge for roulette, but that makes no mathematical sense to me. In most other casino games, many of the players simply play poorly, which effectively increases the house edge, but that should be impossible in roulette if the wheel is in good condition.
 
Jul 2008
5,233
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Western Canada
It does have to do with the player's limited bankroll. If both the player and the house had unlimited funds then hold percentage should equal the house edge. However, because the player's bankroll is limited, a losing streak will have a non linear ratchet effect. Once you've lost all your money you're out of the game, and have no chance to win it back. This is why the house also makes money on games like blackjack which are very slightly in the player's favour (if you're a good player).

In other words, to determine the hold percentage, you not only have to take into account the house edge, but you also have to calculate the probability of players having long enough losing streaks to lose all of their money, and the average amount that is lost when this happens.
 
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skipjack

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Dec 2006
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I don't see the logic in that. If the player loses their limited bankroll on a particular day, that deprives the casino of further takings from that player on that day, but if the player has a winning streak, the house has to continue paying out to that player until the end of the day or the end of the winning streak. If the losing player is always replaced by another player, that effectively removes the limited bankroll concept.
 

mathman

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May 2007
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House edge in slot machines are very much higher. From what I remember it is in double figures.
 
Jul 2008
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If a gambler with a limited bankroll plays long enough he will eventually encounter a losing streak long enough to lose all of his money. This is a classic problem known as "Gambler's Ruin."

If you consider the extreme case where every gambler plays long enough to lose all of his money, then the house's hold percentage will be 100%, and the house edge only determines the average number of rounds it takes for each gambler to lose all of his money. The mathematical relationship between hold percentage and house edge would have to include a factor that accounts for when an average player will quit playing if he quits before losing all his money.
 

skipjack

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Dec 2006
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I doubt that the player has a slight theoretical edge in blackjack unless (a) card counting is possible, and (b) the cards used by the house aren't effectively shuffled.

I disagree about "Gambler's Ruin", because most versions (all versions?) of that refer to gamblers who stake more when they're winning more, but don't stake less when they're not. If the model used supposes that the house has an unlimited amount of money, the gambler will lose. In reality, though, the house funds are high, but not unlimited, and so the house protects itself by setting a limit on each stake, but this also protects the gambler from "Gambler's Ruin", because the gambler can't speed up the game to circumvent the stake limit. From the point of view of the house, going bust on some particular day is impossible if their funds exceed the maximum individual payout multiplied by the maximum number of stakes that can be placed during that day, so it might seem they benefit anyway as various gamblers go bust. However, this doesn't apply if each losing gambler is replaced by another gambler, because this means that the gamblers as a group don't go bust and don't increase their combined stakes in proportion to their combined winnings. The house edge is usually mathematically determined by how the casino operates. I've never seen a mathematical calculation of the casino's hold percentage.
 
Jul 2008
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Do you agree or disagree with this statement?
"A gambler with a limited bankroll, playing against the house with an infinite bankroll, will eventually encounter losing streak sufficient to bankrupt him."

The original definition of gambler's ruin included the changing bet scenario, but more recent treatments of the problem simplify it to a constant size bet. The only way I can see that a gambler could play forever would be to limit the bet always to some fraction of the remaining bankroll. It might get smaller and smaller, but it would never disappear (in theory). However, at some point it would conflict with the minimum size of bet that the house allows. If that's one dollar, and you have exactly one dollar left, do you risk it, or do you go home? I think that calculation of the house's hold percentage would also have to take into account human psychology.
 

skipjack

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Dec 2006
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Doesn't "eventually" require the player to have an unlimited number of opportunities to play? That's more unrealistic than the house having an infinite bankroll, especially for games such as roulette (with a real wheel).
 

mathman

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The house makes its money because most players have a moderate bankroll for gambling. The house will have a slight advantage in roulette against players who decide to leave at their own discretion. However those players quitting because they lost their bankroll add much more to the house winnings.