{"id":993,"date":"2011-05-31T20:00:59","date_gmt":"2011-05-31T20:00:59","guid":{"rendered":"http:\/\/gwae.apps-1and1.com\/?p=993"},"modified":"2011-05-31T20:00:59","modified_gmt":"2011-05-31T20:00:59","slug":"a-different-sort-of-breakpoint-calculation","status":"publish","type":"post","link":"https:\/\/www.lasvegasadvisor.com\/gambling-with-an-edge\/a-different-sort-of-breakpoint-calculation\/","title":{"rendered":"A Different Sort of Breakpoint Calculation"},"content":{"rendered":"

In last week’s column, we looked at the standard way of calculating royal flush breakpoints (BPs) on progressives. If you didn’t see that column, I suggest you review that one first before tackling this one. https:\/\/www.lasvegasadvisor.com\/bob_dancer\/2011\/0524.cfm<\/p>\n

Assume you are playing 8\/5 Double Double Bonus on the progressives at the M with a 4% meter and you’re trying to work out how to play A\u2666 K\u2663 T\u2663 7\u2665 6\u2665. If you’ve played DDB at all, you know that at low levels of the royal flush you hold A\u2666 by itself, and sometime later, when the royal flush gets high enough, the correct play becomes K\u2663 T\u2663. The problem, of course, is to calculate the level of royal flush where the value of A\u2666 and K\u2663 T\u2663 are equal. That will be our BP.<\/p>\n

The first thing to discuss is how this is a different type of problem than what we had last week and why the formula we used before won’t work here. In last week’s problem, we were comparing the value of A\u2660 J\u2665 (which never changes as the royal rises) with the value of A\u2660 T\u2660 (which does change with the royal.) Can you see how this is different than what we are doing this week?<\/p>\n

This week, we’re comparing the value of K\u2663 T\u2663 (which rises with the royal at exactly the same speed as A\u2660 T\u2660 did last week) with the value of A\u2666 (which ALSO rises with the royal, albeit at a slower pace.) When both parts of the equations are moving with the royal, that makes it a bit more complicated than when one part is moving and one part is stationary.<\/p>\n

It probably isn’t obvious to some why the value of K\u2663 T\u2663 changes at a different rate than A\u2666, so let me address that. There are 16,215 different combinations we could draw to K\u2663 T\u2663. Only one of them is A\u2663 Q\u2663 J\u2663, giving us the royal flush. Drawing four cards to A\u2666 gives us 178,365 combinations, only one of which leads to the royal flush.<\/p>\n

As it happens,178,365 is exactly 11 times as big as 16,215, which means that the value of K\u2663 T\u2663 grows 11 times as fast as the value of A\u2666. Another way to say this is that it is 11 times easier to draw three perfect cards than it is to draw four perfect cards.<\/p>\n

Let’s use Video Poker for Winners to put some numbers on this. We’re looking at the change in values of both K\u2663 T\u2663 and A\u2666 as the value of the royal flush increases from 8,000 coins to 12,000 coins. Note that at 8,000 coins, the value of A\u2666 is higher, so that’s what we hold. At 12,000 coins, the value of K\u2663 T\u2663 is higher. We can conclude that somewhere between 8,000 coins and 12,000 coins, the correct play switches.<\/p>\n\n\n\n\n\n
<\/td>\n8,000-coin
\nRoyal<\/td>\n		12,000-coin
\nRoyal<\/td>\n		Change<\/td>\n<\/tr>\n
K\u2663 T\u2663<\/td>\n2.32810<\/td>\n2.57480<\/td>\n0.24670<\/td>\n<\/tr>\n
A\u2666<\/td>\n2.33315<\/td>\n2.35555<\/td>\n0.02240<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

You might have noticed that the change in K\u2663 T\u2663 was 11 times as big as the change in A\u2666, allowing for rounding.<\/p>\n

\u0394RF = (2.33315 \u2014 2.32810) * 4,000 \/ (10\/11) * (2.57480 \u2014 2.32810) = 90 (rounded)<\/p>\n

Let’s look at the terms. The (2.33315 \u2014 2.32810) term represents how far K\u2663 T\u2663 and A\u2666 are apart when the royal flush pays 8,000 coins. The (2.57480 \u2014 2.32810) term represents how much the value of K\u2663 T\u2663 changes in 4,000 coins. Since A\u2666 increased by one-eleventh as much during that same 4,000-coin meter rise, we only count 10\/11 of the change in value of K\u2663 T\u2663 towards catching up with original deficit.<\/p>\n

When we multiply this all out, we get a value of 90 for \u0394RF, which means that the BP is 8,090. If we plug in the value of 8,090 into VPW and look at the hand in question, we get 2.33365 for both the value of K\u2663 T\u2663 and A\u2666. So 8,090 is our BP.<\/p>\n

We don’t have to use this type of calculation very often. It’s only when there are two different royal flush combinations in the same hand which change in values at different rates as the royal flush goes up.<\/p>\n

Consider this hand in Double Bonus: A\u2665 T\u2665 7\u2663 5\u2666 3\u2660. At low royal levels we hold the A\u2665 and at high royal levels we hold A\u2665 T\u2665. Do we figure this BP using last week’s formula or this week’s?<\/p>\n

It might be surprising to learn we use last week’s formula. At different levels of the royal flush, the value of A\u2665 by itself is fixed. It seems wrong to say that the value of A\u2665 by itself doesn’t increase as the royal increases until we realize that when we’re calculating the value of A\u2665 in this particular hand, we’re assuming that we’re discarding the T\u2665 (along with the 7\u2663 5\u2666 3\u2660.) Since we’re throwing away the T\u2665, it’s impossible to get a heart royal, and since we’re holding on to the A\u2665, it’s impossible to get a royal in any other suit.<\/p>\n\"Facebook\"<\/a>\"twitter\"<\/a>\"youtube\"<\/a>\"instagram\"<\/a>","protected":false},"excerpt":{"rendered":"

In last week’s column, we looked at the standard way of calculating royal flush breakpoints (BPs) on progressives. If you didn’t see that column, I suggest you review that one first before tackling this one. https:\/\/www.lasvegasadvisor.com\/bob_dancer\/2011\/0524.cfm<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_exactmetrics_skip_tracking":false,"_exactmetrics_sitenote_active":false,"_exactmetrics_sitenote_note":"","_exactmetrics_sitenote_category":0,"spay_email":"","jetpack_publicize_message":"","jetpack_is_tweetstorm":false},"categories":[6],"tags":[],"yoast_head":"\nA Different Sort of Breakpoint Calculation - Gambling With An Edge<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.lasvegasadvisor.com\/gambling-with-an-edge\/a-different-sort-of-breakpoint-calculation\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A Different Sort of Breakpoint Calculation - Gambling With An Edge\" \/>\n<meta property=\"og:description\" content=\"In last week’s column, we looked at the standard way of calculating royal flush breakpoints (BPs) on progressives. 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